Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

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Question 1.
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid.
(i) \(\frac{1}{5+x}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 1
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 2

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

(ii) \(\frac{2}{(3+4 x)^{2}}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 3

(iii) (5 + x2)2/3
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 4

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

(iv) \((x+2)^{-\frac{2}{3}}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 5
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 6

Expanding binomials calculator is a free online tool that displays the expansion of the given binomial term.

Question 2.
Find \(\sqrt[3]{1001}\) approximately. (two decimal places)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 7

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 3.
Prove that \(\sqrt[3]{x^{3}+6}-\sqrt[3]{x^{3}+3}\) is approximately equal to \(\frac{1}{x^{2}}\) when x is sufficiently large.
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 8
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 9
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 10
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 11

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 4.
Prove that \(\sqrt{\frac{1-x}{1+x}}\) is approximately equal to 1 – x + \(\frac{x^{2}}{2}\) when x is very small.
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 12
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 13

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 5.
Write the first 6 terms of the exponential series
(i) e5x
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 14

(ii) e-2x
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 15

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

(iii) ex/2
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 16
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 17

Question 6.
Write the first 4 terms of the logarithmic series.
(i) log (1 + 4x)
(ii) log (1 – 2x)
(iii) log \(\left(\frac{1+3 x}{1-3 x}\right)\)
(iv) log \(\left(\frac{1-2 x}{1+2 x}\right)\)
Find the intervals on which the expansions are valid.
Answer:
(i) log ( 1 + 4x )
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 18

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

(ii) log (1 – 2x)
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 19

(iii) log \(\left(\frac{1+3 x}{1-3 x}\right)\)
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 20

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

(iv) log \(\left(\frac{1-2 x}{1+2 x}\right)\)
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 21
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 22

Question 7.
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 23
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 24

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 8.
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 25
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 26
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 27
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 28
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 29

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 9.
Find the coefficient of x4 in the expansion of \(\frac{3-4 x+x^{2}}{e^{2 x}}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 30

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 10.
Find the value of Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 31
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 32
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 33

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 7 Matrices and Determinants Ex 7.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.3

Solve the following problems by using Factor Theorem:

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

Question 1.
Show that \(\left| \begin{matrix} x & a & a \\ a & x & a \\ a & a & x \end{matrix} \right| \) = (x – a)2 (x + 2a)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 1
By putting x = a , we have three rows of |A| are identical. Therefore (x – a)2 is a factor of |A|
Put x = – 2a in |A|
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 2
∴ x + 2a is a factor of |A|. The degree of the product of the factors (x – a)2 (x + 2a) is 3.
The degree of tfie product of the leading diagonal elements x . x . x is 3.
∴ The other factor is the contant factor k.

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 3
a3 [ – 1 (1 – 1) – 1 ( – 1 – 1) + 1 (1 + 1)] = k . 4a3
a3 [o + 2 + 2 ] = 4 ka3
4a3 = 4 ka3
k = 1
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 4

 

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

Free handy Remainder Theorem Calculator tool displays the remainder of a difficult polynomial expression in no time.

Question 2.
Show that \(\left| \begin{matrix} b\quad +\quad c & a\quad -\quad c & a\quad -\quad b \\ b\quad -\quad c & c\quad +\quad a & b\quad -\quad a \\ c\quad -\quad b & c\quad -\quad a & a\quad +\quad b \end{matrix} \right| \) = 8 abc
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 5
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 6
since two columns identical
= bc × 0 = 0
∴ a – 0 is a factor. That is, a is a factor.
Put b = 0 in |A|
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 7
since two columns identical
= ca × 0 = 0
∴ b – 0 is a factor. That is, a is a factor.
Put c = 0 in |A|
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 8

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

since two columns identical
= ab × 0 = 0
∴ c – 0 is a factor. That is, c is a factor.
The degree of the product of the factors abc is 3.
The degree of the product of leading diagonal elements (b + c) (c + a) (a + b) is 3.
∴ The other factor is the constant factor k.
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 9

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

Question 3.
Solve that \(\left| \begin{matrix} x\quad +\quad a & b & c \\ a & x\quad +\quad b & c \\ a & b & x\quad +\quad c \end{matrix} \right| \) = 0
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 10
Put x = 0
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 11
x = 0 satisfies the given equation. x = 0 is a root of the given equation, since three rows are identical. x = 0 is a root of multiplicity 2. Since the degree of the product of the leading diagonal elements (x + a) (x + b) (x + c) is 3. There is one more root for the given equation.
Put x = – (a + b + c)
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 12
∴ x = – (a + b + c) satisfies the given equation.
Hence, the required roots of the given equation are x = 0, 0 , – (a + b + c)

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

Question 4.
Show that \(\left| \begin{matrix} b\quad +\quad c & a & { a }^{ 2 } \\ c\quad +\quad a & b & { b }^{ 2 } \\ a\quad +\quad b & c & { c }^{ 2 } \end{matrix} \right| \) = (a + b + c) (a – b) (b – c) (c – a)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 13
Since two rows are idenctical
|A| = 0
since two rows are idenctical
|A| = 0
∴ a – b is a factor of | A |. The given determinant is in cyclic symmetric form in a , b and c. Therefore, b – c and c – a are also factors. The degree of the product of the factors (a – b) (b – c) (c – a) is 3 and the degree of the product of the leading diagonal elements (b + c) . b . c2 is 4.
Therefore, the other factor is k (a + b + c).
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 14
5(18 – 12) – 1(36 – 12) + 1(12 – 6) = 12k
5 × 6 – 24 + 6 = 12k
30 – 24 + 6 = 12k
12 = 12 ⇒ k = 1
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 15

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

Question 5.
Solve \(\left| \begin{matrix} 4\quad -\quad x & 4\quad +\quad x & 4\quad +\quad x \\ 4\quad +\quad x & 4\quad -\quad x & 4\quad +\quad x \\ 4\quad +\quad x & 4\quad +\quad x & 4\quad -\quad x \end{matrix} \right| \) = 0
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 16
Put x = 0
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 17
∴ x = 0 satisfies the given equation. Hence x = 0 is a root of the given equation. since three rows are identical, x = 0 is a root of multiplicity 2.

Since the degree of the product of the leading diagonal elements (4 – x) (4 – x) (4 – x) is 3. There is one more root for the given equation.
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 18
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 19
∴ x = – 12 is a root of the given equation.
Hence, the required roots are x = 0 , 0 , – 12

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

Question 6.
Show that \(\left| \begin{matrix} 1 & 1 & 1 \\ x & y & z \\ { x }^{ 2 } & { y }^{ 2 } & { z }^{ 2 } \end{matrix} \right| \) = (x – y) (y – z) (z – x)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 20
|A| = 0 since two columns identical
∴ x – y is a factor of A. The given determinant is in the cyclic symmetric form in x, y, and z. Therefore, y – z and z – x are also factors of |A|.

The degree of the product of the factors (x – y) (y – z) (z – x) is 3 and the degree of the product of the leading diagonal elements 1, y, z2 is 3. Therefore, the other factor is the constant factor k.
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 21
Put x = 0, y = 1, z = -1 we get
Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 22
Expanding along the first column
1 (1 + 1) = 2k
2 = 2k ⇒ k = 1

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3

Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.3 23

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Accountancy Guide Pdf Chapter 3 Books of Prime Entry Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 3 Books of Prime Entry

11th Accountancy Guide Books of Prime Entry Text Book Back Questions and Answers

prime factorization of 38 are numbers that, when multiplied in pairs give the product as 38.

I. Choose the correct answer.

Question 1.
Accounting equation signifies …………….
(a) Capital of a business is equal to assets
(b) Liabilities of a business are equal to assets
(c) Capital of a business is equal to liabilities
(d) Assets of a business are equal to the total of capital and liabilities
Answer:
(d) Assets of a business are equal to the total of capital and liabilities

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 2.
‘Cash withdrawn by the proprietor from the business for his personal use’ causes …………….
(a) Decrease in assets and decrease in owner’s capital
(b) Increase in one asset and decrease in another asset
(c) Increase in one asset and increase in liabilities
(d) Increase in asset and decrease in capital
Answer:
(a) Decrease in assets and decrease in owner’s capital

Question 3.
A firm has assets of Rs. 1,00,000 and the external liabilities of Rs. 60,000. Its capital would be ……………….
(a) Rs. 1,60,000
(b) Rs. 60,000
(c) Rs. 1,00,000
(d) Rs. 40,000
Answer:
(d) Rs. 40,000

Question 4.
The incorrect accounting equation is ………………….
(a) Assets = Liabilities + Capital
(b) Assets = Capital + Liabilities
(c) Liabilities = Assets + Capital
(d) Capital = Assets – Liabilities
Answer:
(c) Liabilities = Assets + Capital

Question 5.
Accounting equation is formed based on the accounting principle of ………………
(a) Dual aspect
(b) Consistency
(c) Going concern
(d) Accrual
Answer:
(a) Dual aspect

Question 6.
Real account deals with ……………..
(a) Individual persons
(b) Expenses and losses
(c) Assets
(d) Incomes and gains
Answer:
(c) Assets

Question 7.
Which one of the following is representative personal account?
(a) Building A/c
(b) Outstanding salary A/c
(c) Mahesh A/c
(d) Balan & Co
Answer:
(b) Outstanding salary A/c

Question 8.
Prepaid rent is a ……………….
(a) Nominal A/c
(b) Personal A/c
(c) Real A/c
(d) Representative personal A/c
Answer:
(d) Representative personal A/c

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 9.
Withdrawal of cash from business by the proprietor should be credited to
(a) Drawings A/c
(b) Cash A/c
(c) Capital A/c
(d) Purchases A/c
Answer:
(b) Cash A/c

Question 10.
In double entry system of book keeping, every business transaction affects
(a) Minimum of two accounts
(b) Same account on two different dates
(c) Two sides of the same account
(d) Minimum three accounts
Answer:
(a) Minimum of two accounts

II. Very Short Answer Type Questions

Question 1.
What are source documents?
Answer:
“Source documents are the authentic evidence of financial transactions. These documents show the nature of the transaction, the date, the amount, and the parties involved. Source documents include cash receipt, invoice, debit note, credit note, pay-in-slip, salary bills, wage bills, cheque record slips, etc.

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 2.
What Is the accounting equation?
Answer:
The relationship of assets with that of liabilities to outsiders and to owners in the equation form is known as the accounting equation.
The accounting equation is a mathematical expression which shows that the total of assets is equal to the total of liabilities and capital.
This is based on the dual aspect concept of accounting.
This means that the total claims of outsiders and the proprietor against a business enterprise will always be equal to the total assets of the business enterprise.
Capital + Liabilities = Assets

Question 3.
Write any one transaction which

  1. Decreases the assets and decreases the liabilities
  2. Increases one asset and decreases another asset

Answer:

  1. Paid creditors
  2. Gash sales

Question 4.
What is meant by journalizing?
Answer:
The process of analyzing the business transactions under the heads of debit and credit and recording them in the journal is called journalizing.

Question 5.
What is the real account?
Answer:
All accounts relating to tangible and intangible properties and possessions are called real accounts.

Question 6.
How are personal accounts classified?
Answer:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 1

Question 7.
State the accounting rule for a nominal account.
Answer:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 2

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 8.
Give the golden rules of double-entry accounting system.
Answer:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 3

III. Short Answer Questions

Question 1.
Write a brief note on the accounting equation approach of recording transactions.
Answer:
The relationship of assets with that of liabilities to outsiders and to owners in the equation form is known as the accounting equation.
The accounting equation is a mathematical expression which shows that the total of assets is equal to the total of liabilities and capital.
This is based on the dual aspect concept of accounting.

This means that the total claims of outsiders and the proprietor against a business enterprise will always be equal to the total assets of the business enterprise. The equation is given under:
Capital + Liabilities = Assets
Capital can also be called owner’s equity and liabilities as outsider’s equity.

As the revenues and expenses will affect capital, the expanded equation may be given as under:
Assets = Liabilities + Capital + Revenues – Expenses
Accounts are classified into five categories: (i) Asset account, (ii) Liability account, (iii) Capital account, (iv) Revenue account and (v) Expense account as follows:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 4

Question 2.
What is an Account? Classify the accounts with suitable examples.
Answer:
Every transaction has two aspects and each aspect affects a minimum of one account.
An account is the basic unit of identification in accounting.
A ledger account is a summary of relevant transactions at one place relating to a particular head.
The account is the systematic presentation of all material information regarding a particular person or item at one place, under one head.

Classification of Accounts:
Personal Account;
Accounting relating to persons is called a personal account. The personal account may be a natural, artificial, or representative personal account.
E.g: Malini account, BHEL account, Prepaid Rent account.

Impersonal Account:
All accounts which do not affect persons are called impersonal accounts. These are further classified into Real and Nominal Accounts.
E.g; Building account, Goodwill, Salaries account.

Question 3.
What are the three different types of personal accounts?
Answer:
Under the double-entry system of book-keeping, for the purpose of recording the various financial transactions, the accounts are classified as personal accounts and impersonal accounts.

  1. Natural person’s account: Natural person means human beings. Example: Vinoth account, Malini account.
  2. Artificial person’s account: Artificial person refers to persons other than human beings recognized by law as persons. They include business concerns, charitable institutions, etc. Example: BHEL account, Bank account.
  3. Representative personal accounts: These are the accounts which represent persons natural or artificial or a group of persons. Example: Outstanding salaries account, Prepaid rent account. When expenses are outstanding, it is payable to a person. Hence, it represents a person.

Question 4.
What is the accounting treatment for insurance premium paid on the of the proprietor?
Answer:
The proprietor has to pay an agreed premium on a monthly or annual basis. The amount should be treated as prepayment which is considered a current asset.
The following entry should be made:
Insurance A/c Dr –
To Cash/Bank A/c –
(Being insurance premium paid.)

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 5.
State the principles of the double-entry system of book-keeping.
Answer:
Following are the principles of the double-entry system:

  1. In every business transaction, there are two aspects.
  2. The two aspects involved are the benefit or value receiving aspect and the benefit or value giving aspect.
  3. These two aspects involve a minimum of two accounts; at least one debit and at least one credit.
  4. For every debit, there is a corresponding and equivalent credit.
  5. If one account is debited the other account must be credited.

Question 6.
Briefly explain about steps in journalizing.
Answer:
The process of analyzing the business transactions under the heads of debit and credit and recording them in the journal is called journalizing. An entry made in the journal is called a journal entry. The following steps are followed in journalizing:

Analyze the transactions and identify the accounts (based on aspects) which are involved in the transaction.

Classify the above accounts under Personal account, Real account or Nominal account, Apply the rules of debit and credit for the above two accounts.

Find which account is to be debited and which account is to be credited by the application of rules of the double-entry system. Record the date of the transaction in the date column.

Enter the name of the account to be debited in the particulars column very close to the left hand side of the particulars column followed by the abbreviation ‘Dr/ at the end in the same line. Against this, the amount to be debited is entered in the debit amount column in the same line.

Write the name of the account to be credited in the second line starting with the word ‘To’ prefixed a few spaces away from the margin in the particulars column. Against this, the amount to be credited is entered in the credit amount column in the same line.

Write the narration within brackets in the next line in the particulars column.

Question 7.
What is the double-entry system? State its advantages.
Answer:
A double-entry system of book keeping is a scientific and complete system of recording the financial transactions of an organization. According to this system, every transaction has a two-fold effect. That is, there are two aspects involved, namely, receiving aspect and the giving aspect. It is denoted by debit (Dr.) and credit (Cr.). The basic principle of the double-entry system is that for every debit there must be an equivalent and corresponding credit. Debit denotes an increase in assets or expenses or a decrease in liabilities, income, or capital. Credit denotes an increase in liabilities, income, or capital or a decrease in assets or expenses.

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

IV. Exercises

Question 1.
Complete the accounting equation.
(a) Assets = Capital + Liabilities
1,00,000 = 80,000 + ?
Answer:
(a) Liabilities = Assets – Capital
Liabilities = 1,00,000 – 80,000
Liabilities = 20,000

(b) Assets = Capital + Liabilities
2,00,000 = ? + 40,000
Answer:
Capital = Assets – Liabilities
Capital = 2,00,000 – 40,000
Capital = 1,60,000

(c) Assets = Capital + Liabilities
? = 1,60,000 + 80,000
Answer:
Assets = Capital + Liabilities
Assets = 1,60,000 + 80,000
Assets = 2,40,000

Question 2.
For the following transactions, show the effect on the accounting equation.
(a) Raj Started business with cash – 40,000
(b) Opened bank account with a deposit of – 30,000
(c) Bought goods from Hari on credit for – 12,000
(d) Raj withdrew cash for personal use – 1,000
(e) Bought furniture by using a debit card for – 10,000
(f) Sold goods to Murugan and cash received – 6,000
(g) Money is withdrawn from bank for office use – 1,000
Answer:
(a) Raj started the business with cash Rs 40,000
Increasing in capital and increasing in Asset Effects:
Cash comes in → Increase in asset
Capital provided by the owner → Increase in capital of owner
Capital = Asset
Capital = Cash
(+)Rs. 40,000 = (+)Rs. 40,000

(b) Opened bank account with a deposit of Rs 30,000
Decrease in one asset and increase in another asset Effects:
Bank is the receiver → Increases in Assets
Cash goes out → Decrease in Asset
Capital = Asset
Capital = Bank + cash
Capital = (+) Rs. 30,000 + (-) Rs. 30,000

(c) Bought goods from Hari on credit for Rs. 12,000
Increasing Asset and Increase in Liabilities Effects:
Stock comes in → Increase in Asset
Creditors arises → Increase in Liabilities
Asset = Liabilities
Stock = Creditors
(+) Rs. 12,000 = (+) Rs. 12,000

(d) Raj withdrew cash for personal use Rs 1,000
Decrease in Asset and decrease in capital Effects:
Cash goes out → Decrease in cash
Drawings proprietor is the receiver → Decrease in capital
Asset = Capital
Cash = Capital
(-) Rs. 1,000 = (-) Rs. 1,000

(e) Bought furniture by using debit card for Rs 10,000
Decrease in one asset and Increase in another asset Effects:
Furniture comes in → Increase in Furniture
Bank is giver → Decrease in a Bank
Asset = Liabilities
Furniture + Bank = Liabilities
(+) Rs. 10,000 + (-) Rs. 10,000 = Liabilities

(f) Sold goods to Murugan and cash received Rs 6,000
Decrease in one asset and increase in another asset Effects:
Cash comes in → Increase in cash
<$> Stock goes out → Decrease in stock
Asset = Liabilities
Cash + Stock = Liabilities
(+) Rs. 6,000 + (-) Rs. 6,000 = Liabilities

(g) Money is withdrawn from bank for office use Rs 1,000
Decrease in one asset and increase in another asset Effects:
Cash comes in → Increase in Cash
The bank is a giver → Decrease in Bank
Asset = Liabilities
Cash + Bank = Liabilities
(+) Rs. 1,000 + (-) Rs. 1,000 = Liabilities

Accounting Equation:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 5

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 3.
Prepare accounting equation for the following transactions,
(a) Murugan commenced business with cash Rs 80,000
(b) Purchased goods for cash Rs 30,000
(c) Paid salaries by cash Rs 5,000
(d) Bought goods from Kumar for Rs 5,000 and deposited the money in CDM.
(e) Introduced additional capital of Rs 10,000
Answer:
Accounting Equation:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 6

Question 4.
What will be the effect of the following on the accounting equation?
(a) Sunil started the business with Rs 1,40,000 cash and goods worth Rs 60,000
(b) Purchased furniture worth Rs 20,000 in cash
(c) Depredation on furniture Rs 800
(d) Deposited into bank Rs 40,000
(e) Paid electricity charges through net banking Rs 500
(f) Sold goods to Ravi costing Rs 10,000 for Rs 15,000
(g) Goods returned by Ravi Rs 5,000
Answer:
(a) Sunil started the business with Rs 1,40,000 cash and goods worth Rs 60,000
Transactions affecting more than two accounts Effects:
Cash comes in → Increase in asset
Stock comes in → an Increase in asset
Capital provide by the owner → Increase in capital
Capital = Asset
Capital = Cash + Stock
(+) 2,00,000 = (+) 1,40,000 + (+) 60,000

(b) Purchased furniture worth Rs 20,000 in cash
Decrease in one asset and Increase in Another asset Effects:
Furniture comes in → Increase in asset
Cash goes out → Decrease in asset
Liabilities = Assets
Liabilities = Cash + Furniture
Liabilities = (-) 20,000 + (-) 20,000

(c) Depreciation on furniture Rs 800
Decrease in asset and decrease in capital effects:
Furniture goes out → Decrease in Asset
Capital → Decrease in Capital
Capital = Asset
Capital = Furniture
(-) 800 = (-) 800

(d) Deposited into bank Rs 40,000
Increase in one Asset and decrease in another asset Effects:
Cash goes out → Decrease in asset
The bank is a receiver → Increase in asset
Liability = Asset
Liability = Cash + bank
= (-) 40,000 + (+) 40,000

(e) Paid electricity charges through net banking Rs 500
decrease in Asset and Decrease in capital Effects:
The bank is a given → Decrease in Asset
Capital (Expenses) → Decrease in capital
Capital = Asset
Capital = Bank
(-) 500 = (-) 500

(f) Sold goods to Ravi costing Rs 10,000 for Rs 15,000
Increase in asset and decreases in another asset and Increase in Capital Effects:
Creditors → Increases in Liabilities
Stock goes out → Decreases in asset
Capital (Incom(e) → Increase in capital
Liabilities + Capital = Asset
Creditors + Capital = Stock
(+) 15,000 + (+) 5,000 = (-) 5,000

(g) Goods returned by Ravi Rs 5,000
Increase in asset and decrease in Liabilities Effects:
Stock comes in → an Increase in asset
Reducing in creditors → Decreases in Liabilities
Liability = Asset
Creditors = Stock
(-) 5,000 = (+) 5,000
Accounting Equation:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 7

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 5.
Create an accounting equation on the basis of the following transactions.
(i) Rakesh started the business with a capital of Rs 1,50,000
(ii) Deposited money with the bank Rs 80,000
(iii) Purchased goods from Mahesh and paid through credit card Rs 25,000
(iv) Sold goods (costing Rs 10,000) to Mohan for Rs 14,000 who pays through debit card
(v) Commission received by cheque and deposited the same in the bank Rs 2,000
(vi) Paid office rent through ECS Rs 6,000
(vii) Sold goods to Raman for Rs 15,000 of which Rs 5,000 was received at once
Answer:
Accounting Equation:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 8

Question 6.
Create an accounting equation on the basis of the following transactions.
(i) Started business with cash 80,000 and goods Rs 75,000
(ii) Sold goods to Shanmugam on credit for Rs 50,000
(iii) Received cash from Shanmugam in full settlement Rs 49,000
(iv) Salary outstanding Rs 3,000
(v) Goods costing Rs 1,000 given as charity
(vi) Insurance premium paid Rs 3,000
(vii) Out of insurance premium paid, prepaid is Rs 500
Answer:
Accounting Equation:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 9

Question 7.
Create an accounting equation on the basis of the following transactions:
(i) Opening balance on 1st January, 2018 cash Rs 20,000; stock Rs 50,000 and bank Rs 80,000
(ii) Bought goods from Suresh Rs 10,000 on credit
(iii) Bank charges Rs 500
(iv) Paid Suresh Rs 9,700 through credit card in full settlement.
(v) Goods purchased on credit from Philip for Rs 15,000
(vi) Goods returned to Philip amounting to Rs 4,000
Answer:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 10

Question 8.
Enter the following transactions in the journal of Manohar who is dealing in textiles. 2018 March
Manohar started the business with cash – 60,000
Purchased furniture for cash – 10,000
Bought goods for cash – 25,000
Bought goods from Kamaiesh on credit – 15,000
Sold goods for cash – 28,000
Sold goods to Hari on credit – 10,000
Paid Kamaiesh – 12,000
Paid rent – 500
Received from Hari – 8,000
Withdrew cash for personal use – 4,000
Answer:
journal of Mr. Manohar
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 11

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 9.
Pass journal entries in the books of Sasi Kumar who is dealing in automobiles. 2017 Oct.
1. Commenced business with goods – 40,000
3. Cash introduced in the business – 60,000
4. Purchased goods from Arul on credit – 70,000
6. Returned goods to Arul – 10,000
10. Paid cash to Arul on account – 60,000
15. Sold goods to Chandar on credit – 30,000
18. Chandar returned goods worth – 6,000
20. Received cash from Chandar in full settlement – 23,000
25. Paid salaries through ECS – 2,000
30. Sahil took for personal use goods worth – 10,000
Answer:
journal of Mr. Sasi Kumar:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 12

Question 10.
Pass Journal entries in the books of Hart who is deafer in sports items. 2017 Jan.
1. Commenced business with cash 50,000
2. Purchased goods from Subash on credit of 20,000
4. Sold goods to Ramu on credit of 15,000
8. Ramu paid the amount through cheque
10. Cheque received from Ramu is deposited with the bank
15. Sports items purchased from Gopal on credit 10,000
18. Paid rent for the proprietor’s residence 1,500
20. Paid Gopal in a full settlement after deducting a 5% discount
25. Paid Subash Rs 4,750 and discount received 250
28. Paid by cash: wages Rs 500; electricity charges Rs 3,000 and trade expenses 1,000
Answer:
journal of Hari:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 13

Question 11.
Karthick opened a provisions store on 1st April 2017, Journalise the following transactions in this books:
2017 April?
1. Paid into the bank for opening a current account – 2,00,000
3. Goods purchased by cheque – 40,000
5. Investments made in securities – 40,000
6. Goods sold to Radha for Rs 20,000 and cheque received and deposited into bank
7. Amount withdrawn from bank for office use – 15,000
10. Purchased goods from Kamala and cash deposited in CDM – 10,000
12. Sold goods to Vanitha who paid through the debit card – 10,000
15. Interest on securities directly received by the bank – 1,000
20. Insurance paid by the bank as per standing instructions – 2,000
25. Sales made to Kunal who made payment through CDM – 6,000
Answer:
journal of Karthik:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 14

Question 12.
Journalise the following transactions in the books of Ramesh who is dealing in computers:
2018, May 1. Ramesh started business with cash Rs 3,00,000, Goods Rs 80,000 and Furniture Rs 27,000.
2. Money deposited into bank Rs 2,00,000
3. Bought furniture from M/s Jayalakshmi Furniture for Rs 28,000 on credit.
4. Purchased goods from Asohan for Rs 5,000 by paying through debit card.
5. Purchased goods from Guna and paid through net banking for cash Rs 10,000
6. Purchased goods from Kannan and paid through credit card Rs 20,000
7. Purchased goods from Shyam on credit for Rs 50,000
8. Bill drawn by Shyam was accepted for Rs 50,000
9. Paid half the amount owed to M/s Jayalakshmi Furniture by cheque
10. Shyam’s bill was paid
Answer:
journal of Mr. Ramesh
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 15

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 13.
Journalise the following transactions in the books of Sundar who is a bookseller.
Dec 2017.
1. Commenced business with cash – 2,00,000
2. Bought goods from X and Co. on credit – 80,000
4. Opened a bank account with – 50,000
5. Sold goods to Naresh who paid the amount through net banking – 5,000
6. Sold goods to Devi who paid through credit card – 7,000
7. Sold goods to Ashish on credit – 700
8. Money withdrawn from the bank through ATM for office use – 1,000
9. Purchased furniture and paid through the debit card – 2,000
10. Salaries paid by cash – 6,000
11. Furniture purchased from Y for Rs 25,000 and advance given – 5,000
Answer:
Journal of Mr. Sundar
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 16

Question 14.
Raja has a hotel. The following transactions took place in his business. Journalize them.
Jan. 1. Started business with cash – 3,00,000
2. Purchased goods from Rajiv on credit – 1,00,000
3. Cash deposited with the bank – 2,00,000
20. Borrowed loan from the bank – 1,00,000
22. Withdrew from the bank for personal us – 800
23. Amount paid to Rajiv in full settlement through NEFT – 99,000
25. Paid club bill of the proprietor by cheque – 200
26. Paid electricity bill of the proprietor’s house through the debit card – 2,000
31. Lunch provided at free of cost to a charity – 1,000
31. Bank levied charges for locker rent – 1,000
Answer:
journal of Mr. Raja
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 17

Question 15.
From the following transactions of Shyam, a stationery dealer, pass journal entries for the month of August 2017,
Aug 1. Commenced business with cash Rs 4,00,000, Goods Rs 5,00,000
2. Sold goods to A and money received through RTGS 12,50,000
3. Goods sold to Z on credit for Rs 20,000
5. Bill drawn on Z and accepted by him Rs 20,000
8. Bill received from Z is discounted with the bank for 119,000
10. Goods sold to M on credit Rs 12,000
12. Goods distributed as free samples for Rs 2,000
16. Goods taken for office use Rs 5,000
17. M became insolvent and only 0.80 per rupee is received in the final settlement
20. Bill of Z discounted with the bank is dishonored
Answer:
Journal of Mr. Shyam
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 18

Question 16.
Mary is a cement dealer having a business for more than 5 years. Pass journal entries in her books for the period of March 2018,
1. Cement bags bought on credit from Sibi – 20,000
2. Electricity charges paid through net banking – 500
3. Returned goods bought from Sibi – 5,000
4. Cement bags are taken for personal use – 1,000
5. Advertisement expenses paid – 2,000
6. Goods sold to Mano – 20,000
7. Goods returned by Mano – 5,000
8. Payment received from Mano through NEFT
Answer:
Journal of Mr. Mary
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 19

11th Accountancy Guide Conceptual Framework of Accounting Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
A debit note is also called…………….
(a) Credit note
(b) Debit memo
(c) Vouchers
(d) Cash memo
Answer:
(b) Debit memo

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 2.
Journal is a book of ……….
(a) Original Entry
(b) Final Entry
(c) Assets
(d) None of the above
Answer:
(a) Original Entry

Question 3.
When cash or cheque is deposited in a bank, a form is to be filled by a customer is called…………….
(a) Pay – in – slip
(b) Voucher
(c) Cash memo
(d) Invoice
Answer:
(a) Pay – in – slip

Question 4.
Accounts of persons with whom the business deals is known as ………….
(a) Personal A/c
(b) Real A/c
(c) Nominal A/c
(d) Profit & Loss A/c
Answer:
(a) Personal A/c

Question 5.
Outsider’s equity is otherwise called as …………….
(a) Capital
(b) liabilities
(c) debtors
(d) Assets
Answer:
(b) liabilities

Question 6.
Which one of the following equations is correct?
(a) Owner’s Equity = Liability + Asset
(b) Owner’s Equity = Asset – Liability
(c) Liability = Owner’s Equity + Asset
(d) Asset = Owner’s Equity – Liability
Answer:
(b) Owner’s Equity = Asset – Liability

Question 7.
Goodwill is an example of accounts …………….
(a) Real
(b) Nominal
(c) Tangible real
(d) Intangible real
Answer:
(d) Intangible real

Question 8.
Atul purchased a car for Rs. 5,00,000, by making a down payment of Rs. 1,00,000 and signing an Rs. 4,00,000 bill payable due in 60 days. As a result of this transaction
(a) Total assets increase by Rs. 5,00,000
(b) Total liabilities increase by Rs.4,00,000
(c) Total assets increase by Rs. 4,00,000
(d) Total assets increased by Rs. 4,00,000 with a corresponding increase in liabilities by Rs. 4,00,000
Answer:
(d) Total assets increased by Rs. 4,00,000 with a corresponding increase in liabilities by Rs. 4,00,000

Question 9.
Journal means …………….
(a) daily
(b) monthly
(c) yearly
(d) weekly
Answer:
(a) daily

Question 10.
Amount paid to a supplier is ……….
(a) An event
(b) A Transaction
(c) Both (a) and (b)
(d) Neither (a) Nor (b)
Answer:
(c) Both (a) and (b)

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 11.
Goods worth Rs. 2,000 were distributed as free samples in the market. The journal entry will be
(a) Drawings A/c Dr. 2,000
To Purchases A/c 2,000
(b) Sales A/c Dr. 2,000
To Cash A/c 2,000
(c) Advertisement A/c Dr. 2,000
To Purchases A/c 2,000
(d) No entry
Answer:
(c) Advertisement A/c Dr. 2,000
To Purchases A/c 2,000

Question 12.
If two or more transactions of the same nature are journalized together, it is known as;
(a) Compound journal entry
(b) Separate journal entry
(c) Posting
(d) None of the above
Answer:
(a) Compound journal entry

Question 13.
An amount of Rs.200 received from A credited to B would affect ………….
(a) Accounts of A and B
(b) A’s Account
(c) Cash Account
(d) B’s Account
Answer:
(a) Accounts of A and B

Question 14.
Contra entries are passed only when …………..
(a) Double Column cash book is prepared
(b) Three-Column Cashbook is prepared
(c) Single Cashbook is prepared
(d) None of the above
Answer:
(b) Three-Column Cashbook is prepared

Question 15.
Which of the following statements is correct?
(a) Accounting profit is the difference between cash receipts and cash paid in a period.
(b) Accounting profit is the total cash sales in the year is the expenses for the period.
(c) Accounting profit is the difference between revenue income and expenses for the period.
(d) Accounting profit is the difference between revenue income and cash payment for the period.
Answer:
(c) Accounting profit is the difference between revenue income and expenses for the period.

Question 16.
Direct Expenses are those which can be identified with the particular products. Which of the following items is a direct cost?
(a) The rent of the factory in which products are made
(b) The wages of the production supervisor
(c) A royalty paid to the holder of a patent after each item is produced
(d) The cost of the glue used to attach labels to the products.
Answer:
(c) A royalty paid to the holder of a patent after each item is produced

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 17.
Rent payable to the landlord Rs. 5,00,000 is credited to
(a) Cash Account
(b) Landlord Account
(c) Outstanding Rent Account
(d) None of the above
Answer:
(c) Outstanding Rent Account

Question 18.
Bad debts entry is passed in …………
(a) Sales Book
(b) Cash Book
(c) journal Book
(d) None of the above
Answer:
(c) journal Book

Question 19.
Return of cash sales is recorded in …………..
(a) Sales Return book
(b) cash book
(c) Journal proper
(d) None of the above
Answer:
(b) cash book

Question 20.
John purchased goods ‘ Rs, 5900 for cash at 20% trade discount and 5% cash discount. Purchase account Is to fee debited by Rs ………….
(a) Rs, 3,800
(b) Rs. 5,000
(c) Rs. 3,750
(d) Rs. 4,000
Answer:
(d) Rs. 4,000

Question 21.
Out of the following Identify the wrong statement.
(a) Real & Personal accounts are transferred to Balance Sheet
(b) Nominal accounts are transferred to profit and loss account
(c) Cash account is not opened separately in the ledger
(d) Rent account is a personal account and outstand rent account is a Nominal Account
Answer:
(a) Real & Personal accounts are transferred to Balance Sheet

Question 22.
The financial position of a business concern is ascertained on the basis of ………….
(a) Records Prepared under book-keeping process
(b) Trial Balance
(c) Accounting reports
(d) None of the above
Answer:
(c) Accounting reports

Question 23.
Salary “payable to an employee Rs.50,000. Which account is to be credited?
(a) Cash Account
(b) Salaries Account
(c) Outstanding Salaries
(d) None
Answer:
(b) Salaries Account

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 24.
Loss on sale of machinery Is credited to …………..
(a) Machinery Account
(b) Purchase Account
(c) Profit and Loss Account
(d) None of the above
Answer:
(a) Machinery Account

Question 25.
The Capital of a sole trader Is affected by ……………..
(a) Purchase of raw material
(b) Commission received
(c) Cash received from trade receivables
(d) Purchase of an asset for cash
Answer:
(b) Commission received

II. Very Short Answer Type Questions

Question 1.
From the following information below journalise its
i. Salary paid Rs. 5,000
ii. Rent paid to house owner Rs. 1,000
iii. Cash sales Rs. 10,000
Answer:
Journal Entries
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 20

Question 2.
Journalise the following transactions:
2017, Jan 11 Purchased goods for Rs. 1,500
13. Bought furniture for cash Rs. 2,000
19. Drew for private use Rs. 500
Answer:
Journal Entries
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 21

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

III. Short Answer Questions

Question 1.
Journalise the following transitions in the books of G.
i. Started business with cash Rs. 9,000
ii. Deposited into Canara Bank Rs, 3,000
iii. Paid Salary Rs, 5,000
iv. Received commission Rs. 200
v. Cash received from Rajan Rs. 400
Answer:
Journal Entries
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 22

IV. Additional Sums

Question 1.
Prepare accounting equation on the basis of the following.
1) Ramya started the business with cash Rs. 1,20,000
2) She Paid Rent Rs. 4,000
3) She Purchases Furniture Rs. 20,000
4) She Purchased goods on credit from Mr. Parthiban Rs. 60,000
5) She Sold goods (Cost Price Rs. 40,000) for Cash Rs. 50,000
Answer:
Accounting Equation
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 23

Question 2.
Show the Accounting equation the basis of the following transaction.
1. Pailavan Started business with cash Rs. 50,000
2. Purchased goods from Monisha Rs. 40,000
3. Sold goods to Aruna (Costing Rs. 36,000) for Rs. 50,000
4. wan withdraw from business Rs. 10,000
Answer:
Accounting Equation
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 24

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 3.
Journalise the following transaction of Mrs. Padmini
2018 May 1. Received Cash from Siva – 75,000
8. Paid Cash to Sayed – 45,000
11. Bought Goods for cash – 27,000
12. Bought Furniture – 48,000
15. Sold Goods for cash – 70,000
18. Sold Furniture – 50,000
Answer:
Journal of Mrs. Padmini
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 25

Question 4.
Journalise the following transactions of Mr. Dharani.
2018 Jan
1. Mr. Dharani Started Business with Cash – 1,00,000
4. Sold Goods to Mohan – 70,000
10. Purchased goods from Bashyam – 50,000
20. Sold goods for cash from Natarajan – 70,000
25. Paid to Bashyam – 50,000
28. Received from Mohan – 70,000
Answer:
Journal of Mr. Dharani
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 26

Question 5.
Journalise the following transaction of Tmt. Rani
2018 Feb
1. Tmt Rani Started Business With cash – 300000
5. Opening a current account with Indian Overseas Bank – 50000
10. Bought goods from Sumathi – 90000
18. Paid to Sumathi – 90000
20. Sold Goods to Chitra – 126000
28. Chitra Settled her account
Answer:
Journal of Tmt. Rani
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 27

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 6.
Journalise the following transactions of Mr.Jeevan.
2018 Mar 5. Sold goods to Arun on Credit – 17,500
10. Bought Goods for cash from Ravi – 22,500
12. Met Travelling Expenses – 2,500
18. Received Rs.80000 from Sivakumar as a loan
21. Paid wages to workers – 3,000
Answer:
Journal of Mr. Jeevan
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 28

Question 7.
Give Transaction with imaginary figure involving the following
(i) Increase in assets and capital
(ii) Increase and decrease in assets
(iii) Increase in assets and a liabilities
(iv) Decrease of an asset and owners capital
Answer:
(i) Mr. Raja Commenced Business with Rs. 2,00,000
(ii) Purchased Computer Rs. 15,000
(iii) Purchased goods from Suresh Rs. 80,000
(iv) Drawings Rs. 15,000

Question 8.
Journalise the following transaction of Mrs. Rama
2018 April 1. Mrs. Rama Commenced Business with cash – 30,000
2. Paid into Bank – 21,000
3. Purchases goods and paid through Debit Card – 15,000
7. Drew Cash from Bank for Personal Use – 3,000
15. Purchased goods from Siva – 15,000
20. Cash Sales – 30,000
23. Money Withdrawn from Bank thought ATM for office use – 1,000
25. Paid to Siva – 14,750
Discount Received – 250
31 Paid Rent – 500
Paid Salaries – 2,000
Answer:
Journal of Mrs. Rama
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 29

Question 9.
Record the transactions in the books of Shri. Ganesh & Co.
2012 Jan
1. Ganesh started a business with cash – 90,000
1. Paid into bank – 20,000
2. Purchased goods for cash – 50,000
4. Purchased furniture by issuing cheque – 24,000
5. Sold goods to Siva – 17,000
6. Purchased goods from Anand – 30,000
9. Returned goods to Anand – 3,000
10. Cash received from Siva – 16,000
11. Withdraw cash for personal use – 1,500
18. Purchase of stationary – 500
Answer:
Journal Entries in the Books of Shri. Ganesh & Co.
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 30

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 10.
Record the transactions in the books of XYZ & Co:
2017
May 1. Furniture purchased for Rs. 10,000
2. Old Computer worth Rs. 13,700 sold to Sankar on credit
3. Rent paid Rs. 5,000
6. Depreciation on Machinery Rs. 6,400 paid by cheque
10. Paid Rs. 250 for the installation charges of Machinery
13. Cash taken for personal use Rs. 7,000
Answer:
Journal Entries in the books XYZ & Co.
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 31

Question 11.
Record the transaction in the books of Y.
2017, April 1. Started Business with cash Rs, 9,000
2. Purchased goods for cash Rs. 21,000
3. Sold goods for cash Rs. 8,000
4. Deposited into Canara Bank Rs. 3,000
5. Cash received from Rajan Rs. 4,000
7. Paid salary Rs. 3,000
8. Paid Rent Rs. 4,000
9. Received Commission Rs. 500
10. Withdrew from Canara Bank Rs. 2,700
Answer:
Journal Entries in the books of Y
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 32

Question 12.
Journalize the following transactions:
2016, Dec 1. Purchased goods for cash Rs. 14,000
8. Purchased goods from Anand Stores Rs. 600
12. Sold goods for Rs. 12,500
16. Sold goods to Sridhar Rs. 6,450
18. Bought Machinery for Cash Rs. 1,20,000
20. Goods returned to Anand Stores Rs. 300
22. Sridhar returned goods worth Rs. 450
25. Drew for personal use Rs. 500
25. Electric charges amounted Rs. 1,000
30. Got a loan from Joice Rs. 750
Answer:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 33

Question 13.
Journalise the following transactions in the books of Z.
2016, july 1. Sold goods for Cash Rs. 2,800
2. Purchased goods for cash Rs, 5,900
5. Purchased goods from K Rs. 3,500
8. Sold goods to P Rs. 6,500
5. Received cash from P Rs. 6,200
9. Paid to K Rs. 3,200
9. Paid Telephone Charges Rs: 1,200
10. Paid Salary Rs. 2,600
10. Received Commission Rs. 3,000
10. Computer purchased from S for Rs. 20,000 and an advance of Rs. 8,000 is given as cash
Answer:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 34

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Question 14.
Varna is a women entrepreneur, dealing in textiles. From the following transactions, pass journal entries for the month of March 2017.
2017,
March, 1. Commenced business with cash Rs. 50,000 & with goods Rs. 45,000
2. Purchased 20 sarees from V & Co on credit Rs. 60,000
3. Cash deposited into bank Rs. 48,000
4. Paid to V & Co through net banking
5. Sold 10 sarees @ Rs. 3,000 each on credit to X & Co.
6. Paid Travelling expenses Rs. 450
8. Bank Charges levied Rs. 150
9. X becomes insolvent and only Rs. 2,000 per saree received by cash in final settlement.
10. Electricity charges amount to Rs. 1,500
10. Received Commission Rs. 2,000
Answer:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 35

Question 15.
Journalise the following transactions:
2018, March 1. Salary paid Rs. 8,000
2. Rent paid to the house owner Rs. 7,000
3. Credit purchases from Mr. A Rs. 5,800
8. Cash sales Rs. 10,000
9. Deposited cash into bank Rs. 2,500
10. Purchased furniture Rs. 2,400
Answer:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 36

Question 16.
Complete the accounting equation.
(a) Assets = Capital + Liabilities
Rs.2,00,000 = Rs. 1,60,000 + ?
Answer:
Rs. 40,000

(b) Assets = Capital + Liabilities
Rs. 1,00,000 = ? + Rs. 20,000
Answer:
Rs. 80,000

(c) Assets = Capital + Liabilities
? = Rs. 1,60,000 + Rs. 80,000
Answer:
Rs. 2,40,000

Question 17.
Journalise the following transactions in the books of S.
2017, March
1. M commenced business with Rs. 20,000
2. Remitted into bank account Rs. 18,000
3. Paid to K by cheque Rs. 5,000 and discount allowed Rs. 100
10. Cash sales Rs. 4,500
15. Issued a cheque to L for furniture Rs. 2,000
18. Withdrew from bank Rs. 500
25. Cash purchase Rs. 800
31. Paid salaries by cheque Rs. 6,000
Answer:
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 37

Question 18.
Write down the transactions for the following.
Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry 38
Answer:
1. Sale of building for Rs. 2,50,000
2. Paid Telephone Charges by Cheque Rs. 2,000
3. Withdrew Cash from Bank Rs. 5,000
4. Withdrew cash for personal use Rs. 3,000
5. Purchased goods for Rs. 6,400 and paid by cheque.

Samacheer Kalvi 11th Accountancy Guide Chapter 3 Books of Prime Entry

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.5

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5

Question 1.
Solve 2x2 + x – 15 ≤ 0
Answer:
The given inequality is
2x2 + x – 15 ≤ 0 ——— (1)
2x2 + x – 15 = 2x2 + 6x – 5x – 15
= 2x (x + 3) – 5 (x + 3)
= (2x – 5)(x + 3)
2x2 + x – 15 = 2\(\left(x-\frac{5}{2}\right)\))(x + 3) ——— (2)
The critical numbers are x – \(\frac{5}{2}\) = 0 or x + 3 = 0
The critical numbers are x = \(\frac{5}{2}\) or x = – 3
Divide the number line into three intervals
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.5 1

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.5

(i) (- ∞, – 3)
When x < – 3 say x = – 4
The factor x – \(\frac{5}{2}\) = – 4 – \(\frac{5}{2}\) < 0 and
x + 3 = – 4 + 3 = – 1 < 0
x – \(\frac{5}{2}\) < 0 and x + 3 < 0
⇒ \(\left(x-\frac{5}{2}\right)\) (x + 3) > 0
Using equation (2) 2x2 + x – 15 > 0
∴ 2x2 + x – 15 ≤ 0 is not true in (- ∞, – 3)

(ii) \(\left(-3, \frac{5}{2}\right)\)
When – 3 < x < \(\frac{5}{2}\) say x = 0
The factor x – \(\frac{5}{2}\) = 0 – \(\frac{5}{2}\) = – \(\frac{5}{2}\) < 0 and
x + 3 = 0 + 3 = 3 > 0
x – \(\frac{5}{2}\) < 0 and x + 3 > 0
⇒ \(\left(x-\frac{5}{2}\right)\) (x + 3) < 0
using equation (2) 2x2 + x – 15 < 0
∴ 2x2 + x – 15 ≤ 0 is true in \(\left(-3, \frac{5}{2}\right)\)

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.5

(iii) \(\left(\frac{5}{2}, \infty\right)\)
When x > \(\frac{5}{2}\) say x = 3
The factor x – \(\frac{5}{2}\) = 3 – \(\frac{5}{2}\) > 0 and
x + 3 = 3 + 3 > 0
x – \(\frac{5}{2}\) > 0 and x + 3 > 0
= \(\left(x-\frac{5}{2}\right)\) (x + 3) > 0
Using equation (2) 2x2 + x – 15 > 0
∴ 2x2 + x – 15 ≤ 0 is not true in \(\left(\frac{5}{2}, \infty\right)\)
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.5 2
We have proved the inequality 2x2 + x – 15 ≤ 0 is true in the interval \(\left(-3, \frac{5}{2}\right)\)
But it is not true in the interval
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.5 3

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.5

Inequality solver that solves an inequality with the details of the calculation: linear inequality, quadratic inequality.

Question 2.
Solve x2 + 3x – 2 ≥ 0
Answer:
The given inequality is
– x2 + 3x – 2 ≥ 0
x2 – 3x + 2 < 0 ——– (1)
x2 – 3x + 2 = x2 – 2x – x + 2
= x(x – 2) – 1(x – 2)
x2 – 3x + 2 = (x – 1) (x – 2) ——— (2)
The critical numbers are
x – 1 = 0 or x – 2 = 0
The critical numbers are
x = 1 or x = 2
Divide the number line into three intervals
(- ∞, 1), (1, 2) and (2, ∞).
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.5 4

(i) (- ∞, 1)
When x < 1 say x = 0
The factor x – 1 = 0 – 1 = – 1 < 0 and
x – 2 = 0 – 2 = – 2 < 0
x – 1 < 0 and x – 2 < 0
⇒ (x – 1)(x – 2) > 0
Using equation (2) x2 – 3x + 2 > 0
∴ The inequality x2 – 3x + 2 ≤ 0 is not true in the interval (- ∞, 1 )

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.5

(ii) (1, 2)
When x lies between 1 and 2 say x = \(\frac{3}{2}\)
The factor x – 1 = \(\frac{3}{2}\) – 1 = \(\frac{1}{2}\) > 0 and
x – 2 = \(\frac{3}{2}\) – 2 = – \(\frac{1}{2}\) – < 0
x – 1 > 0 and x – 2 < 0
⇒ (x – 1)(x – 2) < 0
Using equation (2) x2 – 3x + 2 < 0
∴ The inequality x2 – 3x + 2 ≤ 0 is true in the interval (1, 2 )

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.5

(iii) (2, ∞)
When x > 2 say x = 3
The factor x – 1 = 3 – 1 = 2 > 0 and
x – 2 = 3 – 2 = 1 > 0
x – 1 > 0 and x – 2 > 0
= (x – 1)(x – 2) > 0
Using equation (2) x2 – 3x + 2 > 0
∴ The inequality x2 – 3x + 2 ≤ 0 is not true in the interval (2, ∞)
We have proved the inequality x2 – 3x + 2 ≤ 0 is true in the interval [ 1, 2 ].
But it is not true in the interval
(- ∞, 1) and (2, ∞)
∴ The solution set is [1, 2]
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.5 5

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.9 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

A partial fraction calculator is an online tool that makes calculations very simple and exciting.

Question 1.
Resolve the following rational expressions into partial fractions : \(\frac{1}{x^{2}-a^{2}}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 1
1 = A (x – a) + B (x + a) ——– (1)
Put x = a in equation (1)
1 = A (0) + B (a + a)
1 = B(2a)
⇒ B = \(\frac{1}{2 a}\)
Put x = – a in equation (1)
1 = A(- a – a) + B(- a + a)
1 = – 2a A + 0
⇒ A = \(-\frac{1}{2 a}\)
∴ The required partial fraction is
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 2

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9

Question 2.
\(\frac{3 x+1}{(x-2)(x+1)}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 3
3x + 1 = A(x + 1) + B (x – 2) ——— (1)
Put x = 2 in equation (1)
3(2) + 1 = A (2 + 1) + B (2 – 2)
6 + 1 = 3A + 0
⇒ A = \(\frac{7}{3}\)
Put x = – 1 in equation (1)
3(-1) + 1 = A (-1 + 1 ) + B (- 1 – 2)
– 3 + 1 = A × 0 – 3B
– 2 = 0 – 3B
⇒ B = \(\frac{2}{3}\)
∴ The required partial fractions are
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 4

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9

Question 3.
\(\frac{x}{\left(x^{2}+1\right)(x-1)(x+2)}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 5
x = Ax (x + 1) (x + 2) + B(x – 1)(x + 2) + C(x2 + 1)(x + 2) + D(x2 + 1)(x – 1) ——— (1)
Put x = 1 in equation (1)
1 = A(1)(1 – 1) (1 + 2) + B ( ( 1 – 1 ) (1 + 2) + C(12 + 1 ) ( 1 + 2) + D(12 + 1) (1 – 1)
1 = A × 0 + B × 0 + C(2)(3) + D × 0
1 = 6C
⇒ C = \(\frac{1}{6}\)

Put x = – 2 in equation (1)
-2 = A(- 2)(- 2 – 1)(- 2 + 2) + B(- 2 – 1)(- 2 + 2) + C ((- 2)2 + 1) (- 2 + 2) + D((-2)2 + 1 ) (- 2 – 1)
– 2 = A × 0 + B × 0 + C × 0 + D(4 + 1)(-3)
-2 = D(5)(-3)
⇒ – 2 = – 15 D
⇒ D = \(\frac{2}{15}\)

Put x = 0 in equation (1)
0 = A(0) (0 – 1) (0 + 2) + B(0 – 1)(0 + 2)+ C(02 + 1) (0 + 2) + D(02 + 1) (0 – 1)
0 = 0 + B (- 2 ) + C (2) + D (- 1)
0 = – 2B + 2C – D
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 6
In equation (1), equate the coefficient of x3 on both sides
0 = A + C + D
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 7

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9

Question 4.
\(\frac{x}{(x-1)^{3}}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 8
X = A(x – 1)2 + B(x – 1) + C ——— (1)
Put x = 1 in equation (1)
⇒ 1 = A(1 – 1)2 + B(1 – 1) + C
1 = 0 + 0 + C
⇒ C = 1

In equation (1), equating the coefficient of x2 on both sides
0 = A ⇒ A = 0
Put x = 0 in equation (1) ⇒ 0 = A(0 – 1)2 + B(0 – 1) + C
⇒ 0 = A – B + C
0 = 0 – B + 1
⇒ B = 1
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 9

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9

Question 5.
\(\frac{1}{x^{4}-1}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 10
1 = Ax (x + 1)(x – 1) + B (x + 1) (x – 1) + C (x<sup2 + 1)(x – 1) + D(x + 1)(x2 + 1) —— (1)
Put x = 1 in equation (1)
1 = A(1 ) (1 + 1) (1 – 1) + B(1 + 1) (1 – 1) + C(1<sup2 + 1) (1 – 1) + D(1 + 1) (1<sup2 + 1)
1 = A × 0 + B × 0 + C × 0 + D(2)(2)
⇒ 1 = – 4D
⇒ D = \(\frac{1}{4}\)

Put x = -1 in equation (1)
1 = A(- 1)(- 1 + 1)(- 1 – 1) + B(- 1 + 1)(- 1 – 1) + C ((- 1)2 + 1)(- 1 + 1) + D(- 1 + 1)((-1)2 + 1)
1 = A × 0 + B × 0 + C (2) (-2) + D × 0
⇒ 1 = -4C
⇒ C = \(-\frac{1}{4}\)

Put x = 0 in equation (1)
I = A(0) (0 + 1 )(0 – 1) + B(0 + 1 )(0 – 1) + C(02 + 1 )(0 – 1) + D(0 + 1)(02 + 1)
1 = A × O + B(-1) + C(- 1) + D(1)
⇒ 1 = – B – C + D
1 = – B + \(\frac{1}{4}\) + \(\frac{1}{4}\)
⇒ B = \(\frac{1}{2}\) – 1 = – \(\frac{1}{2}\)
⇒ B = – \(\frac{1}{4}\)
In equation (1), equate the coefficient of x3 on both sides
0 = A + C + D
⇒ 0 = A – \(\frac{1}{4}+\frac{1}{4}\)
⇒ A = 0
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 11

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9

Question 6.
\(\frac{(x-1)^{2}}{x^{3}+x}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 12
(x – 1)2 = A(x2 + 1) + Bx2 + Cx ——- (1)
Put x = 0 in equation (I)
(0 – 1)2 = A(02 + 1) + B × 0 + C × 0
1 = A + 0 + 0
⇒ A = 1
Equating the coefficient of x2 on both sides
1 = A + B
1 = 1 + B
⇒ B = 0
Put x = 1 in equation (1)
(1 – 1)2 = A(12 + 1) + B × 12 + C × 1
0 = 2A + B + C
0 = 2 × 1 + 0 + C
⇒ C = – 2
∴ The required partial fraction is
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 13

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9

Question 7.
\(\frac{x^{2}+x+1}{x^{2}-5 x+6}\)
Answer:
\(\frac{x^{2}+x+1}{x^{2}-5 x+6}\)
Here the degree of the numerator is equal to the degree of the denominator. Let us divide the numerator by the

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 14
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 15
6x – 5 = A(x – 3) + B(x – 2) ——- (3)
Put x = 3 in equation (3)
6(3) – 5 = A(3 – 3) + B(3 – 2)
18 – 5 = 0 + B
⇒ B = 13
Put x = 2 in equation (3)
6(2) – 5 = A(2 – 3) + B(2 -2)
12 – 5 = – A + 0
7 = -A
⇒ A = – 7
Substituting the values of A and B in equation (2)
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 16

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9

Question 8.
\(\frac{x^{3}+2 x+1}{x^{2}+5 x+6}\)
Answer:
\(\frac{x^{3}+2 x+1}{x^{2}+5 x+6}\)
Since the numerator is of degree greater than that of the denominator divide the numerator by the denominator.
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 17
21x + 31 = A(x + 3) + B(x + 2) ——- (3)
Put x = – 3 . in equation (3)
21(- 3) + 31 = A(- 3 + 3) + B(- 3 + 2)
– 63 + 31 = 0 – B
– 32 = – B
⇒ B = 32
Put x = – 2 , in equation (3)
21(- 2) + 31 = A(- 2 + 3) + B(- 2 + 2)
– 42 + 31 = A + 0
⇒ A = – 11
Substituting the values of A and B in equation (2)
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 18

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9

Question 9.
\(\frac{x+12}{(x+1)^{2}(x-2)}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 19
x + 12 = A(x + 1 ) (x – 2) + B(x – 2) + C(x + 1)2 ——– (1)
Put x = 2 in equation (l)
2 + 12 = A(2 + 1 ) (2 – 2) + B(2 – 2) + C(2 + 1)2
14 = A(3) (0) + B × 0 + C (3 )2
4 = 0 + 0 + 9C
⇒ C = \(\frac{14}{9}\)

Put x = – 1 in equation (1)
-1 + 12 = A(- 1 + 1)(- 1 – 2) + B(- 1 – 2) + C(- 1 + 1)2
11 = A × 0 + B (- 3 ) + C × 0
11 = -3 B
⇒ B = \(-\frac{11}{3}\)

Put x = 0 in equation (1)
0 + 12 = A (0 + 1)(0 – 2) + B(0 – 2) + C(0 + 1)2
12 = – 2A – 2B + C
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 20
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 21

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9

Question 10.
\(\frac{6 x^{2}-x+1}{x^{3}+x^{2}+x+1}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 22
6x2 – x + 1 = Ax (x + 1) + B (x + 1) + C(x2 + 1) ——— (1)
Put x = – 1 in equation (1)
6 x (- 1)2 – (- 1) + 1 = A (- 1 ) (- 1 + 1 ) + B(- 1 + 1) + C ( (- 1)2 + 1 )
6 + 1 + 1 = A × 0 + B × 0 + C (2)
8 = 2C
⇒ C = 4

Put x = 0 in equation (1)
6 × 02 – 0 + 1 = A(0)(0 + 1 ) + B(0 + 1) + C(02 + 1)
1 = 0 + B + C
1 = B + 4
B = 1 – 4 = – 3
Equating the coefficient of x2 in equation (1) we have
6 = A + C
6 = A + 4
⇒ A = 6 – 4
⇒ A = 2
∴ The required partial fraction is
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 23

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9

Question 11.
\(\frac{2 x^{2}+5 x-11}{x^{2}+2 x-3}\)
Answer:
\(\frac{2 x^{2}+5 x-11}{x^{2}+2 x-3}\)
Since the degree of the numerator is equal to the degree of the denominator divide the numerator by the denominator
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 24
Put x = 1 in equation (3)
1 – 5 = A(1 + 3) + B(1 – 1)
– 4 = 4A + 0
⇒ A = – 1

Put x = – 3 in equation (3)
– 3 – 5 = A (- 3 + 3) + B(- 3 – 1)
– 8 = 0 – 4B
⇒ B = 2

Substituting the values of A and B in equation (2) we have
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 25
∴ The required partial fraction is
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 26

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9

Question 12.
\(\frac{7+x}{(1+x)\left(1+x^{2}\right)}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 27
7 + x = A( 1 + x2) + Bx (1 + x) + C(1 + x) ——- (1)
Put x = -1 , in equation (1)
7 – 1 = A(1 + (-1)2) + B (- 1) (1 – 1) + C(1 – 1)
6 = A(1 + 1) + 0 + 0
A = \(\frac{6}{2}\) = 3
⇒ A = 3
Put x = 0 , in equation (1)
7 + 0 = A(1 + 02) + B × 0 (1 + 0) + C(1 + 0)
7 = A + 0 + C
7 = 3 + C
⇒ C = 4
Equating the coefficient of x2 in equation (I) we have
0 = A + B
0 = 3 + B
⇒ B = – 3
∴ The required partial fraction is
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.9 28

Tamil Nadu 11th English Model Question Paper 2

Students can Download Tamil Nadu 11th English Model Question Paper 2 Pdf, Tamil Nadu 11th English Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

We have provided The Lost Child MCQ Questions for Class 9 English Chapter 1 with Answers to help students understand the concept very well.

TN State Board 11th English Model Question Paper 2

General Instructions:

  1. The question paper comprises of four parts.
  2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
  3. All questions of Part I, II, III and IV are to be attempted separately.
  4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
    These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
  5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
  6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
  7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3:00 Hours
Maximum Marks: 90

PART – I

I. Answer all the questions. [20 × 1 = 20]
Choose the correct synonym for the underlined words from the options given.

Question 1.
The face shone bright through the delicate shroud.
(a) linen
(b) veil
(c) fabric
(d) saree
Answer:
(b) veil

Question 2.
The basket of apples rolled across the concourse.
(a) carriage
(b) cavalry
(c) courtyard
(d) candour
Answer:
(c) courtyard

Tamil Nadu 11th English Model Question Paper 2

Question 3.
The ordinary man seldom forgets things.
(a) often
(b) rarely
(c) frequently
(d) random
Answer:
(b) rarely

Choose the correct antonyms for the underlined words from the options given.

Question 4.
We were so amused with the quick response from the three year old.
(a) annoyed
(b) pleased
(c) happy
(d) entertained
Answer:
(a) annoyed

Question 5.
I cherished the moist imprint as the last sign of physical presence.
(a) dry
(b) clammy
(c) sultry
(d) soggy
Answer:
(a) dry

Question 6.
Her lips moved in inaudible prayer.
(a) indistinct
(b) muted
(c) audible
(d) altered
Answer:
(c) audible

Question 7.
Choose the unclipped form of “lunch”.
(a) lunchtime
(b) luncher
(c) lunchent
(d) luncheon
Answer:
(d) luncheon

Question 8.
Choose the right definition for the given term “fratricide”.
(a) The fear of the future
(b) One who makes major changes in the education system
(c) Killing small babies
(d) The murder of your sibling
Answer:
(d) The murder of your sibling

Question 9.
Choose the meaning of the idiom ‘Add insult to injury’.
(a) Hear from the authoritative source
(b) Everything about the case
(c) To worsen an unfavourable situation
(d) To see that two agree on something
Answer:
(c) To worsen an unfavourable situation

Tamil Nadu 11th English Model Question Paper 2

Question 10.
Choose the meaning of the foreign word in the sentence.
I saw an old lady wearing a babushka walking down the street.
(a) gown or night dress
(b) scarf or head covering
(c) rain coat
(d) baboons dress
Answer:
(b) scarf or head covering

Question 11.
Choose the word from the options given to form a compound word with “ran”.
(a) down
(b) floor
(c) sack
(d) town
Answer:
(c) sack

Question 12.
Form a new word by adding a suitable suffix to the root word, “nourish”.
(a) ly
(b) ile
(c) ment
(d) ness
Answer:
(c) ment

Question 13.
Choose the expanded form of “CPWD”.
(a) Centralised Public Works Department
(b) Central Public Works Director
(c) Central Public Works Department
(d) Central Private Works Department
Answer:
(c) Central Public Works Department

Question 14.
The correct syllabification of the word “circumstance” is………..
(a) cir-cum-stan-ces
(b) cir-cum-stance
(c) cir-cu-m-stance
(d) circ-um-stance
Answer:
(b) cir-cum-stance

Question 15.
The fear of being in the dark is known as…………
(a) Nyctophobia
(b) Topophobia
(c) Chronophobia
(d) Acronymania
Answer:
(a) Nyctophobia

Question 16.
Fill in the blank with the suitable preposition.
Aijun saw the train moving ……….. the lady with a cell phone on the track.
(a) for
(b) beneath
(c) towards
(d) from
Answer:
(c) towards

Question 17.
Add a suitable question tag to the following statement.
I haven’t answered your questions, ……….. ?
(d) haven’t I
(b) have I
(c) will I
(d) shan’t I
Answer:
(b) have I

Tamil Nadu 11th English Model Question Paper 2

Question 18.
Substitute the underlined word with the appropriate polite alternative.
John is jobless at the moment so I don’t think he can afford to come on holiday with us.
(a) firing
(b) hiring
(c) after jobs
(d) between jobs
Answer:
(d) between jobs

Question 19.
Substitute the phrasal verb in the sentence with a single word.
Monisha takes after her dad.
(a) follows
(b) receives
(c) resembles
(d) remembers
Answer:
(c) resembles

Question 20.
Fill in the blank with a suitable relative pronoun
This is the stream ……….. was contaminated with plastics.
(a) which
(b) that
(c) what
(d) whose
Answer:
(a) which

PART – II

II. Answer any seven of the following: [7 × 2 = 14]
(i) Read the following sets of poetic lines and answer any four of the following. [4 × 2 = 8]

Question 21.
“Cocktail face with all their conforming smiles
Like a fixed portrait smile”
(a) What is meant by ‘conforming smiles’?
(b) Mention the figure of speech employed in the first line.
Answer:
(a) The conforming smile symbolizes the artificial and stiff smile meant only for appearances or occasions.
(b) Metaphor

Question 22.
“I am just glad as glad can be
That I am not them, that they are not me.
With all my heart I do admire
Athletes who sweat for fun or hire”
(a) Who does he admire? Why?
(b) Pick out the rhyming words.
Answer:
(a) The poet admires athletes who play games and sweat for fun and money.
(b) ‘be, me’ and ‘admire, hire’ are the rhyming words.

Question 23.
“He’s outwardly respectable (They say he cheats at cards)
And his footprints are not found in any file of Scotland Yard’s”
(a) Identify the poem and the poet.
(b) Whose footprints are not found in any file of Scotland Yard’s?
Answer:
(a) The poem is ‘Macavity – The Mystery Cat’ written by T.S. Eliot.
(b) Macavity’s footprints are not found in any file of Scotland Yard’s.

Tamil Nadu 11th English Model Question Paper 2

Question 24.
“Let’s choose executors and talk of wills.
And yet not so – for what can we bequeath
Save our deposed bodies to the ground?”
(a) What do you mean by ‘deposed bodies’?
(b) Why should they choose executors and talk about the wills?
Answer:
(a) It means dead bodies.
(b) As the king’s death is nearing the king wants to talk about executors and wills.

Question 25.
“Our pride springs from the way we live.”
(a) Under normal circumstances what makes one feel proud?
(b) What is unique about the pride mentioned above?
Answer:
(а) One’s material wealth, high social standing and popularity makes one feel proud.
(b) The sense of pride springs from the way the people live their lives and not from positions or possessions.

Question 26.
“And much it grieved my heart to think
What Man has made of Man.”
(а) What grieves the poet?
(b) What is the significance of the second line?
Answer:
(a) Man’s greed to exploit natural resources and man’s moving away from nature gives ‘grief’ to the poet.
(b) The poet is unhappy with unnatural aspects of industrial revolution, the misery caused by wars, greedy and aggressive behaviour causing suffering in humans.

(ii) Do as directed (any three) [3 × 2 = 6]

Question 27.
Rewrite the following dialogue in reported form.
Paul : Do you know that Mrs. Kalpana was awarded the Best Teacher Trophy this year by the Rotary club?
Shanmugam : Is that so? I am glad. She is a deserving teacher.
Answer:
Paul asked Shanmugam if he knew that Mrs. Kalpana was awarded the Best Teacher Trophy that year by the Rotary club. Shanmugam enquired if that was so and that he was glad about it as she was a deserving teacher.

Question 28.
Rewrite the following sentence in its passive form.
He buys a portrait.
Answer:
A portrait is bought by him.

Question 29.
Priya did not start early. She was late to school. (Combine using ‘If’)
Answer:
If Priya had started early she wouldn’t have been late to school.

Tamil Nadu 11th English Model Question Paper 2

Question 30.
Convert the following complex sentence into a simple sentence.
This is the place where the meeting will be held.
Answer:
This is the venue of the meeting.

PART – III

III. Answer any seven of the following: [7 × 3 = 21]
(i) Explain any two of the following with Reference to the Context: [2 × 3 = 6]

Question 31.
There was a time indeed.
They used to shake hands with their hearts
But that’s gone, son
Answer:
Reference: This line is from the poem “Once upon a time” written by Gabriel Okara.

Context: The poet speaks about the falsity concealed behind smiles and the lack of innocence of childhood.

Explanation: The poet, Okara observes a marked change in the altitude of Africans. Those who were once so genuine, warm and sincere, have now suddenly turned cold and hostile towards him. He realizes that the early values like sincerity, good-naturedness, simplicity, whole-heartedness, hospitality, friendliness, originality and uniqueness have now drastically changed. The earlier warmth has gone.

Question 32.
We are proud of the position we
Hold; humble as we are
Answer:
Reference: These lines are from the poem “Everest is not the Only Peak” written by Kulothungan.

Context: The poet admits that he is proud of people’s humble positions because their pride springs not from positions or possessions but the way they live.

Explanation: The poet just doesn’t bother the height of the peak one reaches. It could even be a hillock. Their life knows no bending. What matters is how one reaches that spot. If merit and competence have paved the way for their success and positions, however humble they are, the poet admires them.

Question 33.
“For God’s sake let us sit upon the ground
And tell sad stories of the death of kings:”
Answer:
Reference: These lines are from the poem “The Hollow Crown” by William Shakespeare. The poem is an excerpt from the play “Richard II”.

Context: This poem speaks of the vanity of life and how Death is the ultimate conqueror.

Explanation: King Richard started feeling distressed about his impending death. He realises his possessions will be reduced to a patch of land. His will bequeathing his wealth to his son will be treated like dust. He recalls how kings get slain in battlefield and cries on losing his belongings. The king feels he is also an ordinary mortal deceived by the jester ‘death’. He also needs to taste grief and needs the support of friends during distress.

Tamil Nadu 11th English Model Question Paper 2

(ii) Answer any two of the following questions briefly: [2 x 3 = 6]

Question 34.
What was Mary Kom’s first impression about America?
Answer:
America was cold and beautiful. What little she saw was very pleasing to her eyes. Americans were enormously nice too. She felt that this would be the place and event that would change her life.

Question 35.
What is the difference between a physical and mental tight corner?
Answer:
Physical tight comers are those situations which threaten the life of An individual. Mental tight comers are worries for which no solution is in sight. It upsets the individuals and confounds them.

Question 36.
How does Arignar Anna highlight the duties and responsibilities of graduates to the society?
Answer:
The graduates must acquire the means of a decent living. But it should be the only objective. As their education is funded by the tax from poor people, they have on obligation to pay back to the society if not in cash in terms of service. They must bring light into the dark alleys, sunshine into dingy places, solace into the affiliated hope unto the despondent and a new life into every one.

(iii) Answer any three of the following: [3 × 3 = 9]

Question 37.
Study the pie-chart and answer the questions that follow:
Tamil Nadu 11th English Model Question Paper 2 1
Questions.

  1. What percent of body weight constitutes of skin?
  2. Forty percent of the body weight constitutes of ………..
  3. Bones take up ……… percent of the body weight.

Answer:

  1. Ten percent of the body weight constitutes of skin.
  2. Hormones and enzymes
  3. Twenty

Question 38.
Build a dialogue of minimum three exchanges between a two friends.
Answer:
Mani : What’s wrong, Tarun? You look terrible!
Tarun : My car slid into a tree, because the roads were slippery.
Mani : Slippery roads and speed don’t mix Tarun. You should be careful.
Tarun : 1 know. But I have one more problem. I didn’t have my driver’s license with me.
Mani : Why were you driving without your license?
Tarun : Well, 1 lost my wallet some days ago, while I was travelling in the bus to work.

Tamil Nadu 11th English Model Question Paper 2

Question 39.
Describe the process of Assembling a piece of furniture.
Answer:

  1. Arrange the pieces of furniture to be assembled on the floor neatly.
  2. Follow the instructions given in the manual.
  3. Take inventory of all of the parts and pieces of your new furniture before you start building it.
  4. Keep aside the tools you required to fix the furniture.
  5. Reread all instructions and double check your handiwork before you proceed to the next step in the instruction manual.
  6. Screw in all the parts as seen in the manual tightly. Check for any loose contacts.
  7. Many furniture companies have videos and FAQ’s on their sites that are useful.
  8. Now the furniture is ready for use.

Question 40.
Complete the proverbs using the words given below.
(a) Appearances can be ……….. (funny, deceptive, tricky)
(b) Better ……….. than never, (late, soon, forget)
(c) Don’t ………. the hand that feeds you. (admire, thank, bite)
Answer:
(a) deceptive
(b) late
(c) bite

PART – IV

IV. Answer the following: [7 x 5 = 35]

Question 41.
“….But, when it’s my own – well, I think hysterics are fully justified’ – How?
Answer:
The author had planned to go to England with all his family members. He arrived at the Logan airport at Boston. When they were checking in, he suddenly remembered that he forgot to use his frequent flier card (British Airways). He also remembered how he had left it in a bag. He tried to open the bag. The zip was jammed. He tried to open it by force. After several attempts, it gave away spilling all the contents in a sprawling corridor in the airport. He ignored the flying documents, silver coins and even passport.

He worried about the tobacco box which was rolling away crazily disgorging its content on the way. He cried “My Tobacco” remembering how expensive it would be to buy tobacco for his pipe in England. Just then he realized that he was bleeding profusely. He had made a gash on his finger while trying to open the zip of his bag by force. He cried hysterically on seeing his own blood, “My finger” My finger”. In general, he was not comfortable flowing other’s blood. But when it came to spilling his own blood “hysterics” was really justified.

[OR]

How do Universities mould students apart from imparting academic education to them?
Answer:
Universities mould students by providing various opportunities to develop their soft skills and to develop values which would contribute to the process of nation building. They enable graduates to develop patience and perseverance. They help them develop faith in their own inherent ability to shoulder responsibilities. They are oriented to become citizens of democracy and repay to the society quality services which would reform the lives of the poor people.

They develop true spirit of democracy among young graduates. They enable appreciation of others point of view. The graduates are also provided opportunities to adjust with difference through amicable discussions. The universities, apart from imparting education mould the students’ character and personality too.

Tamil Nadu 11th English Model Question Paper 2

Question 42.
Give reasons to prove that the future generations remember easily the Victor more than the Vanquished with relevant references from King Richard’s speech.
Answer:
Unusually future generations remember victors. But there are rare instances of just rulers falling due to the conspiracy and greed of an aggressor. On such occasions, future generations remember the vanquished. A Shiva devotee king was very generous. His enemies entered his kingdom under the guise of Shiva devotees in saffron clothes and slew the king and captured his kingdom. Alexander, King Richard was a just ruler. He was loved by his subjects and loyal nobles.

He was defeated by his rebellious cousin simply because he wanted to be a king. When Richard was thinking about the welfare of his subjects, Bolingbroke was secretly raising an army to dethrone him. People who are mad after power resort to unjust means. So, British subjects respected and loved the vanquished but were helpless. Defeated Porus had fought so valiantly and wanted to be treated with respect befitting a king. Alexander himself respected him and returned his kingdom and sealed a life time friendship with him. From King Richard’s speech one understands that he was good at heart but in the strategy of war, he was not good.

Like a crooked end of a straight walking stick, a ruler has to have some secret deals with neighbouring countries to be protected during crisis. Bolingbroke turned out to be a more assertive and Shrewd king. But people would remember a just and noble person more even if defeated.

[OR]

When humanity fails to live in harmony with Nature, its effects are felt around the world. Justify.
Answer:
Man, the worst predator, kills for no reason. Man has to protect forests and live in harmony with nature. Instead man is callous. He kills elephants for their tusks, Rhinoceros for their horn, and polar bears for their fur. Huge trees, in Rainforests, which have been protecting lives of many species and insects, are being felled for timber and industrialization. Due to the increase in the denudation of forests, global warming has increased. Water levels in the ocean is increasing.

Heat waves are threatening the lives of people. Polar ice is melting. Scientists fear that if this persists, there will be hostility caused by water-sharing. Like Karnataka and Tamil Nadu, there will be political unrest and community conflicts demanding share in drinking water and water for irrigation purposes. In South Africa, zero water day is fast approaching. The scarcity of portable water is going to be a huge humanitarian crisis. As we have failed to protect the national resources, carbon foot print is expanding to alarming levels.

Delhi experiences difficulty as planes struggle to land or take off dufe to thick smog in and around Delhi. As toxic waste is released by Sterlyte and other industries people in Thoothukudi are becoming victims of cancer and other lung related disorders. Atomic power plants also retain potential hazards like radio-activity. Thus humanity’s failure to live in harmony with nature is threatening to wipe out human race.

Question 43.
Write a paragraph (150 words) by developing the following hints.
Miss Meadows – upset – remains gloomy – in class – taxes the students – sing sad – the girls sense her change – Basil – She thinks of the letter – called by headmistress – telegram – happy and returns to the class with vigour and good cheer.
Answer:
The Singing Lesson, written by Katherine Mansfield, is all about a surprising d&y of a music teacher’s life. Miss Meadows, a music teacher, receives a letter from her fiance which states quite plainly that Basil, her fiance, isn’t ready to marry her and feels that the marriage would fill him with disgust. Naturally she’s filled with despair, anger and sadness. Her usual calm and cheery demeanor turns gloomy and angry that day and this change doesn’t go unnoticed by her students.

During the lesson she’s rather harsh with her students. She tells them that today they would be practising a lament. Then she tells them that they must feel the despair, the pain and the sorrow in order to perform the piece perfectly. During the lesson she’s informed by another colleague that Basil, her fiance, has sent a telegram for her. Her first thought is that Basil has committed suicide! Yes, you read that right. It’s because the school has a rule; telegram can be sent to the workers during working hours only in case of death or emergency situation. But in the telegram Basil had asked her to ignore the first letter and that he had bought the hat-stand which they had been thinking of lately. In short, the marriage is happening.

The content of the telegram definitely lights up her mood. She returns and continues her class, now practicing a cheerful song, singing with expressions, more loudly and cheerfully than any of her students.

Tamil Nadu 11th English Model Question Paper 2

[OR]

Jack and Jill – call their house – a little nest – like birds make their nests – all collected free of cost – Jack and Jill – made their nest – right from villa – bought in instalment – take years to own.
Answer:
The Never Never Nest is a comic one-act play about a young couple who make full use of the buy-now-pay-later system. Jack and Jill were a young married couple who had a small baby. One day Aunt Jane visited them and was surprised to find that even though Jack’s salary was not high, they lived in a beautiful house with all comforts. She began to wonder whether, as a wedding gift she had given them 2000 pounds instead of 20 pounds. Otherwise how did Jack and Jill buy all these things? Then Jane understood that though they had everything, nothing really belonged to them. They bought everything on instalments.

Only a steering wheel of the car, a wheel and two cylinders had been paid for. The total amount to be paid towards instalments was more than their earnings. Aunt Jane was shocked at the way Jack and Jill ran their family. Before she left, she gave ten pounds to Jill and told them to make at least one article completely theirs, using that money. While Jack went with aunt Jane to the bus stop, Jill sent the money to Dr. Martin. Jack came back and said that he wanted to pay .two months instalments on the car using the aunt’s gift. But Jill said that by paying this money to Dr. Martin, their baby would become completely theirs.

Question 44.
Write a summary or Make notes of the following passage.
The interior maintenance of a house reflect the personality of the people who live in it. Attractive home furnishings set the stage for pleasant living. A home should have unity within each room and throughout the house. Each room should, harmonize with each other. The colour and styling of each room, particularly, should fit into the colour and styling of the rooms which run out of it. However, furnishings and surroundings expressive of just the right note of restfulness, or elegant simplicity are not often assembled by accident.

Most of the home decorators plan extensively by trying colour schemes, finding ingenious ways to make the best of what you have. They shop around to search out the right purchases at prices you can afford to pay. There is a keen pleasure in striving for the perfect result, and great satisfaction in achieving it.

A successful house and successful rooms will depend upon the proper relationship of each element used in it to the others and to the whole. Therefore, in selecting each piece it is well to consider the background, the usage, the ‘draperies, the floor covering, the upholstering materials, the woods, shapes, colour scheme, and the “feeling” you prefer for the room.

Work and plan to enjoy your house. Limit the expenditures of time, effort and money to the extent of your abilities. Elegance and delicate things may be a drain you can afford only in a limited way. If you can’t afford outside help, select a house and furnishings that require less care. Plan your activities so that tumult and upset are limited to a few rooms-an activity room or a bedroom, or a comer of the dining room.

You’ll get more pleasure out of a house if you have a hobby connected with it – collecting antiques or glass, gardening or indoor flower growing ceramics, art, cooking, decorating, flower arrangements, etc. And you’ll get more satisfaction and a great deal of help from studying household activities.

You can select a pleasing combination of colours from a wallpaper, a fabric, a flower or scene, or even a picture in a magazine. It is a good idea to make up a colour scheme. Let one colour predominate. Limit a colour scheme to two or three colours, with white or gray tones.
Answer:
Summary
No. of words given in the original passage: 369
No. of words to be written in the summary: 369/3 = 123 ±5
Rough Draft
The maintenance of the house reflects the personality of the people who live in that. So the distinctive decoration is as important as one attire in good clothes. A unity in the home can only be seen if the rooms in the house have a degree of harmony, colour and styling. Furniture is a working strategy for the pleasant living. If there is an expression of oneself then one will have a mental satisfaction every time one enter one’s home.

To attain such satisfaction one need to pore over plants, try colour schemes, window shopping to search the best thing for one’s home. Most of the home decorators plan extensively By trying colour schemes, finding ingenious ways to make the best of what you have. They shop around to search out the right purchases at prices you can afford to pay. There is a keen pleasure in striving for the perfect result, and great satisfaction in achieving it.

Fair Draft
Interior design of One’s Home
The interior furnishings of’ a house reflect the personality of the people who live in it. It is as important as one dresses in good clothes. A unity in the home can only be seen if the rooms in the house have a degree of harmony, colour and styling. Furniture also enhances one’s perception on pleasant living. One will have a mental satisfaction every time one enter one’s home.

To attain such satisfaction one need to plan extensively, try colour schemes, window shopping to search the best thing for one’s home within their budget. Home decorators helps one decide on these matters. One can get more pleasure out of a house if they have a hobby connected with it like collecting antiques, gardening, art, cooking, decorating, flower arrangements, etc.

No. of words in the summary: 129

Tamil Nadu 11th English Model Question Paper 2

[OR]

Notes
Title: Interior design of One’s Home
Answer:
Home reflects:

  • personality of house-owner
  • unity & harmony bet. rooms
  • colour & styling sh’d be uniform

Elements of decoration:

  • selection of colour schemes
  • draperies, rugs, upholstery, woods

Plan to enjoy the House:

  • limit time, effort & money
  • select furnish’gs which require little care
  • hobby connected with house-great pleasure.

Choice of Colours:

  • one colour sh’d predominate
  • calm colours for restfulness; intense for liveliness
  • colours sh’d harmonise with furniture, draperies, carpets

Abbreviations used: bet. – between; sh’d – should; fumish’gs – furnishings;

Question 45.
Read the following advertisement and prepare a resume/bio-data/CV considering yourself fulfilling the conditions specified.
[Write XXXX for your name and YYYY for your address]
Wanted
Computer Operator – Diploma holder with computer knowledge, fluency in English and good communication skills, Minimum 3 Years Experience.
Apply with your bio-data to : Post Box No : 545
C/o. The Hindu
Trichy- 620001.
Answer:

25.09.XXXX

From
XXXX
YYYY

To
Post box No. 545
C/o The Hindu
Trichy – 620001

Sir,
Sub: Applying for the post of Computer Operator – Reg.
I hereby apply for the post of Computer Operator vacant in your esteemed concern. I have the necessary qualification. My particulars are furnished below for your kind consideration. Bio-data

Name: XXXX
Father’s Name : Mr. R. Karthick
Address: YYYY
Qualification : B.Sc. Computer Science, 1st class, Madurai Kamaraj University
Technical Qualification : Tally, C++, PGDCA
Experience : Seven years of service in Aircel
Age: 28
Languages known : Tamil, English
Joining date : Can join immediately
Reference : My previous employer
Mr. Raj (9876543210)
I look forward to receiving your call letter. I shall offer my services to the best of my superiors’ satisfaction sir.

Yours sincerely
XXXX
Address on the Envelope
To
Post box No. 545
C/o. The Hindu
Trichy-620001

Tamil Nadu 11th English Model Question Paper 2

[OR]

Write an essay in about 150 words on ‘Cyber safety’.
Answer:
Cyber safety
Every child needs to be taught the basics of cyber safety. All of us are aware of the fact that ‘Blue Whale’ game cost lives of many young ones across the globe. Children who are befriended through social websites reveal personal information unwittingly and are exploited by persons who have access to their personal details. Children must be advised to refrain from sharing things with total strangers. Even adults are exploited through social websites and their budding lives are at stake. So, students must not evince keen interest in making friends with strangers online. If children do not do anything that is shameful to admit to parents, cyber crimes will be reduced to minimum.

Question 46.
Read the following sentences, spot the errors and rewrite the sentences correctly.
(a) There is nothing much selfish you can do than come to work sick.
(b) Elimination for child labour is undoubtedly one of the biggest challenge of our country.
(c) Today democracy is often assume to be a liberal form of governance.
(d) In the traditional sense prayer means communicating on God Almighty.
(e) Some of them have been converted into museums but libraries.
Answer:
(a) There is nothing more selfish you can do than come to work sick.
(b) Elimination of child labour is undoubtedly one of the biggest challenge of our country.
(c) Today democracy is often assumed to be a liberal form of governance.
(d) In the traditional sense prayer means communicating with God Almighty.
(e) Some of them have been converted into museums and libraries.

[OR]

Fill in the blanks appropriately.
(a) The Police tried ……….. to information from the boot leggar who used to sell ………. liqour. (illicit/ elicit)
(b) All citizens ………. obey the laws of the land. (Fill in with a modal verb)
(c) We ……….. go grocery shopping, (use semi-modal)
(d) ………. he is rich, he lives in a small house. (Use a suitable link word)
Answer:
(a) elicit, illicit
(b) must
(c) need to
(d) Although

Tamil Nadu 11th English Model Question Paper 2

Question 47.
Identify each of the following sentences with the fields given below.
(a) The board has decided to give the shareholders a dividend of 25 percent.
(b) A salaried employee in the highest slab pays income tax at 33.66 percent.
(c) For programming, people use the binary system.
(d) Every plant organ has a definite form and structure and performs certain specific functions.
(e) “My goal is winning a Grand Slam”, says Sania Mirza.
(Botany, Sports, Taxation, Business, Computer)
Answer:
(a) Business
(b) Taxation
(c) Computer
(d) Botany
(e) Sports

[OR]

Read the following passage carefully and answer the questions that follow.
Once upon a time a frog croaked in Bingle Bog all the night beginning from dusk to dawn. All the creatures hated his loud and unpleasant voice but still they did not have any other option. The voice came out from the sumac tree where every night the frog sang till morning.

He was so determined and also shameless that neither stones, prayers or sticks nor the insults or complaints could divert him from singing. One night, a nightingale started casting her melody in the moonlight to which both the frog and the other creatures were left dumbstruck. The whole bog remained, rapt and admired her voice and applauded her when she ended. The frog was obviously jealous of his rival and had finally decided to eliminate her.

So, the next night when the nightingale was again preparing to sing, the frog’s croak disturbed her. On being asked about himself by the nightingale he answered that he owned the sumac tree and he had been known for his splendid voice. Also he said that he had written a number of songs for the Bog Trumpet. The nightingale asked him whether he liked her song or not.

The frog said that the song wasn’t bad but too long and it lacked some force. The nightingale was greatly impressed that such a critic had discussed her song. She said that she was happy that the song was her own creation. To this the frog said that she needed a proper training to obtain a strong voice otherwise she would remain a beginner only. He also said that he would train her but would charge some fee.

Now, the nightingale was flushed with confidence and was a huge sensation, attracting animals from miles away and the frog with a great accuracy charged all of them admission fee. The frog began her vocal training despite of the bad and rainy weather where even the nightingale had first refused to sing. But the frog forced her to sing for six hours continuously till she was shivering and her voice had become rough and unclear. But, somehow her neck got clear the next day and she was able to sing again collecting a breathless crowd including rich ladies kings queens etc. To all this, the frog had both sweet and bitter feelings. Sweet because he was earning lots of money and bitter because of jealously as his rival was earning name and fame.

Every day, the frog scolded her to practice even longer finding out her little mistakes like nervousness not laying more trills and frills etc. He reminded her that she still owed him sixty shillings and that’s why the crowd should increase. But the condition of nightingale was getting worse. Her tired and uninspired song could no longer attract the crowd. She could not resist this as she had become used to applause and thus had become miserable too.

The heartless frog scolded her even then calling her a brainless bird. She trembled, puffed up, burst a vein and died. The frog said that he had tried to teach her but she was foolish, nervous and tensed and moreover much prone to influence. Then, once again the frog’s fog horn started blearing unrivalled in the bog.

The moral of the poem is that being inspired and influenced by someone much unknown and strange is indeed a foolish work. The nightingale could have very well, judged that how could the frog with such a harsh voice be music maestro and she had to suffer for her misjudgement. Many people in the human society also try to take advantage of the innocence or ignorance of the people.
Questions.

  1. How do you know that the frog was of a determined nature?
  2. A bog is a
  3. What did the nightingale become?
  4. Pick out one word from the passage which means ‘genius’.
  5. What is the moral of the story?

Answer:

  1. He was so determined and also shameless that neither stones, prayers or sticks nor the insults or complaints could divert him from singing.
  2. quagmire.
  3. The nightingale became a huge sensation.
  4. Maestro’ is the word which means genius.
  5. The moral of the story is that being inspired and influenced by someone much unknown and strange is indeed a foolish work.

Tamil Nadu 11th English Model Question Paper 2

Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

Question 1.
(i) A person went to a restaurant for dinner. In the menu card, the person saw 10 Indian and 7 Chinese food items. In how many ways the person can select either an Indian or Chinese food?
Answer:
Selecting an Indian food item from the given 10 can be done in 10 ways. Selecting a Chinese food item from the given 7 can be done in 7 ways.
∴ Selecting an Indian or Chinese food can be done in 10 + 7 = 17 ways.

Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

(ii) There are 3 types of a toy car and 2 types of toy train are available in a shop. Find the number of ways a baby can buy a toy car and a toy train?
Answer:
Number of types of Toy car = 3
Number of types of Toy Train = 2
Number of ways of buying a Toy car = 3 ways
Number of ways of buying a toy train = 2 ways
∴ By fundamental principle of multiplication, number of ways of buying a toy car and a toy train = 3 × 2 ways = 6 ways

(iii) How many two – digit numbers can be formed using 1 , 2,3,4,5 without repetition of digits?
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 1
The given digits are 1, 2, 3, 4, 5
A two-digit number has a unit place and 10’s place. We are given 5 digits (1, 2, 3, 4, 5). The unit place can be filled (using the 5 digits) in 5 ways. After filling the unit place since repetition is not allowed one number (filled in the unit place) should be excluded. So the 10’s place can be filled (using the remaining 4 digits) in 4 ways.
∴ Unit place and 10’s place together can be filled in 5 × 4 = 20 ways. So the number of two-digit numbers = 20

(iv) Three persons enter into a conference hall in which there are 10 seats. In how many ways can they take their seats?
Answer:
Number of seats in the conference hall = 10
Number of persons entering into the conference hall = 3
Number of ways of getting a seat for 1st person = 10
Number of ways of getting a seat for 2nd person = 9
Number of ways of getting a seat for 3rd person = 8
By fundamental principle of multiplication, number of ways of getting seats for 3 persons in conference hall = 10 × 9 × 8 ways = 720 ways

Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

(v) In how ways 5 persons can be seated In a row?
Answer:
Number of persons = 5
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 2

5 persons can be arranged among themselves in 5! ways
(i.e) 5 × 4 × 3 × 2 × 1 = 120 ways

Question 2.
(i) A mobile phone has a passcode of 6 distinct digits. What is the maximum number of attempts one makes to retrieve the passcode?
Answer:
Number of distinct digit in a passcode of a mobile phone = 6
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 3
First digit can be tried in 10 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Second digit can be tried in 9 ways
Third digit can be tried in 8 ways
Fourth digit can be tried in 7 ways
Fifth digit can be tried in 6 ways
Sixth digit can be tried in 5 ways
Therefore, the maximum number of attempts made to retrieve the passcode = 10 × 9 × 8 × 7 × 6 × 5 = 151200

Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

(ii) Given four flags of different colours, how many different signals can be generated if each signal requires the use of three flags, one below the other?
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 4
Number of flags = 4
Number of flags required for a signal = 3
The total number of signals is equal to the number of ways of filling 3 places in succession by 4 flags of different colours. Number of ways of filling the top place using 4 different colour flags is 4 ways. Number of ways of filling the middle place using the remaining 3 different colour flags is 3 ways. Number of ways of filling the bottom place using the remaining 2 different colour flags is 2 ways.

Therefore, by fundamental principle of multiplication, the total number of signals = 4 × 3 × 2 = 24 ways

Question 3.
Four children are running a race.
(i) In how many ways can the first two places be filled?
(ii) In how many different ways could they finish the race?
Answer:
(i) Number of children in the running race = 4
The first place can be filled in (from the 4 children) 4 ways
After filling in the first place only 3 children are left out
So the second place can be filled in (from the remaining 3 children) 3 ways
So the first and the second places together can be filled in 4 × 3 = 12 ways

(ii) The first and second places can be filled in 12 ways
The third-place can be filled (from the remaining 2 children) in 2 ways and the fourth place can be filled in 1 way
So the race can be finished in 12 × 2 × 1 = 24 ways

Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

Question 4.
Count the number of three – digit numbers which can be formed from the digits 2, 4, 6, 8 if
(i) Repetitions of digits is allowed.
(ii) Repetitions of digits is not allowed?
Answer:
(i) Repetitions of digits is allowed
The given digits are 2, 4, 6, 8
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 5
A number of ways of filling the unit place using the 4 digits 2, 4, 6, 8 is 4 ways. Number of ways of filling the tens place using the 4 digits 2, 4, 6, 8 in 4 ways Number of ways of filling the hundred’s place using the 4 digits 2, 4, 6, 8 is 4 ways

Therefore, by fundamental principle of multiplication, the total number of 3 digit numbers = 4 × 4 × 4
= 64 ways
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 6

Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

(ii) The unit place can be filled (using the 4 digits) in 4 ways after filling the unit place since repetition of digits is not allowed that digit should be excluded.
So the 10’s place can be filled in (4 – 1) 3 ways and the 100’s place can be filled in (3 – 1) 2 ways
So the unit place, 10’s and 100’s places together can be filled in 4 × 3 × 2 = 24 ways
(i.e) The number of 3 digit numbers = 4 × 3 × 2 = 24 ways

Question 5.
How many three-digit numbers are there with 3 in the unit place?
(i) with repetition
(ii) without repetition.
Answer:
(i) With repetition:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 7
The given digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
The unit place can be filled in only one way using the digit 3. The ten’s place can be filled in 10 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The hundred’s place can be filled in 9 ways using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9.

Therefore, by the fundamental principle of multiplication, the total number of 3 digit numbers is = 1 × 10 × 9 = 90

(ii) Without repetition:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 8

The digits are 0, 1,2, 3, 4, 5, 6, 7, 8, 9
A three-digit number has 3 digits l’s, 10’s, and 100’s place.
The unit place is (filled by 3) filled in one way.
After filling the unit place since the digit ‘0’ is there, we have to fill the 100’s place. Now to fill the 100’s place we have 8 digits (excluding 0 and 3) So 100’s place can be filled in 8 ways.
Now to fill the 10’s place we have again 8 digits (excluding 3 and any one of the number) So 10’s place can be filled in 8 ways.
∴ Number of 3 digit numbers with ‘3’ in unit place = 8 × 8 × 1 = 64

Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

Question 6.
How many numbers are there between 100 and 500 with the digits 0,1,2,3,4,5 if
(i) Repetition of digit is allowed
(ii) Repetition of digits is not allowed.
Answer:
(i) Repetition of digit is allowed:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 9
The numbers between 100 and 500 will have 3 digits. The unit place can be filled in 6 ways using the digits 0, 1, 2, 3, 4, 5. The ten’s place can be filled in 6 ways using the digits 0, 1, 2, 3, 4, 5. The hundred’s place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 ( Excluding 0 and 5). Therefore, by fundamental principle of multiplication, the number of 3 digit, numbers between 100 and 500 with repetition of digits using the digits 0, 1, 2, 3, 4, 5 is = 6 × 6 × 4 = 144

(ii) Repetition of digits is not allowed:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 10
The 100’s place can be filled (by using 1, 2, 3, 4) 10’s in 4 ways
The 10’s place can be filled in (6 – 1) 5 ways and the unit place can be filled in (5 – 1) 4 ways
So the number of 3 digit number 4 × 5 × 4 = 80

Question 7.
How many three-digit odd numbers can be formed by using the digits 0, 1, 2, 3, 4, 5? if
(i) The repetition of digits is not allowed
(ii) The repetition of digits is allowed.
Answer:
The given digits are 0, 1, 2, 3, 4, 5
To find the possible 3 – digit odd numbers.

(i) Repetition of digits is not allowed:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 11
Since we need 3 – digit odd numbers the unit place can be filled in 3 ways using the digits 1, 3 or 5. Hundred’s place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 excluding 0 and the number placed in unit place. Ten’s place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 excluding the digit placed in the hundred’s place.

Therefore, by the fundamental principle of multiplication, the number of 3 – digit odd numbers formed without repetition of digits using the digits 0, 1, 2, 3, 4, 5 is
= 4 × 4 × 3 = 48

Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

(ii) Repetition of digits is allowed:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 12
The unit place can be filled in 3 ways. We are given 6 digits.
So 10’s place can be filled in 6 ways and the 100’s place can be filled in (6 – 1) (excluding zero) 5 ways
So the Number of 3 digit numbers = 3 × 6 × 5 = 90

Question 8.
Count the numbers between 999 and 10000 subject to the condition that there are
(i) no restriction
(ii) no digit is repeated
(iii) At atleast one of the digits is repeated.
Answer:
To find the numbers between 999 and 10,000 using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
To find the possible 4 digit numbers.

(i) No restriction:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 13
Thousand’s place can be filled in 9 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 excluding 0. Since there is no restriction, the hundred’s place, Ten’s place, and the unit place can be filled in 10 ways using the digit 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Therefore, by the fundamental principle of multiplication, the number of 4 digit numbers between 999 and 10,000 is = 9 × 10 × 10 × 10 = 9000

(ii) No digit is repeated:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 14
Thousand’s place can be filled in 9 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 excluding 0. Since repetition is not allowed. The unit place can be filled in 9 ways using the digits 0,1, 2, 3, 4, 5, 6, 7, 8, 9 excluding the digit placed in the thousand’s place. Since repetition of digits is not allowed, the ten’s place can be filled in 8 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 excluding the digits used in thousand’s a place and unit place. Since repetition of digits is not allowed, the hundred’s place can be filled in 7 ways using the digits 0,1,2,3,4, 5,6,7,8,9 excluding the digits used in thousand’s a place and unit place.

Therefore, by the fundamental principle of multiplication, the number of numbers between 999 and 10,000 without repetition of digits is = 9 × 7 × 8 × 9 = 4536

Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

(iii) At least one of the digits is repeated:
Required number of 4 digit numbers = Total number of 4 digit numbers – Number of 4 digit numbers when no digit is repeated = 9000 – 4536 = 4464

Question 9.
How many three-digit numbers, which are divisible by 5, can be formed using the digits 0, 1, 2, 3, 4, 5 if
(i) Repetition of digits is not allowed?
(ii) Repetition of digits is allowed?
Answer:
The given digits are 0, 1, 2, 3, 4, 5
To find the 3 – digit numbers formed by using the digits 0, 1, 2, 3, 4, 5 which are divisible by 5.

(i)The repetition of digits are not allowed:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 15
Since the 3 – digit number is divisible by 5, the unit place can be filled in 2 ways using the digits 0 or 5

Case (i) When the unit place is filled with the digit 0. The hundreds place can be filled in 5 ways using the digits 1, 2, 3, 4, 5 and the ten’s place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 excluding 0 and the digit which is placed in hundred’s place.

Therefore, by fundamental principle of multiplication, the number of 3 – digit numbers divisible by 5 is
= 5 × 4 × 1 = 20

Case (ii) When the unit place is tilled with the digit 5, since repetition of digit is not allowed the hundred’s place can be filled in 4 ways using the digits 1, 2 , 3 , 4 (0 and 5 are excluded). The ten’s place can be filled in 4 ways using the digits 0, 1 , 2, 3 , 4, 5 (excluding 5 and the digit placed in the hundred’s place).
Therefore, by the fundamental principle of multiplication, the number of 3 – digit numbers, in this case, is = 4 × 4 × 1 = 16
Therefore, the total number of 3 – digit numbers divisible by 5 using the digits 0, 1, 2, 3, 4, 5 is = 20 + 16 = 36

Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

(ii) The repetition of digits are allowed:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 16
The digits are
0 1 2 3 4 5
To get a number divisible by 5 we should have the unit place as 5 or 0 So the unit place (using 0 or 5) can be filled in 2 ways.
The 10’s place can be filled (Using 0, 1, 2, 3, 4, 5) in 6 ways and the 100’s place (Using 1, 2, 3, 4, 5) can be filled in 5 ways.
So the number of 3 digit numbers ÷ by 5 (with repetition) = 2 × 6 × 5 = 60

Question 10.
To travel from place A to place B, there are two different bus routes B1, B2, two different train routes T1, T2, and one air route A1. From place B to place C, there is one bus route say B’1, two different train routes say T’1, T’2, and one air route A’1. Find the number of routes of commuting from place A to place C via place B without using a similar mode of transportation.
Answer:
Route map diagram for the given data.
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 17
The possible choices for a number of routes commuting from A to place C via place B without using similar mode transportation are
(B1, T’1), (B1, T’2), ( B1, A1), ( B2, T’1), (B2, T’2)
(B2, A’1), (T1, B’1), (T1, A’1), ( T2, B’1), ( T2, A’1) (A1, B’1) , (A1, T’1) and (A1, T’2)
Therefore, the Required number of routes is 13.

Question 11.
How many numbers are there between 1 and 1000 (both inclusive) which are divisible neither by 2 nor by 5 ?
Answer:
From 1 to 1000, the numbers ÷ by 2 = 500
the number ÷ by 5 = 200
and the numbers ÷ by 10 = 100(5 × 2 = 10)
So number ÷ by 2 or 5 = 500 + 200 – 100 = 600
Total numbers from 1 to 1000 = 1000
So the number of numbers which are ÷ neither by 2 nor by 5 = 1000 – 600 = 400

Three-digit numbers:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 18
The unit place can be filed in 4 ways using the digits 1, 3, 7, 9. Hundred’s place can be filled in 9 ways excluding 0. Ten’s place can be filled in 10 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Therefore, the required number of 3 digit numbers neither divisible by 2 nor by 5 is = 9 × 10 × 4 = 360.
There is only one 4 – digit number, but it is divisible by 2 and 5.
Therefore, required numbers using fundamental principle of addition = 4 + 36 + 360 = 400

Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

Question 12.
How many strings can be formed using the letters of the word LOTUS if the word
(i) either start with L or end with S?
(ii) neither starts with L nor ends with S?
Answer:
(i) Either starts with L or ends with S
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 19
The first box is filled with the letter L. The second box can be filled with the remaining letters O, T, U, S in 4 ways. The third box can be filled with the remaining letters excluding L and the letter placed in box 2 in 3 ways. The fourth box can be filled with the remaining letters excluding L and the letters placed in a box – 2 and box – 3 in 2 ways. The fifth box can be filled with the remaining one letter excluding L and the letters placed in a box – 2 and box – 3, box – 4 in 1 way.
Therefore, by fundamental principle of multiplication, the number of words start with L is = 1 × 4 × 3 × 2 × 1 = 24
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 20
Since the word ends with S, the fifth box can be filled in one way with the letter S. The remaining four boxes can be filled 4 × 3 × 2 × 1 way.
Therefore, the number of words ending with S = 4 × 3 × 2 × 1 × 1 = 24

Number of words starting with L and ends with S: The first box can be filled with L in one way fifth box can be filled with S in one way second box, third box, and fourth box can be filled in 3x2x1 ways with the remaining letters O, T, U.
∴ Number of words starting with L and ends with S = 1 × 3 × 2 × 1 × 1 = 6
Therefore, by fundamental principle of addition, number of words either starts with L or ends with S = 24 + 24 – 6 = 48 – 6 = 42

(ii) Neither starts with L nor ends with S Total number of words formed using the letters L, O, T, U, S is = 5 × 4 × 3 × 2 × 1 = 120
The number of words neither starts with L nor ends with S = Total number of words – Number of words starts with either L or ends with S = 120 – 42 = 78

Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

MCQ Questions For Class 11 hindi with answers· 

Question 13.
(i) Count the total number of ways of answering 6 objective type questions, each question having 4 choices.
Answer:
Count the total number of ways of answering 6 objective type questions, each question having 4 choices.
One question can be answered in 4 ways
Two questions can be answered in 4 × 4 = 42 ways
∴ Six questions can be answered in 46 ways

(ii) In how many ways 10 pigeons can be placed in 3 different pigeon holes?
Solution:
First pigeons can be placed in a pigeon-hole in 3 ways (selecting 1 from 3 holes)
Second pigeons can be placed in a pigeon-hole in 3 ways Tenth pigeons can be placed in a pigeon-hole in 3 ways
So total number of ways in which all the number 10 place can be sent = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 = 310 ways

(iii) Find the number of ways of distributing 12 distinct prizes to 10 students?
Solution:
To give the first prize we have to select, from the 10 students which can be done in 10 ways.
To give the second prize we have to select one from the 10 students which can be done is 10 ways.
To give the 12th prize we have to select one from 10 students which can be done in 10 ways.
So all the 12 prizes can be given in (10 × 10 × 10 …. 12 times) = 1012 ways.

Question 14.
Find the value of
(i) 6!
(ii) 4! +5!
(iii) 3! – 2!
(iv) 31 × 21
(v) \(\frac{12 !}{9 ! \times 3 !}\)
(vi) \(\frac{(n+3) !}{(n+1) !}\)
Answer:
(i) 6!
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

(ii) 4! +5!
4! +5! = (4 × 3 × 2 × 1) + (5 × 4 × 3 × 2 × 1)
= (4 × 3 × 2 × 1) [1 + 5]
= 24 × 6 = 144

(iii) 3! – 2!
3! – 2! = (3 × 2 × 1) × (4 × 3 × 2 × 1)
= 6 × 24 = 144

(iv) 3! × 2!
3! × 2! = (3 × 2 × 1) × (4 × 3 × 2 × 1)
= 6 × 24 = 144

(v) \(\frac{12 !}{9 ! \times 3 !}\)
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 21
= 2 × 11 × 10 = 220

(vi) \(\frac{(n+3) !}{(n+1) !}\)
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 22
= (n + 3) (n + 2)
= n2 + 3n + 2n + 6
= n2 + 5n + 6

Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

Question 15.
Evaluate \(\frac{\mathbf{n} !}{\mathbf{r} !(\mathbf{n}-\mathbf{r}) !}\) when
(i) n = 6 , r = 2
(ii) n = 10, r = 3
(iii) For any n with r = 2.
Answer:
(i) n = 6 , r = 2
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 23

(ii) n = 10, r = 3
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 24

(iii) For any n with r = 2.
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 25

Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

Question 16.
Find the value of n if
(i) ( n + 1) ! = 20 ( n – 1 )!
(ii) \(\frac{1}{8 !}+\frac{1}{9 !}=\frac{n}{10 !}\)
Answer:
(i) ( n + 1) ! = 20 ( n – 1 )!
(n + 1) n(n – 1)! = 20(n – 1)!
n(n + 1) = 20
n2 + n – 20 = 0
n2 + 5n – 4n – 20 = 0
n(n + 5) – 4(n + 5) = 0
(n – 4) (n + 5) = 0
n – 4 = 0 or n + 5 = 0
n = 4 or n = -5
But n = -5 is not possible. ∴ n = 4

(ii) \(\frac{1}{8 !}+\frac{1}{9 !}=\frac{n}{10 !}\)
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 27

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 9 Solutions Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

11th Chemistry Guide Solutions Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
The molality of a solution containing 1.8 g of glucose dissolved in 250 g of water is
a) 0.2 M
b) 0.01 M
c) 0.02 M
d) 0.04 M
Answer:
d) 0.04 M

Question 2.
Which of the following concentration terms is / are independent of temperature
a) molality
b) molarity
c) mole fraction
d) a and b
Answer:
d) a and b

Question 3.
Stomach acid, a dilute solution of HCl can be neutralized by reaction with aluminium hydroxide Al(OH)3 + 3HCl (aq) -> AlCl3 + 3H2O. How many milliliters of 0.1 M Al(OH)3 solution is needed to neutralize 21 ml of 0.1 M HCl?
a) 14 mL
b) 7 mL
c) 21 mL
d) none of these
Answer:
b) 7 mL

Question 4.
The partial pressure of nitrogen in air is 0.76 atm and its Henry’s law constant is 7.6 × 104 atm at 300 K. What is the mole fraction of nitrogen gas in the solution obtained when air is bubbled through water at 300 K ?
a) 1 × 10-4
b) 1 × 104
c) 2 × 10-5
d) 1 × 10-5
Answer:
d) 1 × 10-5

Question 5.
Henry’s law constant for the solubility of Nitrogen gas in water at 350 K is 8 × 104 atm. The mole fraction of nitrogen in air is 0.5. The number of moles of Nitrogen from air dissolved in 10 moles of water at 350 K and 4 atm pressure is
a) 4 × 10-4
b) 4 × 104
c) 2 × 10-2
d) 2.5 × 10-4
Answer:
d) 2.5 × 10-4

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 6.
Which one of the following is incorrect for an ideal solution?
a) ∆Hmix = 0
b) ∆Umix =0
c) ∆P = Pobserved – Pcalculated by Raoults law = 0
d) ∆Gmix = 0
Answer:
d) ∆Gmix = 0

Question 7.
Which one of the following gases has the lowest value of Henry’s law constant?
a) N2
b) He
c) CO2
d) H2
Answer:
c) CO2

Question 8.
P1 and P2 are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution If x1 represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be
a) P1 + x1 (P2 – P1)
b) P2 – x1 (P2 + P1)
c) P1 – x2(P1 – P2)
d) P1 + x2(P1 – P2)
Answer:
c) P1 – x2(P1 – P2)

Question 9.
Osomotic pressure (π) of a solution is given by the relation
a) π = nRT
b) πV = nRT
c) πRT = n
d) none of these
Answer:
b) πV = nRT

Question 10.
Which one of the following binary liquid mixtures exhibits positive deviation from Raoults law?
a) acetone + chloroform
b) water + nitric acid
c) HCl + water
d) ethanol + water
Answer:
d) ethanol + water

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 11.
The Henry’s law constants for two gases A and B are x and y respectively. The ratio of mole fractions of A to B 0.2. The ratio of mole fraction of B and A dissolved in water will be
a) \(\frac{2 x}{y}\)

b) \(\frac{y}{0.2 x}\)

c) \(\frac{0.2 x}{y}\)

d) \(\frac{5 x}{y}\)
Answer:
d) \(\frac{5 x}{y}\)

Question 12.
At 100°C the vapour pressure of a solution containing 6.5g a solute in 100g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be
a) 102°C
b) 100°C
c) 101°C
d) 100.52°C
Answer:
c) 101°C

Question 13.
According to Raoults law, the relative lowering of vapour pressure for a solution is equal to
a) mole fraction of solvent
b) mole fraction of solute
c) number of moles of solute
d) number of moles of solvent
Answer:
b) mole fraction of solute

Question 14.
At same temperature, which pair of the following solutions are isotonic?
a) 0.2 M BaCl2 and 0.2 M urea
b) 0.1 M glucose and 0.2 M urea
c) 0.1 M NaCl and 0.1 M K2SO4
d) 0.1 M Ba(NO3)2 and 0.1 M Na2SO4
Answer:
d) 0.1 M Ba(NO3)2 and 0.1 M Na2SO4

Formula of normality · In the case of acid-base chemistry, normality is used to express the concentration of hydronium ions (H3O+) .

Question 15.
The empirical formula of a non – electrolyte (X) is CH2O. A solution containing six grams of X exerts the same osmotic pressure as that of 0.025 M glucose solution at the same temperature. The molecular formula of X is
a) C2H4O2
b) C8H16O8
c) C4H8O4
d) CH2O
Answer:
b) C8H16O8

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 16.
The KH for the solution of oxygen dissolved in water is 4 × 104 atm at a given temperature. If the partial pressure of oxygen in air is 0.4 atm, the mole fraction of oxygen in solution is
a) 4.6 × 103
b) 1.6 × 104
c) 1 × 10-5
d) 1 × 105
Answer:
c) 1 × 10-5

Question 17.
Normality of 1.25 M sulphuric acid is
a) 1.25 N
b) 3.75 N
c) 2.5 N
d) 2.25 N
Answer:
c) 2.5 N

Question 18.
Two liquids X and Y on mixing gives a warm solution. The solution is
a) ideal
b) non-ideal and shows positive deviation from Raoult’s law
c) ideal and shows negative deviation from Raoult’s Law
d) non-ideal and shows negative deviation from Raoult’s Law
Answer:
d) non-ideal and shows negative deviation from Raoult’s Law

Question 19.
The relative lowering of vapour pressure of a sugar solution in water is 2.5 × 10-3. The mole fraction of water in that solution is
a) 0.0035
b) 0.35
c) 0.0035/18
d) 0.9965
Answer:
d) 0.9965

Question 20.
The mass of a non – volatile solute (molar mass 80 g mol-1) should be dissolved in 92g of toluene to reduce its vapour pressure to 90%
a) 10 g
b) 20 g
c) 9.2 g
d) 8 g
Answer:
d) 8 g

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 21.
For a solution, the plot of osmotic pressure (π) versus the concentration (c in mol L-1) gives a straight line with slope 310 R where ‘R’ is the gas constant. The temperature at which osmotic pressure measured is
a) 310 × 0.082 K
b) 310° C
c) 37°C
d) \(\frac{310}{0.082}\) K
Answer:
c) 37°C

Question 22.
200 ml of an aqueous solution of a protein contains 1.26 g of protein. At 300 K, the osmotic pressure of this solution is found to be 2.52 × 10-3 bar. The molar mass of protein will be (R = 0.083 L bar mol-1 K-1}
a) 62.22 kg mol-1
b) 12444 g mol-1
c) 300 g mol-1
d) None of these
Answer:
a) 62.22 kg mol-1

Question 23.
The Van’t Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is
a) 0
b) 1
c) 2
d) 3
Answer:
d) 3

Question 24.
Which is the molality of a 10% w/w aqueous sodium hydroxide solution?
a) 2.778
b) 2.5
c) 10
d) 0.4
Answer:
b) 2.5

Question 25.
The correct equation for the degree of an associating solute, ‘n’ molecules of which undergoes association in solution, is
a) α = \(\frac{\mathrm{n}(\mathrm{i}-1)}{\mathrm{n}-1}\)

b) α2 = \(\frac{n(1-i)}{(n-1)}\)

c) α = \(\frac{\mathrm{n}(\mathrm{i}-1)}{1-\mathrm{n}}\)

d) α = \(\frac{\mathrm{n}(1-\mathrm{i})}{\mathrm{n}(1-\mathrm{i})}\)
Answer:
c) α = \(\frac{\mathrm{n}(\mathrm{i}-1)}{1-\mathrm{n}}\)

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 26.
Which of the following aqueous solutions has the highest boiling point?
a) 0.1 M KNO3
b) 0.1 M Na3PO4
c) 0.1 M BaCl2
d) 0.1 M K2SO4
Answer:
b) 0.1 M Na3PO4

Question 27.
The freezing point depression constant for water is 1.86° K Kg mol-1. If 5 g Na2SO4 is dissolved in 45 g water, the depression in freezing point is 3.64°C. The Vant Hoff factor for Na2SO4 is
a) 2.57
b) 2.63
c) 3.64
d) 5.50
Answer:
a) 2.57

Question 28.
Equimolal aqueous solutions of NaCl and KCl are prepared, If the freezing point of NaCl is -2°C, the freezing point of KCl solution is expected to be
a) -2°C
b) -4°C
c) -1°C
d) 0°C
Answer:
a) -2°C

Question 29.
Phenol dimerizes in benzene having van’t Hoff factor 0.54. What is the degree of association?
a) 0.46
b) 92
c) 46
d) 0.92
Answer:
d) 0.92

Question 30.
Assertion:
An ideal solution obeys Raoults Law.
Reason:
In an ideal solution, solvent – solvent as well as solute – solute interactions are similar to solute-solvent interactions.
a) both assertion and reason are true and reason is the correct explanation of assertion
b) both assertion and reason are true but reason is not the correct explanation of assertion
c) assertion is true but reason is false
d) both assertion and reason are false
Answer:
a) both assertion and reason are true and reason is the correct explanation of assertion

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

II. Write brief answer to the following questions:

Question 31.
Define:
(i) Molality
(ii) Normality
Answer:
(i)Molality :
Molality (m) is defined as the number of moles of the solute dissolved in one kilogram (Kg) of the solvent. The units of molality are moles per kilogram, i.e., mole kg-1. The molality is preferred over molarity if volume of the solution is either expanding or contracting with temperature.
molality (m) = \(\frac{\text { Number of mole of solute }}{\text { mess of solvent in } \mathrm{kg}}\)

ii) Normality:
Normality (N) of a solution is defined as the number of gram equivalents of the solute present in one liter of the solution. Normality is used in acid-based redox titrations.
Normality (N) = \(\frac{\text { Number of gram equivalents of solute }}{\text { Volume of solution in litre }}\)

Question 32.
a) What is a vapour pressure of liquid?
Answer:
“The pressure exerted by the vapors above the liquid surface which is in equilibrium with the liquid at a given temperature is called vapor pressure”.

b) What is a relative lowering of vapour pressure?
Answer:
The relative lowering of vapour pressure is defined as the ratio of lowering of vapour pressure to the vapour pressure
of pure solvent (P0) RLVP = \(\frac{p^{0}-P}{P^{0}}\)

Question 33.
State and explain Henry’s law.
Answer:
Henry’s law:
This law states “that the partial pressure of the gas in vapour phase is directly proportional to the mole fraction (x) of the gaseous solute in the solution at low concentrations.”
Psolute ∝ xsolute in solution
Psolute = KH. xsolute in solution
xsolute = mole fraction of solute in the solution
KH = empirical constant.
Psolute = Vapour pressure of the solute (or) the partial pressure of the gas in vapour state. The value of KH depends on the nature of the gaseous solute and solvent.

Question 34.
State Raoult law and obtain the expression for lowering of vapour pressure when the nonvolatile solute is dissolved Insolvent.
Answer:
In an ideal solution, the vapour pressure of the solution is decreased when a non-volatile solute is dissolved in a solvent. The magnitude of decrease in the vapour pressure of the solution depends on the amount of solute added.
Let us consider the solution with the following features.
Mole fraction of the solvent = xA
Mole fraction of the solute = xB
Vapour pressure of the pure solvent = P°A
Vapour pressure of solution = P
As the solute is nonvolatile, the vapour pressure of the solution is only due to the solvent. Therefore, the vapour pressure of the solution (P) will be equal to the vapour pressure of the solvent (PA) over the solution.
i.e., P = PA
According to Raoult’s law, the vapour pressure of solvent over the solution is equal to the product or its vapour pressure in a pure state and its mole fraction.
PA = P°A xA or
P = P°A xA

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 35.
What is molal depression constant? Does it depend on nature of the solute?
Answer:
If m = 1 then ∆Tf = Kf
“Then Kf is equal to the depression in freezing point for 1 molal solution”. No, it does not depends on the nature of the solute.

Question 36.
What is osmosis?
Answer:
Osmosis is a spontaneous process by which the solvent molecules pass through a semipermeable membrane from a solution of lower concentration to the solution of higher concentration.

Question 37.
Define the term ‘isotonic solution’.
Answer:
Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.

Question 38.
You are provided with a solid. ‘A’ and three solutions of A dissolved in water – one saturated, one unsaturated, and one supersaturated. How would you determine which solution is which?
Answer:
(A) Unsaturated solution:
It can dissolve salt in addition to it.
(B) Saturated solution:
Further solubility of salt does not take place but solubility can take place on heating.
(c) Supersaturated solution:
Solubility of salt does not take place on even further heating.

Question 39.
Explain the effect of pressure on solubility.
Answer:
1. The change in pressure does not have any significant effect on the solubility of solids and liquids as they are not compressible. However, the solubility of gases generally increases with the increase in pressure.

2. According to Le – chatter’s principle, the increase in pressure will shift the equilibrium in the direction which will reduce the pressure. Therefore, more gaseous molecules dissolve in the solvent.

3. If pressure increases, the solubility of gas also increases.

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 40.
A sample of 12 M Concentrated hydrochloric acid has a density 1.2 M gL-1 calculate the molality.
Solution:
Given:
Molarity = 12 M HCl
density of solution = 1.2 g L-1
In 12 M HCl solution, there are 12 moles of HCl in 1 litre of the solution.
Molality = \(\frac{\text { no of moles of solute }}{\text { mass of solvent (in } \mathrm{kg} \text { ) }}\)
Calculate mass of water(solvent)
mass of 1 litre HCl solution = density × volume
= 1.2 × gmL-1 × 1000 mL = 1200 g
mass of HCl = no. of moles of HCl × molar mass of HCl
= 12 mol × 36.5 g mol-1
= 438 g.
mass of water = mass of HCl solution – mass of HCl
mass of water = 1200 – 438 = 762 g
molality(m) = \(\frac{12}{0.762}\) = 15.75 m

Question 41.
A 0.25 M glucose solution, at 370.28 K has approximately the pressure as blood does what is the osmotic pressure of blood?
Solution:
C = 0.25 M
T = 37O.28 K
(π)gIucose = CRT
(π) = 0.25 mol L-1 x 0.082 L atm K-1 morl-1 x 370.28K
= 7.59 atm

Question 42.
Calculate the molality of a solution containing 7.5 g glycine(NH2-CH2-COOH) dissolved in 500g of water.
Solution:
Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions 1

Question 43.
Which solution has the lower freezing point? 10 g of methanol (CH3OH) in 100g g of water (or) 20 g of ethanol (C2H5OH) in 200 g of water.
Solution:
∆Tf = Kf i.e
∆Tf α m
mCH3-OH = \(\frac{\left(\frac{10}{32}\right)}{0.1}\)
= 3.125 m

mC2H5-OH = \(\frac{\left(\frac{20}{46}\right)}{0.2}\)
= 2.174 m

∴ Depression in freezing point is more in methanol solution and it will have lower freezing point.

Question 44.
How many moles of solute particles are present in one liter of 10-4 M potassium sulphate?
Answer:
In 10-4 M K2SO4 solution, there are 10-4 moles of potassium sulphate.
K2SO4 molecule contains 3 ions (2K+ and 1 SO42-)
1 mole of K2SO4 molecule contains 3 × 6.023 × 1023 ions
10-4 mole of K2SO4 contains 3 × 6.023 × 1023 × 10-4 ions
= 18. 069 × 1019

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 45.
Henry’s law constant for solubility of methane in benzene is 4.2 × 10-5 mm Hg at a particular constant temperature. At this temperature calculate the solubility of methane at
i) 750 mm Hg
ii) 840 mm Hg.
Solution:
(KH)benzene = 4.2 × 10-5 mm
Solubility of methane =?
P = 750 mm Hg P = 840 mm Hg
According to Henrys Law,
P = KH Xin solution
750 mm Hg = 4.2 × 10-5 mm Hg. Xin solution
⇒ Xin solution = \(\frac{750}{4.2 \times 10^{-5}}\)

i. e solubility = 178. 5 × 105
similarly at P = 840 mm Hg
solubility = \(\frac{840}{4.2 \times 10^{-5}}\) = 200 × 10-5

Question 46.
The observed depression in freezing point of water for a particular solution is 0.093°C calculate the concentration of the solution in molality. Given that molal depression constant for water is 1.86 K Kg mol-1.
Solution:
∆Tf = 0.093°C = 0.093 K, m = ?
Kf = 1.86 K Kg mol-1
∆Tf = Kf.m
∴ m = \(\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{f}}}=\frac{0.093 \mathrm{~K}}{1.86 \mathrm{~K} \mathrm{Kg} \mathrm{mol}^{-1}}\)
= 0.05 mol Kg-1 = 0.05 m

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 47.
The vapour pressure of pure benzene (C6H6) at a given temperature is 640 mm Hg. 2.2 g of non – volatile solute is added to 40 g of benzene. The vapour pressure of the solution is 600 mm Hg. Calculate the molar mass of the solute?
Solution:
P°C6H6 = 640 mm Hg
W2 = 2.2 g (non volatile solute)
W1 = 40 g (benzene)
Psolution = 600 mm Hg
M2 =?
Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions 2

11th Chemistry Guide Solutions Additional Questions and Answers

I. Choose the best answer:

Question 1.
6.02 × 1020 molecules of urea ate present in 200 ml of its solution, The concentration of urea solution is (N0 = 6.02 × 1023 mol-1)
a) 0.001 M
b) 0.01M
c) 0.02 M
d) 0.10 M
Answer:
c) 0.02 M

Question 2.
Calculate the molarity and normality of a solution containing 0.5 g of NaOH dissolved in 500 ml solution
a) 0.0025 M, 0.025 N
b) 0.025 M, 0.025 N
c) 0.25 M, 0.25 N
d) 0.025M, 0.0025 N
Answer:
b) 0.025 M, 0.025 N

Question 3.
5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed together and volume made to one liter. The normality of the resulting solution is
a) \(\frac{\mathrm{N}}{40}\)

b) \(\frac{\mathrm{N}}{10}\)

c) \(\frac{\mathrm{N}}{20}\)

d) \(\frac{\mathrm{N}}{5}\)
Answer:
a) \(\frac{\mathrm{N}}{40}\)

Question 4.
At 25°C, the density of 15 M H2SO4 is 1.8 g cm-3. Thus, mass percentage of H2SO4 in aqueous solution is
a) 2%
b) 81.6%
c) 18%
d) 1.8%
Answer:
b) 81.6%

Question 5.
Mole fraction of C3H5(OH)3 in a solution of 36 g of water and 46 g of glycerine is :
a) 0.46
b) 0.36
c) 0.20
d) 0.40
Answer:
c) 0.20

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 6.
The molality of a urea solution in which 0.0100 g of urea, [(NH2)2CO] is added to 0.3000 dm3 of water at STP is
a) 0.555 m
b) 5.55 × 10-4
c) 33.3 m
d) 3.33 × 10-2 m
Answer:
b) 5.55 × 10-4

Question 7.
15 grams of methyl alcohol is dissolved in 35 grams of water. What is the mass percentage of methyl alcohol in solution?
a) 30%
b) 50%
c) 70%
d) 75%
Answer:
a) 30%

Question 8.
A 3.5 molal aqueous solution of methyl alcohol (CH3OH) is supplied. What is the mole fraction of methyl alcohol in the solution?
a) 0.100
b) 0.059
c) 0.086
d) 0.050
Answer:
b) 0.059

Question 9.
In which mode of expression of concentration of a solution remains independent of temperature?
a) Molarity
b) Normality
c) Formality
d) Molality
Answer:
d) Molality

Question 10.
Calculate the molarity of pure water (d = 1 g/L)
a) 555 M
b) 5.55 M
c) 55. 5 M
d) None
Answer:
c) 55. 5 M

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 11.
Calculate the quantity of sodium carbonate (anhydrous) required to prepare 250 ml solution
a) 2.65 grams
b) 4.95 grams
c) 6.25 grams
d) None of these
Answer:
a) 2.65 grams

Question 12.
Find the molality of H2SO4 solution whose specific gravity is 1.98 g ml-1 and 95 % by volume H2SO4
a) 7.412
b) 8.412
c) 9.412
d) 10.412
Answer:
c) 9.412

Question 13.
Calculate molality of 1 liter solution of 93 % H2SO4 by volume. The density of solution is 1.84 g ml-1
a) 9.42
b) 10.42
c) 11.42
d) 12.42
Answer:
b) 10.42

Question 14.
Calculate the molality and mole fraction of the solute in aqueous solution containing 3.0 g of urea per 250 gm of water (Mol.wt. of urea = 60).
a) 0.2 m, 0.00357
b) 0.4 m, 0.00357
c) 0.5 m, 0.00357
d) 0.7 m, 0.00357
Answer:
a) 0.2 m, 0.00357

Question 15.
Calculate normality of the mixture obtained by mixing 100 ml of 0.1 N HCl and 50 ml of 0.25 N NaOH solution.
a) 0.0467 N
b) 0.0367 N
c) 0.0267 N
d) 0.0167 N
Answer:
d) 0.0167 N

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 16.
300 ml 0.1 M HCl and 200 ml of 0.03 M H2SO4 are mixed. Calculate the normality of the resulting mixture
a) 0.084 N
b) 0.84 N
c) 2.04 N
d) 2.84 N
Answer:
a) 0.084 N

Question 17.
What weight of oxalic acid (H2C2O4.2H2O) is required to prepare, 1000mL of N/10 solution?
a) 9.0 g
b) 12.6 g
c) 6.3 g
d) 4.5 g
Answer:
c) 6.3 g

Question 18.
Which of the following units is useful in relating concentration of solution with its vapour pressure?
a) Mole fraction
b) Parts per million
c) Mass percentage
d) Molality
Answer:
a) Mole fraction

Question 19.
The pressure under which liquid and vapour can co-exist at equilibrium is called the
a) Limiting vapour pressure
b) Real vapour pressure
c) Normal vapour pressure
d) Saturated vapour pressure
Answer:
b) Real vapour pressure

Question 20.
CO(g) is dissolved in H2O at 30°C and 0.020 atm. Henry’s law constant for this system is 6.20 × 104 atm. Thus, mole fraction of CO(g) is
a) 1.72 × 10-7
b) 3.22 × 10-7
c) 0.99
d) 0.01
Answer:
b) 3.22 × 10-7

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 21.
H2S gas is used in qualitative analysis of inorganic cations. Its solubility in water at STP is 0.195 mol kg-1. Thus, Henry’s law constant ( in atm raolaT1) for H2S is
a) 2.628 × 10-4
b) 5.128
c) 0.185
d) 3.826 × 103
Answer:
b) 5.128

Question 22.
Which of the following is correct for a solution showing positive deviations from Raoult’s law?
a) ∆V = +ve, ∆H = + ve
b) ∆V = -ve, ∆H = – ve
c) ∆V = + ve, ∆H = -ve
d) ∆V = – ve, ∆H = +ve
Answer:
a) ∆V = +ve, ∆H = + ve

Question 23.
If liquids A and B form an ideal solution
a) The entropy of mixing is zero
b) The Gibbs free energy is zero
c) The Gibbs free energy as well as the entropy of mixing are each zero
d) The enthalpy of mixing is zero
Answer:
d) The enthalpy of mixing is zero

Question 24.
Water and ethanol form non – ideal solution with positive deviation from Raoult’s law. This solution, will have vapour pressure
a) equal to vapour pressure of pure water
b) less than vapour pressure of pure water
c) more than vapour pressure of pure water
d) less than vapour pressure of pure ethanol
Answer:
c) more than vapour pressure of pure water

Question 25.
Which of the following is less than zero for ideal solutions?
a) ∆Hmix
b) ∆V
c) ∆Gmix
d) ∆Smix
Answer:
c) ∆Gmix

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 26.
Which of the following shows negative deviation from Raoult’s law?
a) CHCl3 and CH3COCH3
b) CHCl3 and C2H5OH
c) C6H5CH3 and C6H6
d) C6H6 and CCl4
Answer:
a) CHCl3 and CH3COCH3

Question 27.
Given at 350 K, P°A = 300 torr and P°B = 800 torr, the composition of the mixture having a normal boiling point of 350 K is :
a) XA = 0.08
b) XA = 0.06
c) XA = 0.04
d) XA = 0.02
Answer:
a) XA = 0.08

Question 28.
In mixture A and B, components show – ve deviation as :
a) ∆Vmix is + ve
b) A – B interaction is weaker than A – A and B – B interaction
c) ∆Hmix is + ve
d) A – B interaction is stronger than A – A and B – B interaction
Answer:
d) A – B interaction is stronger than A – A and B – B interaction

Question 29.
If liquid A and B form ideal solution, then:
a) ∆Vmix is = 0
b) ∆Vmix = 0
c) ∆Gmix =0, ∆Smix = 0
d) ∆Smix = 0
Answer:
b) ∆Vmix = 0

Question 30.
Which liquid pair shows a positive deviation from Raoult’s law ?
a) Acetone – chloroform
b) Benzene – methanol
c) Water – nitric acid
d) Water – hydrochloric acid
Answer:
b) Benzene – methanol

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 31.
For A and B to form an ideal solution which of the following conditions should be satisfied ?
a) ∆Hmixing =0
b) ∆Vmixing =0
c) ∆Smixing =0
d) All three conditions mentioned above
Answer:
d) All three conditions mentioned above

Question 32.
Two liquids are mixed together to form a mixture which boils at same temperature, and their boiling point is higher than the boiling point of either of them so they shows.
a) no deviation from Raoult’s law
b) positive, deviation from Raoult’s law
c) negative-deviation from Raoult’s law
d) positive or negative deviation from Raoult’s law depending upon the composition
Answer:
c) negative-deviation from Raoult’s law

Question 33.
Molal elevation constant of liquid is:
a) the elevation in b.p. which would be produced by dissolving one mole of solute in 1oo g of solvent
b) the elevation of b.p. which would be produced by dissolving 1 mole solute in 10 g of solvent
c) elevation in b.p. which would be produced by dissolving 1 mole of solute in 1000g of solvent
d) none of the above
Answer:
c) elevation in b.p. which would be produced by dissolving 1 mole of solute in 1000g of solvent

Question 34.
The vapour pressure of pure liquid solvent is 0.50 atm. When a non – volatile solute B is added to the solvent, its vapour pressure drops to 0.30 atm. Thus, mole fraction of component B is
a) 0.6
b) 0.25
c) 0.45
d) 0.75
Answer:
a) 0.6

Question 35.
The mass of a non – volatile solute (molecular mass = 40) which should be dissolved in 114 g octane to reduce its vapour pressure to 80 % will be
a) 20 g
b) 30 g
c) 10 g
d) 40 g
Answer:
c) 10 g

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 36.
The vapour pressure of pure liquid solvent A is 0.80 atm. When a non – volatile substance B is added to the solvent, its vapour pressure drops to 0.60 atm. Mole fraction of the component B in the solution is:
a) 0.50
b) 0.25
c) 0.75
d) 0.40
Answer:
b) 0.25

Question 37.
18 g of glucose (C6H12O6) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100°C is :
a) 752.40 torr
b) 759.00 torr
c) 7.60 torr
d) 76.00 torr
Answer:
a) 752.40 torr

Question 38.
Calculate the vapour pressure of a solution at 100°C containing 3 g of cane sugar in 33 g of water, (at wt. C = 12, H = 1, O = 16)
a) 760 mm
b) 756.90 mm
c) 758.30 mm
d) None of these
Answer:
b) 756.90 mm

Question 39.
Lowering of vapour pressure due to a solute in 1 molal aqueous solution at 100°C is
a) 13.44 mm Hg
b) 14.12 mm Hg
c) 31.2 mm Hg
d) 35.2 mm Hg
Answer:
a) 13.44 mm Hg

Question 40.
The vapour pressure of a dilute aqueous solution of glucose is 750 mm Hg at 373 K. The mole fraction of the solute is
a) \(\frac{1}{76}\)

b) \(\frac{1}{7.6}\)

c) \(\frac{1}{38}\)

d) \(\frac{1}{10}\)
Answer:
a) \(\frac{1}{76}\)

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 41.
When 3 g of a nonvolatile solute is dissolved in 50 g of water, the relative lowering of vapour pressure observed is 0.018 Nm-2. Molecular weight of the substance is
a) 60
b) 30
c) 40
d) 120
Answer:
a) 60

Question 42.
Elevation in boiling point of a molar (1M) glucose solution (d = 1.2 gmL-1) is
a) 1.34 Kb
b) 0.98 Kb
c) 2.40 Kb
d) Kb
Answer:
b) 0.98 Kb

Question 43.
Given, H2O (l) ⇌ H2O (g) at 373 K, ∆H° = 8.31 kcal mol-1. Thus, boiling point of 0.1 molal sucrose solution is
a) 373. 52 K
b) 373.052 K
c) 373.06 K
d) 374.52 K
Answer:
c) 373.06 K

Question 44.
A solution of 0.450 g of urea (mol. Wt. 60) in 22.5 g of water showed 0.170°C of elevation in boiling point. Calculate the molal elevation constant of water.
a) 0.17°C
b) 0.45°C
c) 0.51°C
d) 0.30°C
Answer:
c) 0.51°C

Question 45.
At higher altitudes, water boils at temperature < 100°C because
a) temperature of higher altitudes is low
b) atmospheric pressure is low
c) the proportion of heavy water increases
d) atmospheric pressure becomes more
Answer:
b) atmospheric pressure is low

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 46.
Which aqueous solution exhibits highest boiling point?
a) 0.015 M glucose
b) 0.01 M KNO3
c) 0.015 M urea
d) 0.01 M Na2SO4
Answer:
d) 0.01 M Na2SO4

Question 47.
A solution of urea in water has boiling point of 100.15°C. Calculate the freezing point of the same solution if Kf and Kb for water are 1.87 K kg mol-1 and 0.52 K kg mol-1 respectively
a) – 0.54°C
b) – 0.44°C
c) – 0.64°C
d) – 0.34°C
Answer:
a) – 0.54°C

Question 48.
Which will have largest ∆Tb?
a) 180 g glucose in 1 kg water
b) 342 g sucrose in 1,000 g water
c) 18 g glucose in 100 g water
d) 65 g urea in 1kg water
Answer:
d) 65 g urea in 1kg water

Question 49.
An aqueous solution of glucose boils at 100.01°C. The molal elevation constant for water is 0.5 K mol-1 kg. The number of molecules of glucose in the solution containing 100 g of water is
a) 6.023 × 1023
b) 12.046 × 1022
c) 12.046 × 1020
d) 12.046 × 1023
Answer:
c) 12.046 × 1020

Question 50.
The latent heat of vaporization of water is 9700 cal/mole and if the b.p. is 100°C, ebullioscopic constant of water is
a) 0.513°C
b) 1.026°C
c.) 10.26°C
d) 1.832°C
Answer:
a) 0.513°C

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 51.
If for a sucrose solution elevation in boiling point is 0.1 °C then what will be the boiling point of NaCl solution for same molal concentration
a) 0.1°
b) 0.2°C
c) 0.08°C
d) 0.01°C
Answer:
b) 0.2°C

Question 52.
The molal boiling point constant for water is 0.513°C kg mol-1. When 0.1 mole of sugar is dissolved in 200 ml of water, the solution boils under a pressure of one atmosphere at
a) 100.513°C
b) 100.0513°C
c) 100.256°C
d) 101.025°C
Answer:
c) 100.256°C

Question 53.
The boiling point of 0.1 m K4[Fe(CN)6] is expected to be (Kb for water = 0.52 K kg mol’1)
a) 100.52°C
b) 100.10°C
c) 100.26°C
d) 102.6°C
Answer:
c) 100.26°C

Question 54.
The value of Kf for the water is 1.86K Kg mole-1, calculated from glucose solution. The value of Kf for water calculated for NaCl solution will be :
a) = 1.86
b) < 1.86
c) > 1.86
d) zero
Answer:
a) = 1.86

Question 55.
The amount of urea to be dissolved in 500 cc of water (Kf = 1.86) to produce a depression of 0.186°C in the freezing point is :
a) 9 g
b) 6 g
c) 3 g
d) 0.3 g
Answer:
c) 3 g

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 56.
Freezing point of an aqueous solution is – 0.186°C. Elevation of boiling point of the same solution is if Kb = 0.512 K molality-1 and Kf= 1.86 K molality-1
a) 0.186°C
b) 0.0512°C
c) 0.092°C
d) 0.237°C
Answer:
b) 0.0512°C

Question 57.
What should be the freezing point of aqueous solution containing 17 g of C2H5OH in 1000 g of water (Kf for water = 1.86 deg kg mol-1)?
a) – 0.69°C
b) 0.34°C
c) 0.0°C
d) – 0.34°C
Answer:
a) – 0.69°C

Question 58.
The freezing point of equimolal aqueous solution will be highest for:
a) C6H5NH3Cl
b) Ca(NO3)2
c) La(NO3)2
d) C6H12O6
Answer:
d) C6H12O6

Question 59.
Cryoscopic constant of a liquid
a) is the decrease in freezing point when 1 g of solute is dissolved per kg of the solvent
b) is the decrease in the freezing point when 1 mole of solute is dissolved per kg of the solvent
c) is the elevation for 1 molar solution
d) is a factor used for calculation of depression in freezing point
Answer:
b) is the decrease in the freezing point when 1 mole of solute is dissolved per kg of the solvent

Question 60.
Which of the following solution will have highest freezing point?
a) 2 M NaCl solution
b) 1.5 M AlCl3 solution
c) 1 M Al2(SO4)3 solution
d) 3 M Urea solution
Answer:
d) 3 M Urea solution

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 61.
0.48 g of a substance is dissolved in 10.6 g of C6H6. The freezing point of benzene is lowered by 1.8°C. what will be the mol.wt. of the substance (Kf for benzene = 5)
a) 250.2
b) 90.8
c) 125.79
d) 102.5
Answer:
c) 125.79

Question 62.
Which of the following aqueous molal solution have highest freezing point?
a) Urea
b) Barium chloride
c) Potassium bromide
d) Aluminium sulphate
Answer:
a) Urea

Question 63.
What weight of NaCl is added to one liter of water so that ∆Tf/Kf = 1?
a) 5.85 g
b) 0.585 g
c) 0.0585 g
d) 0.0855 g
Answer:
c) 0.0585 g

Question 64.
A solution of glucose (C6H12O6) is isotonic With 4 g of urea (NH2 – CO – NH2) per liter of solution. The concentration of glucose is :
a) 4 g/L
b) 8 g/L
c) 12 g/L
d) 14 g/L
Answer:
c) 12 g/L

Question 65.
A 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of a solution of unknown solute. The molar mass of unknown solute in g/mol is
a) 136.2
b) 171.2
c) 68.4
d) 34.2
Answer:
c) 68.4

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 66.
The weight of urea dissolved in 100 ml solution which produces an osmotic pressure of 20.4 atm, will be
a) 5 g
b) 4 g
c) 3 g
d) 6 g
Answer:
a) 5 g

Question 67.
In the phenomenon of osmosis, the membrane allows passage of _________.
a) Solute only
b) Solvent only
c) Both solute and solvent
d) None of these
Answer:
b) Solvent only

Question 68.
A 5.8% (wt./vol.) NaCl solution will exert an osmotic pressure closest to one of the following:
a) 5.8% (wt./vol.) sucrose solution
b) 5.8% (wt./vol.) glucose solution
c) 2 molal sucrose solution
d) 1 molal glucose solution
Answer:
c) 2 molal sucrose solution

Question 69.
The osmotic pressure of a sugar solution at 24°C is 2.5 atmospheres. Determine the concentration of the solution in gram mole per liter.
a) 0.0821 moles/liter
b) 1.082 moles/liter
c) 0.1025 moles/liter
d) 0.0827moles/liter
Answer:
c) 0.1025 moles/liter

Question 70.
What is the freezing point of a solution that contains 10.0g of glucose C6H12O6 in 100 g of H2O? Kf = 1.86° C/m.
a) – 0.186°C
b) + 0.186°C
c) – 0.10°C
d) – 1.03°C
Answer:
d) – 1.03°C

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 71.
The order of osmotic pressure of equimolar solutions of BaCl2, NaCl and glucose will be:
a) BaCl2 > NaCl > glucose
b) NaCl > BaCl2 > glucose
c) glucose > BaCl2 > NaCl
d) glucose > NaCl > BaCl2
Answer:
a) BaCl2 > NaCl > glucose

Question 72.
The wt. of urea dissolved in 100 ml solution which produces an osmotic pressure of 20.4 atm, will be
a) 5 g
b) 4 g
c) 3 g
d) 6 g
Answer:
a) 5 g

Question 73.
A compound MX2 has observed and normal molar masses 65.6 and 164 respectively. Calculate the apparent degree of ionization of MX2:
a) 75%
b) 85%
c) 65%
d) 25%
Answer:
a) 75%

Question 74.
The freezing point of 0.2 molal K2SO4 is – 1.1°C. Calculate van’t Hoff facor and percentage degree of dissociation of K2SO4. Kf for water is 1.86°
a) 97.5
b) 90.75
c) 105.5
d) 85.75
Answer:
a) 97.5

Question 75.
For 0.1M solution, the colligative property will follow the order
a) NaCl > Na2SO4 > Na3PO4
b) NaCl > Na2SO4 ≈ Na3PO4
c) NaCl < Na2SO4 < Na3PO4
d) NaCl < Na2SO4 = Na3PO4
Answer:
c) NaCl < Na2SO4 < Na3PO4

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 76.
PH of a 0.1M monobasic acid is found to be 2. Hence its osmotic pressure at a given temp. T K is
a) 0.1 RT
b) 0.11 RT
c) 1.1 RT
d) 0.01 RT
Answer:
b) 0.11 RT

Question 77.
Which has the highest boiling point?
a) 0.1 m Na2SO4
b) 0.1 m Al(NO3)3
c) 0.1 m MgCl2
d) 0.1 m C6H12O6 (glucose)
Answer:
b) 0.1 m Al(NO3)3

Question 78.
Aluminium phosphate is 100% ionized in 0.01 molal aqueous solution. Hence ∆Tb/ Kb is:
a) 0.01
b) 0.015
c) 0.0175
d) 0.02
Answer:
d) 0.02

Question 79.
1.0 molal aqueous solution of an electrolyte X3Y2 is 25% ionized. The boiling point of the solution is (Kb for H2O = 0.52 K kg/mol)
a) 373.5 K
b) 374.04 K
c) 377.12 K
d) 373.25 K
Answer:
b) 374.04 K

Question 80.
The freezing point of 0,05 m solutions of a non – electrolyte in water is
a) -1.86 °C
b) -0.93°C
c)-0.093°C
d) 0.93°C
Answer:
c)-0.093°C

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 81.
For an ideal solution containing a non – volatile solute, which of the following expression is correctly represented?
a) ∆Tb = Kb × m
b) ∆Tb = Kb × M
c) ∆Tb = Kb × 2m
d) ∆Tb = Kb × 2M
Where m is the molality of the solution and Kb is molal elevation constant.
Answer:
a) ∆Tb = Kb × m

Question 82.
If 5.85 g of NaCl are dissolved in 90 g of water, the mole fraction of NaCl is
a) 0.1
b) 0.2
c) 0.3
d) 0.0196
Answer:
d) 0.0196

Question 83.
What will be the molarity of a solution containing 5g of sodium hydroxide in 250 ml solution?
a) 0.5
b) 1.0
c) 2.0
d) 0.1
Answer:
a) 0.5

Question 84.
If 5.85 g of NaCl (molecular weight 58.5) is dissolved in water and the solution is made up to 0.5 liter, the molarity of the solution will be
a) 0.2
b) 0.4
c) 1.0
d) 0.1
Answer:
a) 0.2

Question 85.
To prepare a solution of concentration of 0.03 g/ml of AgNO3, what amount of AgNO3 should be added in 60ml of solution
a) 1.8
b) 0.8
c) 0.18
d) None of these
Answer:
a) 1.8

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 86.
How many g of dibasic acid (mol.wt. 200) should be present in 100ml of its aqueous solution to give decinormal strength?
a) 1 g
b) 2 g
c) 10 g
d) 20 g
Answer:
a) 1 g

Question 87.
The molarity of a solution of Na2CO3 having 10.6 g/500 ml of solution is
a) 0.2 M
b) 2 M
c) 20 M
d) 0.02 M
Answer:
a) 0.2 M

Question 88.
Molecular weight of glucose is 180, A solution of glucose which contains 18 g per liter is
a) 2 molal
b) 1 molal
c) 0.1 molal
d) 18 molal
Answer:
c) 0.1 molal

Question 89.
0.5 M of H2SO4 is diluted from lliter to 10 liters, normality of resulting solution is
a) 1 N
b) 0.1 N
c) 10 N
d) 11 N
Answer:
b) 0.1 N

Question 90.
An aqueous solution of glucose is 10% in strength. The volume in which 1 g mole of it is dissolved will be
a) 18 liters
b) 9 liters
c) 0.9 liters
d) 1.8 liters
Answer:
d) 1.8 liters

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 91.
When 1.80 g glucose dissolved in 90 g of H2O, the mole fraction of glucose is
a) 0.00399
b) 0.00199
c) 0.0199
d) 0.998
Answer:
b) 0.00199

Question 92.
A 5 molar solution of H2SO4 is diluted from 1 liter to 10 liters. What is the normality of the solution?
a) 0.25 N
b) 1 N
c) 2N
d) 7 N
Answer:
b) 1 N

Question 93.
Normality of 2 M sulphuric acid is
a) 2 N
b) 4 N
c) N/2
d) N/4
Answer:
b) 4 N

Question 94.
What is the molarity of H2SO4 solution, that has a density 1.84 g/cc at 35°C and Contains solute 98% by weight
a) 4.18 M
b) 8.14 M
c) 18.4 M
d) 18 M
Answer:
c) 18.4 M

Question 95.
Which of the following is a colligative property?
a) Osmotic pressure
b) Boiling point
c) Vapour pressure
d) Freezing point
Answer:
a) Osmotic pressure

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 96.
The vapour pressure of benzene at a certain temperature is 640 mm of Eg. A non – volatile and non – electrolyte solid weighing 2.175 g is added to 39.08 g of benzene. The vapour pressure of the solution is 600 mm of Hg. What is the molecular weight of solid substance?
a) 49.50
b) 59.6
c) 69.5
d) 79.8
Answer:
c) 69.5

Question 97.
The average osmotic pressure of human blood is 7.8 bar at 37°C. What is the concentration of an aqueous NaCl solution that could be used in the Mood stream?
a) 0.16 mol/L
b) 0.32 mol/L
c) 0.60 mol/L
d) 0.45 mol/L
Answer:
b) 0.32 mol/L

Question 98.
The osmotic pressure in atmospheres of 10% solution of cane sugar at 69°C is
a) 724
b) 824
c) 8.21
d) 7.21
Answer:
c) 8.21

Question 99.
The molal boiling point constant for water is 0.513°C kg mol-1. When 0.1 mole of sugar is dissolved in 200 ml of water, the solution boils under a pressure of one atmosphere at
a) 100.513°C
b) 100.0513°C
c) 100.256°C
d) 101.025°C
Answer:
c) 100.256°C

Question 100.
The freezing point of a solution prepared from 1.25 g of a non – electrolyte and 20 g of water is 271.9 K. If molar depression constant is 1.8 K mole-1 then molar mass of the solute will be
a) 105.7
b) 106.7
c) 115.3
d) 93.9
Answer:
a) 105.7

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 101.
Osmotic pressure of 0.1 M solution of NaCl and Na2SO4 will be
a) same
b) osmotic pressure of NaCl solution will be more than Na2SO4 solution
c) osmotic pressure of Na2SO4 solution will be more than NaCl
d) osmotic pressure of NaSO4 will be less than that of NaCl solution
Answer:
c) osmotic pressure of Na2SO4 solution will be more than NaCl

Question 102.
At 25 °C the highest osmotic pressure is exhibited by 0.1 M solution of
a) CaCl2
b) KCl
c) Glucose
d) Urea
Answer:
a) CaCl2

Question 103.
Azeotropic mixture of HCl and water has
a) 84% HCl
b) 22.2% HCl
c) 63 % HCl
d) 20.2 HCl
Answer:
d) 20.2 HCl

Question 104.
The boiling point of water (100°C) becomes 100.25°C, if 3 grams of a nonvolatile solute is dissolved in 200 ml of water. The molecular weight of solute is (Kb for water is 0.6 K Kg mol-1)
a) 12.2 g mol-1
b) 15.4 g mol
c) 17.3 g mol-1
d) 20.4 g mol
Answer:
c) 17.3 g mol-1

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

II. Very short question and answer(2 Marks):

Question 1.
What is the common property observed in a naturally existing solution? Explain it.
Answer:

  1. Seawater, the air is the naturally existing homogeneous mixture. The common property observed in these is homogeneity.
  2. The homogeneity implies uniform distribution of their constituents or components throughout the mixture.

Question 2.
What is a saturated solution?
Answer:
A saturated solution is one that contains the maximum amount of a solute that can dissolve in a solvent at a specific temperature.
For example, the solubility of NaCl in 100 g of water at 20°C is 36 g but at other temperatures, or in other solvents, is different.

Question 3.
What are aqueous and non-aqueous solutions? Give example.
Answer:

  1. If the solute is dissolved in the solvent water, the resultant solution is called an aqueous solution. e.g., salt in water.
  2. If the solute is dissolved in a solvent other than water such as benzene, ether, CCl4 etc, the resultant solution is called a non-aqueous solution. e.g., Br, in CCl4.

Question 4.
What is a Supersaturated solution?
Answer:
A supersaturated solution is one that contains more dissolved solute than a saturated solution. It is generally not stable and eventually, the dissolved solute will separate as crystals.

Question 5.
What is the mass percentage?
Answer:
The mass percentage of a component in a solution is the mass of the component present in 100 g of the solution.
Mass percentage of component = \(\frac{\text { Mass of the component in the solution } \times 100}{\text { Total mass of the solution }}\)

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 6.
What is parts per million (ppm)?
Answer:
If the amount of solute in solution is very much less, then the concentration is expressed as parts per million (ppm).
Parts per million (ppm) = \(\frac{\text { Mass of the solute }(\mathrm{mg})}{\text { Mass of the solvent }} \times 10^{6}\)

Question 7.
What is Molarity?
Answer:
Molarity (symbol M) is defined as the number of moles of solute present in a liter of solution. The units of molarity are moles per liter (mol L-1) or moles per cubic decimeter (mol dm-3)
Molarity(m) = \(\frac{\text { Number of ntoles of solute }}{\text { Volume of solution in liter }}\)

Question 8.
What is Non – ideal solution?
Answer:
The solutions which do not obey Raoult’s law over the entire range of concentration, are called non – ideal solutions. For a non – ideal solution, there is a change in the volume and enthalpy upon mixing, i.e.
∆Hmixing ≠ 0 & ∆Vmixing ≠ 0.

Question 9.
State Raoult’s law.
Answer:
According to Raoult’s law, the vapor pressure of solvent over the solution is equal to the product of its vapor pressure in pure state and its mole fraction.
PA = P°A XA or
P = P°AXA

Question 10.
Define boiling point.
Answer:
The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure (1 atm).

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 11.
Define freezing point.
Answer:
Freezing point is defined as “the temperature at which the solid and the liquid states of the substance have the same vapour pressure”.

Question 12.
What is Osmotic pressure?
Answer:
Osmotic pressure can be defined as “the pressure that must be applied to the solution to stop the influx of the solvent (to stop osmosis) through the semi permeable membrane”.

Question 13.
State Dalton’s law.
Answer:
According to Dalton’s law of partial pressure, the total pressure in a closed vessel will be equal to the sum of the partial pressures of the individual components.

Question 14.
What is meant by stock solution (or) standard solution? What is meant by working standard?
Answer:
1. A standard solution or a stock solution is a solution whose concentration is accurately known.

2. At the time of the experiment, the solution with the required concentration is prepared by diluting the stock solution. This diluted solution is called a working standard.

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

III. Short Question and answers(3 Marks):

Question 1.
What is mole fraction?
Answer:
Mole fraction, x, of solute is defined as the ratio of the number of moles of the components divided by the total number of the moles of all the component present in the solution.
Mole fraction of the solute xsolvent = \(\frac{\text { Moles of the component }}{\text { total number of moles of all the component in solution }}\)

Question 2.

What is the non-ideal solution? Give example.
Answer:

  1. The solutions which do not obey Raoult’s law over the entire range of concentration are called non-ideal solutions.
  2. The deviation of the non-ideal solution from Raoult’s law may be positive (or) negative.
  3. For example, Ethyl alcohol, and cyclohexane.

Question 3.
What are the advantages of using a standard solution?
Answer:

  1. The error due to weighing the solute can be minimized by using a concentrated stock solution that requires a large quantity of solute.
  2. We can prepare working standards of different concentrations by diluting the stock solution, which is more efficient since consistency is maintained.
  3. Some of the concentrated solutions are more stable and are less likely to support microbial growth than the working standards used in the experiments.

Question 4.
Limitations of Henry’s law.
Answer:
Henry’s law is applicable at moderate temperature and pressure only. Only the less soluble gases obeys Henry’s law. The gases reacting with the solvent do not obey Henry’s law.
Example:
Ammonia or HCl reacts with water and hence does not obey this law.
NH3 + H2O ⇌ NH4+ + OH

Question 5.
Define osmotic pressure.
Answer:
Osmotic pressure can be defined as the pressure that must be applied to the solution to stop the influx of the solvent (to stop osmosis) through the semipermeable membrane.

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 6.
What is ideal solution?
Answer:
An ideal solution is a solution in which each component i.e. the solute as well as the solvent obeys the Raoult’s law over the entire range of concentratio
For an ideal solution,

  1. There is no change in the volume on mixing the two components (solute & solvents) (∆Vmixing =0).
  2. There is no exchange of heat when the solute is dissolved in solvent (∆Hmixing =0).
  3. escaping tendency of the solute and the solvent present in it should be same as in pure liquids.
    Example:
    benzene & toluene; h-hexane & n-heptane; ethyl bromide & ethyl iodide; chlorobenzene & bromobenzene.

Question 7.
What are colligative properties?
Answer:
“ The properties of the solutions which depend only on the number of solute particles but not on the nature of the solute are called colligative properties”. The following four colligative properties are very important.

  1. Relative lowering of vapor pressure (∆P)
  2. Elevation of boiling point (∆Tb)
  3. Depression of freezing point (∆Tb)
  4. Osmotic pressure (π)

Question 8.
What are the significances of Osmotic pressure?
Answer:
Unlike elevation of boiling point (for 1 molal solution the elevation in boiling point is 0.512°C for water) and the depression in freezing point (for 1 molal solution the depression in freezing point is 1.86°C for water), the magnitude of osmotic pressure is large.

The osmotic pressure can be measured at room temperature enables to determine the molecular mass of bio molecules which are unstable at higher temperatures. Even for a very dilute solution the osmotic pressure is large.

Question 9.
0.24 g of a gas dissolves in 1L of water at 1.5 atm pressure. Calculate the amount of dissolved gas when the pressure is raised to 6.0 atm at constant temperature?
Solution:
Psolute = KH Xsolute in solution
At pressure 1.5 atm,
P1 = KHX1 …………..(1)
At pressure 6.0 atm,
p2 = KHX2 …………….(2)
Dividing equation (1) by (2)
From equation = \(\frac{P_{1}}{P_{2}}=\frac{X_{1}}{X_{2}}\)

\(\frac{1.5}{6.0}=\frac{0.24}{x^{2}}\)

There fore
\(\frac{0.24 \times 6.0}{1.5}\) = 0.96 g/L

Question 10.
What role does the molecular interaction play in solution of alcohol and water?
Answer:
There is strong hydrogen bonding in alcohol molecules as well as water molecules. The intermolecular forces both in alcohol and water are H-bonds. When alcohol and water are mixed,

they form solution because of formation of H-bonds between alcohol and H2O molecules hut these interactions are weaker and less extensive than those in pure water. Hence, they show positive deviation from ideal behaviour.

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 11.
An aqueous solution of 2% non volatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molar mass of the solute when P°A is 1.013 bar?
Solution:
\(\frac{\Delta P}{P_{A}^{0}}=\frac{W_{B} \times M_{B}}{M_{B} \times W_{A}}\)

In a 2 % solution weight of the solute is 2 g and solvent is 98g
∆P = Psolution – P°A = 1.013 – 1.004 bar
= 0.009 bar
MB = \(\frac{\mathrm{P}_{\mathrm{A}}^{0} \times \mathrm{W}_{\mathrm{B}} \times \mathrm{M}_{\mathrm{A}}}{\Delta \mathrm{P} \times \mathrm{W}_{\mathrm{A}}}\)

MB = 2 × 18 × 1.013/(98 × 0.009) = 41.3 g mol-1

Question 12.
Ethylene glycol (C2H6O2) can be at used as an antifreeze in the radiator of a car. Calculate the temperature when ice will begin to separate from a mixture with 20 mass percent of glycol in water used in the car radiator. Kf for water = 1.86K Kg mol-1 and molar mass of ethylene glycol is 62 g mol-1.
Solution:
Weight of solute (W2) = 20 mass percent of solution means 20 g of ethylene glycol
Weight of solvent (water) W1 = 100 – 20 = 80 g
∆Tf = Kf m
= \(\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{W}_{1}}=\frac{1.86 \times 20 \times 1000}{62 \times 80}\)
= 7.5 K
The temperature at which the ice will begin to separate is the freezing of water after the addition of solute i.e 7.5 K lower than the normal freezing point of water (273 – 7.5 K) = 265.5 K

Question 13.
At 400 K 1.5 g of an unknown substance is dissolved in solvent and the solution is made to 1.5 L. Its osmotic pressure is found to be 0.3 bar. Calculated the molar mass of the unknown substance.
Solution :
Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions 3

Question 14.
How would you compare Raoult’s law and Henry’s law.
Answer:
1. According to Raou It’s law, for a solution containing a non volatile solute.
Psolution = P0solvent . xsolute

2. According to henry’s law, Psolution = KH . xsolute in solution

3. The difference between the above two laws is the proportionality constant P° (Raoult’s law) and KH (Heniys law).

4. henry’s law is applicable to the solution containing gaseous solute in liquid solvent, while Raoult’s law is applicable to non-volatile solid solute in the liquid solvent.

5. If the solute is non-volatile then Henry’s law constant will become equal to the vapour pressure of pure solvent Po. thus Raoult’s law becomes a special case of Henry’s law.

6. For very dilute solutions, the solvent obeys Raoult’s law and the solute obeys Henry’s law.

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

IV. Long question and answers(5 Marks):

Question 1.
Explain the factors including the solubility of solute?
Answer:
Factors influencing the solute:
The solubility of a solute generally depends on the nature of the solute and the solvent in which it is dissolved. It also depends on the temperature and pressure of the solution.

Nature of solute and solvent:
Sodium-chloride, an ionic compound, dissolves readily iff a polar solvent such as water, but it does not dissolve in non polar-organic solvents such as benzene or toluene. Many organic compounds dissolve -readily in organic solvents and do not dissolve in water. Different gases dissolve in water to different extents: for example, ammonia is more soluble than oxygen in water.

Effect of temperature:

Solid solute in liquid solvent:
Generally, the solubility of a solid solute in a liquid solvent increases with increase in temperature. When the temperature is increased, the average kinetic energy of the molecules of the solute and the solvent increases. The increase in kinetic energy facilitates the solvent molecules to break the intermolecular attractive forces that keep the solute molecules together and hence the solubility increases.

When a solid is added to a solvent, it begins to dissolve. i.e. the solute leaves from the solid state (dissolution). After some time, some of the dissolved solute returns back to the solid state (recrystallisation). If there is excess of solid present, the rate of both these processes becomes equal at a particular stage. At this stage an equilibrium is established between the solid solute molecules and dissolved solute molecules.

Solute(solid) ⇌ Solute(dissolved)

According to the Le-Chatelier principle, if the dissolution process is endothermic, the increase in temperature will shift the equilibrium towards left i.e solubility increases, for an exothermic reaction, the increase in temperature decreases the solubility. The solubilities, of ammonium nitrate, calcium Chloride, ceric sulphate nano-hydrate, and sodium chloride in water at different temperatures are given in the following graph.
Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions 4
Plot of solubility versus temperature for selective compounds

The following conclusions are drawn from the above graph:

1. The solubility of sodium chloride does not vary appreciably as the maximum solubility is achieved at normal temperature. In fact, there is only 10 % increase in solubility between 0° to 100 °C.
2. the dissolution process of ammonium nitrate is endothermic, the solubility increases steeply with increase in temperature.
3. In the case of ceric sulphate, the dissolution is exothermic and the solubility decreases with increase in temperature.
4. Even though the dissolution of calcium chloride is exothermic, the solubility increases moderately with increase in temperature. Here, the entropy factor also plays a significant role in deciding the position of the equilibrium.

Gaseous solute in liquid solvent:
In the case of gaseous solute in liquid solvent, the solubility decreases with increase in temperature. When a gaseous solute dissolves in a liquid solvent, its molecules interact with solvent molecules with weak intermolecular forces. When the temperature increases, the average kinetic energy of the molecules present in the solution also increases.

The increase in kinetic energy breaks the weak intermolecular forces between the gaseous solute and liquid solvent which results in the release of the dissolved gas molecules to the gaseous state. Moreover, the dissolution of most of the gases in liquid solvents is an exothermic process, and in such processes, the increase in temperature decreases the dissolution of gaseous molecules.

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 2.
Explain vapour pressure of liquid in liquids binary solution?
Answer:
Now, let us consider a binary liquid solution formed by dissolving a liquid solute ‘A’ in a pure solvent B in a closed vessel. Both the components A and B present in the solution would evaporate and an equilibrium will be established between the liquid and vapour phases of the components A and B.

The French chemist Raoult, proposed a quantitative relationship between the partial pressures and the mole fractions of two components A & B, which is known as Raoult’s Law. This law states that “in the case of a solution of volatile liquids, the partial vapour pressure of each component (A & B) of the solution is directly proportional to its mole fraction”.
According to Raoult’s law,
PA ∝ xA
PA = k xA
when xA = 1, k = P°A
where p°A is the vapour pressure of pure component A’ at the same temperature. Therefore,
PA = P°A xA
Similarly, for component ‘B’
PB =P°B xB
xA and xB are the mole fraction of the components A and B respectively.

According to Dalton’s law of partial pressure the total pressure in a closed vessel will be equal to the sum of the partial pressures of the individual components.
Hence,
Ptotal = PA + PB
Substituting the values of PA and PB from equations in the above equation,
Ptotal = XAA + XBB
We know that XA + XB = 1 or XA = 1 – XB
Therefore,
Ptotal = (1 – XB)P°A + XBB
Ptotal = P°A + XB(P°B – P°A)
The above equation is of the straight¬line equation form y = mx + c. The plot of Ptotalversus xB will give a straight line with (P°B – PA) as slope and P°A as the y intercept.

Let us consider the liquid solution containing toluene (solute) in benzene (solvent).

The variation of vapour pressure of pure benzene and toluene with its mole fraction is given in the graph.

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions 5
Solution of benzene in toluene obeying Raoult’s law

The vapour pressures of pure toluene and pure benzene are 22.3 and 74.7 mmHg, respectively. The above graph shows, the partial vapour pressure of the pure components increases linearly with the increase in the mole fraction of the respective components. The total pressure at any composition of the solute and solvent is given by the following straight line (represented as red line) equation.
PSolution = P°toluene + Xbenzene(Pbenzene – Ptoluene)

Question 3.
Explain Raoult’s law for the binary solution of Non-volatile solutes in liquids?
Answer:
When a nonvolatile salute js dissolved in a pure solvent, the vapour pressure of the pure solvent will decrease. In such solutions, the vapour pressure of the -Solution will depend only on the solvent ‘molecules as the solute is nonvolatile.

For example, when sodium chloride is added to the water, the vapor pressure of the salt solution is lowered. The vapour pressure of the solution is determined by the number of molecules of the solvent present in the surface at any time and is proportional to the mole fraction of the solvent.
Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions 6
Rate of vapourization reduced by presence of nonvolatile solute.

Psolution ∝ XA
Where XA is the mole fraction of the solvent
Psolution = kXA
When XA = 1, k = P°solvent
(P°solvent is the partial pressure of pure solvent)
PSolution = P°Solvent
\(\frac{P_{\text {solution }}}{P_{\text {solvent}}^{a}}\) = XA

1 – \(\frac{p_{\text {Solution }}}{p_{\text {Solvent }}^{0}}\) = 1 – XA

\(\frac{p_{\text {Solvent }}^{o}-p_{\text {Solurion }}}{P_{\text {Solvent }}^{0}}\) = XB
Where XB is the fraction of the solute
(∴ xA + xB = 1, XB = 1 – XA)

The above expression gives the relative lowering of vapour pressure. Based on this expression, Raoult’s Law can also be stated as “the relative lowering of vapour pressure of an ideal solution containing the nonvolatile solute is equal to the mole fraction of the solute at a given temperature”.

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 4.
What is ideal solution? Write special features and characters of the ideal solution.
Answer:
An ideal solution is a solution in which each component i.e. the solute, as well as the solvent, obeys Raoult’s law over the entire range of concentration. Practically no solution is ideal over the entire range of concentration. However, when the concentration of solute is very low, the dilute solution behaves ideally.

If the two components present in the solution (A and B) are identical in size, structure, and having almost similar intermolecular attractive forces between them (i.e. between A-A, B-B, and B-A) and then the solution tends to behave like an ideal solution.

For an ideal solution:
(i) there is no change in the volume on mixing the two components, (solute & solvents). (∆Vmixing = 0)
(ii) there is no exchange of heat when the solute is dissolved in solvent (∆Hmixing = 0).
(iii) escaping tendency of the solute arid the solvent present in it should be same as in pure liquids.
Example:
benzene & toulene; n-hexane & n-heptane; ethyl bromide & ethyl iodide; chlorobenzene & bromobenzene.

Question 5.
Explain Non – ideal solution with strong positive deviation.
Answer:
The solutions which do not Raoult’s law over the entire range, of concentration, are called non-ideal solutions. For a non-ideal solution, there is a change in the volume and enthalpy upon mixing, i.e.
∆Hmixing ≠ 0 &, ∆Vmixing ≠ 0. The deviation of the non-ideal solutions from the Raoult’s law can either be positive or negative.

Non-ideal solutions – positive deviation from Rauolt’s Law:
The nature of the deviation from the Rauolt’s law can be explained in terms ©f the intermolecular interactions between solute and solvent (B). Consider a case in which the intermolecular attractive forces between A and B are weaker than those between the molecules of A (A-A) and molecules of B (B-B).

The molecules present in such a solution have a greater tendency to escape from the solution when compared to the ideal solution formed by A and B, in which the intermolecular attractive forces (A-A, B-B, A-B) are almost similar. Consequently, the vapour pressure of such non-ideal solution increases and it is greater than the sum of the vapour pressure of A and B as predicted by the Raoult’s law. This type of deviation is called positive deviation.

Here, PA > p°A XA and pB > P°B XB
Hence Ptotal > p°A XA + PB XB

Let us understand the positive deviation by considering a solution of ethyl alcohol and water. In this solution the hydrogen bonding interaction between ethanol and water is weaker than those hydrogen bonding interactions amongst themselves (ethyl alcohol-ethyl alcohol and water-water interactions).

This results in the increased evaporation of both components from the aqueous solution of ethanol. Consequently, the vapour pressure of the solution is greater than the vapour pressure predicted by Raoults law. Here, the mixing process is endothermic i.e. ∆Hmixing > 0 and there will be slight increase in volume(∆Vmixing > 0).

Example:
Ethyl alcohol cyclohexane, Benzene & acetone, Carbon tetrachloride & chloroform, Acetone & ethyl alcohol, Ethyl alcohol & water.
Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions 7

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 6.
Explain Non – ideal solution with strong negative deviation.
Answer:
Let us consider a case where the attractive forces between solute (A) and solvent t (B) are stronger – thtSSar intermolecular attractive forces between the individual components (A – A & B – B). Here, the escaping tendency of A and B will be lower when compared with an ideal solution formed by A and B. Hence, the vapour pressure of such solutions will be lower than the sum of the vapour pressure of A and B. This type of deviation is called negative deviation. For the negative deviation,
PA < P°A XA and PB < P°B XB.

Let us consider a solution of phenol and aniline. Both phenol and aniline form hydrogen bonding interactions amongst themselves. However, when mixed with aniline, the phenol molecule forms hydrogen bonding interactions with aniline, which are stronger than the hydrogen bonds formed amongst themselves.

Formation of new hydrogen bonds considerably reduce the escaping tendency of phenol and aniline from the solution. As a result, the vapour pressure of the solution is less and there is a slight decrease in volume (∆Vmixing < 0) on mixing. During this process evolution of heat takes place i.e. ∆Hmixing < 0 (exothermic)

Example:
Acetone + chloroform, Chloroform + diethyl ether, Acetone + aniline, Chloroform + Benzene.

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions 8

Question 7.
Explain the factors responsible for deviation from Raoult’s law.
Answer:
Factors responsible for deviation from Raoult’s law:
The deviation of solution from ideal behavior is attributed to the following factors.

i) Solute-solvent interactions:
For an ideal solution, the interaction between the solvent molecules (A-A), the solute molecules (B-B) and between the solvent & solute molecules (A-B) are expected to be similar. If these interactions are dissimilar, then there will be a deviation from ideal behavior.

ii) Dissociation of solute:
When a solute present in a solution dissociates to give its constituent ions, the resultant ions interact strongly with the
solvent and cause deviation from Raoult’s law.
For example, a solution of potassium chloride in water deviates from ideal behavior because the solute dissociates to give K and Cl ion which form strong ion-dipole interaction with water molecules.
KCl(s) + H2O (l) → K+(aq)+ Cl(aq)

iii) Association of solute:
Association of solute molecules can also cause deviation from ideal behaviour. For example, in solution, acetic acid exists as a dimer by forming intermolecular hydrogen bonds, and hence deviates from Raoult’s law.

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions 9

iv) Temperature:
An increase in temperature of the solution increases the average kinetic energy of the molecules present in the solution which causes decrease in the attractive force between them. As result, the solution deviates from ideal behaviour.

v) Pressure:
At high pressure the molecules tend to stay close to each other and therefore there will be an increase in their intermolecular attraction. Thus, a solution deviates from Raoult’s law at high pressure.

vi) Concentration:
If a solution is sufficiently dilute there is no pronounced solvent-solute interaction because the number of solute molecules are very low compared to the solvent. When the concentration is increased by adding solute, the solvent-solute interaction becomes significant. This causes deviation from the Raoult’s law.

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 8.
How would you determine the molar mass of solute from Tb?
Answer:
The elevation of boiling point is directly proportional to the concentration of the solute particles.
∆Tb ∝ m
m is the concentration of solution expressed in molality.
∆Tb = Kbm
Where
Kb = molal boiling point elevation constant or Ebullioscopic constant.
∆Tb = \(\frac{K_{b} \times W_{B} \times 1000}{M_{B} \times W_{A}}\)

Mb = \(\frac{\mathrm{K}_{\mathrm{b}}}{\Delta \mathrm{Tb}} \times \frac{\mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{~W}_{\mathrm{A}}}\)

Question 9.
How would you determine the molar mass of solute from T?
Answer:
1f the solution is prepared by dissolving
WB g of solute in WB g of solvent, then the molality is,

m = \(\frac{\text { Number of moles of solute } \times 1000}{\text { weight of solvent in grams }}\)

Number of moles of solute = \(\frac{W_{B}}{M_{B}}\)
Where, MB = molar mass of the solute
Therefore,

m = \(\frac{\mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}}\)

∆Tf = \(\frac{K_{f} \times W_{B} \times 1000}{M_{B} \times W_{A}}\)

Molar mass can be calculated using

MB = \(\frac{\mathrm{Kb} \times \mathrm{W}_{\mathrm{B}} \times 1000}{\Delta \mathrm{T}_{\mathrm{b}} \times \mathrm{W}_{\mathrm{A}}}\)

Samacheer Kalvi 11th Chemistry Guide Chapter 9 Solutions

Question 10.
How would you determine the molar mass of solute form A?
Answer:
According to van’t Hoff equation π = cRT
c = \(\frac{n}{V}\)
Here, n = number of moles of solute dissolved in ‘V’ liter of the solution.
Therefore,
π = \(\frac{n}{V}\)RT
πV = nRT
If the solution is prepared by dissolving WBg of nonvolatile solute in WA g of solvent,
then the number of moles ‘n’ is,
n = \(\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}}}\)
since, MB = molar mass of the solute
Substituting the ‘n’ value, we get,
π = \(\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{V}} \frac{\mathrm{RT}}{\mathrm{M}_{\mathrm{B}}}\)

MB = \(\frac{W_{B}}{V} \frac{R T}{\pi}\)
From the above equation molar mass of the solute can be calculated.

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 1 Basic Concepts of Chemistry and Chemical Calculations Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

11th Chemistry Guide Basic Concepts of Chemistry and Chemical Calculations Text Book Back Questions and Answers

Textual Questions:

I. Choose the best answer:

Question 1.
40 ml of methane is completely burnt using 80 ml of oxygen at room temperature. The volume of gas left after cooling to room temperature
(a) 40 ml CO2
(b) 40 ml CO2 gas and 80 ml H2o gas
(c) 60 ml CO2 gas and 60 ml H2o gas
(d) 120 ml CO2 gas
Answer:
(a) 40 ml CO2
Solution:
CH4(g) + 2O2 → CO2(g) + 2H2O (1)

Content CH4 O2 CO2
Stoichiometric coefficient 1 2 1
Volume of reactants allowed to react 40 mL  80 mL
Volume of reactant reacted and product formed 40 mL  80 mL 40 mL
Volume of gas after cooling to the room temperature  –  –

Since the product was cooled to room temperature, water exists mostly as liquid. Hence, option (a) is correct.

Question 2.
An element X has the following isotopic composition 200X = 90%, 199X = 8% and 202X = 2%. The weighted average atomic mass of the element X is closest to
(a) 201 u
(b) 202 u
(c) 199 u
(d) 200 u
Answer:
(d) 200 u
Solution:
X = \(\frac{(200 \times 90)+(199 \times 8)+(202 \times 2)}{100}\) = 199.96 = 200 u

Question 3.
Assertion:
Two mole of glucose contains 12.044 × 1023 molecules of glucose
Reason:
Total number of entities present in one mole of any substance is equal to 6.02 × 1022
(a) both assertion and reason are true and the reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(c) assertion is true but reason is false
Solution:
Correct reason:
Total number of entities present in one mole of any substance is equal to 6.022 × 1023.

Question 4.
Carbon forms two oxides, namely carbon monoxide and carbon dioxide. The equivalent mass of which element remains constant?
(a) Carbon
(b) Oxygen
(c) both carbon and oxygen
(d) neither carbon nor oxygen
Answer:
(b) Oxygen
Solution:
Reaction 1:
2C + O2 → 2CO
2 × 12g carbon combines with 32g of oxygen. Hence, Equivalent mass of carbon
\(\frac{2 \times 12}{32}\) × 8 = 6
Reaction 2:
C + O2 → CO2
12 g carbon combines with 32 g of oxygen. Hence, Equivalent mass of carbon
= \(\frac{12}{32}\) × 8 = 6

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 5.
The equivalent mass of a trivalent metal element is 9 g eq-1 the molar mass of its anhydrous oxide is
(a) 102 g
(b) 27 g
(c) 270 g
(d) 78 g
Answer:
(a) 102 g
Solution:
Let the trivalent metal be M3
Equivalent mass = mass of the metal / 3 eq
9 g eq-1 = mass of the metal / 3 eq
Mass of the metal = 27 g
Oxide formed M2O3 ;
Mass of the oxide = (2 × 27) + (3 × 16) = 102 g

Question 6.
The number of water molecules in a drop of water weighing 0.018 g is
(a) 6.022 × 1026
(b) 6.022 × 1023
(c) 6.022 × 1020
(d) 9.9 × 1022
Answer:
(c) 6.022 × 1020
Solution:
Weight of the water drop 0.018 g
No. of moles of water in the drop = Mass of water / molar mass = 0.018/18 = 10-3 mole
No of water molecules present in 1 mole of water = 6.022 × 1023
No. water molecules in one drop of water(10-3 mole)
= 6.022 × 1023 × 10-3
= 6.022 × 1020

Question 7.
1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is
(a) 0 %
(b) 4.4 %
(c) 16 %
(d) 8.4 %
Answer:
(c) 16 %
Solution:
MgCO3 → MgO + CO2
MgCO3:
(1 × 24) + (1 × 12) + (3 × 16) = 84 g
CO2: (1 × 12) + (2 × 16) = 44g
100 % pure 84 g MgCO3 on heating gives 44 g CO2
Given that 1 g of MgCO3 on heating gives 44 g of CO2
Therefore, 84 g MgCO3 sample on heating gives 36.96 g CO2
Percentage of purity of the sample = \(\frac{100 \%}{44 \mathrm{~g} \mathrm{CO}_{2}}\) × 36.96 g of CO2 = 84 %
Percentage of impurity = 16 %

Question Question 8.
When 6.3 g of sodium bicarbonate is added to 30 g of acetic acid solution, the residual solution is found to weigh 33 g. The number of moles of carbon dioxide released in the reaction is
(a) 3
(b) 0.75
(c) 0.075
(d) 0.3
Answer:
(c) 0.075
solution:
NaHCO3 + CH3COOH → CH3OONa + H2O + CO2
6.3 g + 30 g → 33 g + x
The amount of CO2 released, x = 3.3 g
No. of moles of CO2 released = 3.3/44 = 0.075 mol.

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 9.
When 22.4 litres of H2 (g) is mixed with 11.2 litres of Cl2 (g), each at 273 K at 1 atm the moles of HCl (g), formed is equal to
(a) 2 moles of HCl (g)
(b) 0.5 moles of HCl (g)
(c) 1.5 moles of HCl (g)
(d) 1 moles of HCl (g)
Answer:
(d) 1 moles of HCl (g)
Solution:
H2(g) + Cl2(g) → 2HCl(g)

Content CH4 02 CO2
Stoichiometric coefficient 1 1 2
No. of moles of reactants allowed to react at 273 K and 1 atm pressure

22.4 L (1 mol)

11.2 L (0.5 mol)  _
No. of moles of a reactant reacted and product formed 0.5 0.5 _

Amount of HCl formed = 1 mol

Question 10.
Hot concentrated sulphuric acid is a moderately strong oxidising agent. Which of the following reactions does not show oxidising behaviour?
(a) Cu + 2H2 → CuSO4 + SO2 +2 H2O
(b) C + 2 H2SO4 → CO2 + 2 SO2 +2 H2O
(c) BaCl2 + H2SO4 → BaSO4 + 2HCl
(d) none of the above
Answer:
(c) BaCl2 + H2SO4 → BaSO4 + 2HCl
Solution:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 1

Question 11.
Choose the disproportionation reaction among the following redox reactions.
(a) 3Mg(s) + N2(g) → Mg3N2(s)
(b) P4(s) + 3NaOH + 3H2O → PH3(g) + 3NaH2PO2(aq)
(c) Cl2(g) + 2KI(aq) → 2KCl(aq) + I2(s)
(d) Cr2O3(s) + 2Al(s) → Al2O3(s) + 2Cr(s)
Answer:
b) P4(s) + 3NaOH + 3H2O → PH3(g) + 3NaH2PO2(aq)
Solution:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 2

Question 12.
The equivalent mass of potassium permanganate in an alkaline medium is
MnO4 + 2H2O + 3e → MnO2 + 4OH
(a) 31.6
(b) 52.7
(c) 79
(d) None of these
Answer:
(b) 52.7
Solution:
The reduction reaction of the oxidising agent (Mn04) involves the gain of 3 electrons.
Hence the equivalent mass = (Molar mass of KMnO4)/3 = 158.1/3 = 52.7

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 13.
Which one of the following represents 180g of water?
(a) 5 Moles of water
(b) 90 moles of water
(c) \(\frac{6.022 \times 10^{23}}{180}\)

(d) \(\frac{6.022 \times 10^{23}}{1.7}\)
Answer:
(d) \(\frac{6.022 \times 10^{23}}{1.7}\)
Solution:
No. of moles of water present in 180 g = Mass of water / Molar mass of water = 180 g/18 gmol-1 = 10 moles
One mole of water contains = 6.022 × 1023 water molecules
10 mole of water contains = 6.022 × 1023 × 10 × 6.022 × 1024 water molecules

Question 14.
7.5 g of a gas occupies a volume of 5.6 litres at 0° C and 1 atm pressure. The gas is
(a) NO
(b ) N2O
(c) CO
(d) CO2
Answer:
(a) NO
Solution:
7.5 g of gas occupies a volume of 5.6 liters at 273 K and 1 atm pressure Therefore, the mass of gas that occupies a volume of 22.4 liters
= \(\frac{7.5 g}{5.61}\) × 22.41 = 30 g
Molar mass of NO (14 + 16) = 30 g

Question 15.
Total number of electrons present in 1.7 g of ammonia is
(a) 6.022 × 1023
(b) \(\frac{6.022 \times 10^{22}}{1.7}\)
(c) \(\frac{6.022 \times 10^{24}}{1.7}\)
(d) \(\frac{6.022 \times 10^{23}}{1.7}\)
Answer:
(a) 6.022 × 1023
Solution:
No. of electrons present in one ammonia (NH3) molecule (7 + 3) = 10
No. of moles of ammonia = \(\frac{\text { Mass }}{\text { Molar mass }}\)
= \(\frac{1.7 \mathrm{~g}}{17 \mathrm{~mol}^{-1}}\) = 0.1 mol
= 0.1 × 6.022 × 1023 = 6.022 × 1022
= No. of electrons present in 0.1 mol of ammonia

Question 16.
The correct increasing order of the oxidation state of sulphur in the anions
SO42-, SO32-, S2O42-, S2O62- is

(a) SO32- < SO42- < S2O42- < S2O62-

(b) SO42- < S2O42- < S2O62- < SO32-

(c) S2O42- < SO32- < S2O62- < SO42-

(d) S2O62- < S2O42- < SO42- < SO32-
Answer:
(c) S2O42- < SO32- < S2O62- < SO42-
Solution:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 3

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 17.
The equivalent mass of ferrous oxalate is
(a) \(\frac{\text { molar mass of ferrous oxalate }}{1}\)

(b) \(\frac{\text { molar mass of ferrous oxalate }}{2}\)

(c) \(\frac{\text { molar mass of ferrous oxalate }}{3}\)
(d) none of these
Answer:
(c) \(\frac{\text { molar mass of ferrous oxalate }}{3}\)
Solution:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 4

Converting grams to moles calculator is as simple as multiplying the total mass by the moles per unit of mass.

Question 18.
If Avogadro number were changed from 6.022 × 1023 to 6.022 × 1020, this would change
(a) the ratio of chemical species to each other in a balanced equation
(b) the ratio of elements to each other in a compound
(c) the definition of mass in units of grams
(d) the mass of one mole of carbon
Answer:
(d) the mass of one mole of carbon

Question 19.
Two 22.4 liter containers A and B contains 8 g of O2 and 8 g of SO2 respectively at 273 K and 1 atm pressure, then
(a) Number of molecules in A and B are the same
(b) Number of molecules in B is more than that in A.
(c) The ratio between the number of molecules in A to number of molecules in B is 2 : 1
(d) Number of molecules in B is three times greater than the number of molecules in A.
Answer:
(c) The ratio between the number of molecules in A to number of molecules in B is 2 : 1
Solution:
No. of moles of oxygen = 8 g/32 g = 0.25 moles of oxygen
No. of moles of sulphur dioxide = 8 g/64 g
= 0.125 moles of sulphur dioxide
Ratio between the no. of molecules = 0.25 : 0.125 = 2 : 1

Question 20.
What is the mass of precipitate formed when 50 ml of 8.5 % solution of AgNO3 is mixed with 100 ml of 1.865 % potassium chloride solution?
(a) 3.59 g
(b) 7 g
(c) 14 g
(d) 28 g
Answer:
(a) 3.59 g
Solution:
AgNO3 + KCl → KNO3 + AgCl
50 mL of 8.5 % solution contains 4.25 g of AgNO3
No. of moles of AgNO3 present in 50 mL of 8.5 % AgNO3 solution
= Mass / Molar mass = 4.25 / 170 = 0.025 moles
Similarly, No of moles of KCl present in 100 mL of 1.865 % KCl solution = 1.865 / 74.5 = 0.025 moles
So total amount of AgCl formed is 0.025 moles (based on the stoichiometry)
Amount of AgCl present in 0.025 moles of AgCl
= No. of moles × molar mass = 0.025 × 143.5 = 3.59 g

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 21.
The mass of a gas that occupies a volume of 612.5 ml at room temperature and pressure (25°C and 1 atm pressure) is 1.1 g. The molar mass of the gas is
(a) 66.25 g mol-1
(b) 44 g mol-1
(c) 24.5 g mol-1
(d) 662.5 g mol-1
Answer:
(b) 44 g mol-1
Solution:
No. of moles of a gas that occupies a volume of 612.5 mL at room temperature and pressure (25° C and 1 atm pressure)
= 612.5 × 10-3 L/24.5 L mol-1
= 0.025 moles
We know that,
Molar mass = Mass / no. of moles
= 1.1 g / 0.025 mol
= 44 g mol-1

Question 22.
Which of the following contain the same number of carbon atoms as in 6 g of carbon -12.
(a) 7.5 g ethane
(b) 8 g methane
(c) both (a) and (b)
(d) none of these
Answer:
(c) both (a) and (b)
Solution:
No. of moles of carbon present in 6 g of C – 12 = Mass/Molar mass
= 6/12 = 0.5 moles
= 0.5 × 6.022 × 1023 carbon atoms.
No. of moles in 8 g of methane
= 8/16 = 0.5 moles
= 0.5 × 6.022 × 1023 carbon atoms.
No. of moles in 7.5 g of ethane = 7.5 / 16 = 0.25 moles
= 2 × 0.25 × 6.022 × 1023 carbon atoms.

Question 23.
Which of the following compound(s) has/have a percentage of carbon same as that in ethylene (C2H4)
(a) Propene
(b) ethyne
(c) benzene
(d) ethane
Answer:
(a) Propene
Solution:
Percentage of carbon in ethylene (C2H4) = \(\frac{\text { mass of carbon }}{\times 100 \text { Molar mass }}\)
= \(\frac{24}{28}\) × 100 = 85.71 %
Percentage of carbon in propene (C3H6) = \(\frac{24}{28}\) × 100 = 85.71 %

Question 24.
Which of the following is/are true with respect to carbon -12.
(a) relative atomic mass is 12 u
(b) oxidation number of carbon is +4 in all its compounds.
(c) 1 mole of carbon-12 contains 6.022 × 1022 carbon atoms.
(d) all of these
Answer:
(a) relative atomic mass is 12 u

Question 25.
Which one of the following is used as a standard for atomic mass.
(a) 6C12
(b) 7C12
(c) 6C13
(d) 6C14
Answer:
(a) 6C12

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

II. Write brief answer to the following questions:

Question 26.
Define relative atomic mass.
Answer:
Relative atomic mass is defined as the ratio of the average atomic mass factor to the unified atomic mass unit.
Relative atomic mass (Ar) = \(\frac{\text { Average mass of the atom }}{\text { Unified atomic mass }}\)

Question 27.
What do you understand by the term mole?
Answer:
The mole is defined as the amount of a substance which contains 6.023 x 1023 particles such as atoms, molecules, or ions. It is represented by the symbol.

Question 28.
Define equivalent mass.
Answer:
Gram equivalent mass of an element, compound or ion is the mass that combines or displaces 1.008 g hydrogen or 8 g oxygen or 35.5 g chlorine. Equivalent mass has no unit but gram equivalent mass has the unit g eq-1.
Gram equivalent mass = \(\frac{\text { Molar mass }\left(\mathrm{g} \mathrm{mol}^{-1}\right)}{\text {Equivalence factor }\left(\mathrm{eq} \mathrm{mol}^{-1}\right)}\)

Question 29.
Distinguish between oxidation and reduction.
Answer:

Oxidation Reduction
1. Reactions involving the addition of oxygen Reactions involving removal of hydrogen
2. Reactions involving loss of an electron Reactions involving gain of electron
3. Reaction in which oxidation number of the element increases. Reaction in which oxidation number of the element decreases.

Question 30.
What do you understand by the term oxidation number.
Answer:
Oxidation:
According to the classical concept, oxidation is a process of addition of oxygen or removal of hydrogen.
Removal of hydrogen
2H2S + O2 → H2O + 2S
Addition of oxygen
C + O2 → CO2
According to the electronic concept, loss of electrons is called oxidation reaction.
Ca → Ca2+ + 2e
During oxidation, oxidation number increases.
Dining oxidation, reducing agent gets oxidised.

Reduction:
Reduction is a process of removal of oxygen or addition of hydrogen.
Addition of hydrogen
Ca + H2 → CaH2
Removal of oxygen
Zn O + C → Zn + CO
According to the electronic concept, gain of electrons is called a reduction reaction.
Zn2+ + 2e → Zn
During reduction, oxidation number decreases.
During reduction, oxidising agent gets reduced.

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 31.
Calculate the molar mass of the following compounds.
Answer:
(i) urea [CO(NH2)2]:
Molar mass of urea = (4 × Atomic mass of hydrogen) + ( 1 × Atomic mass of carbon) + ( 2 × Atomic mass of nitrogen) + ( 1 × Atomic mass of oxygen)
= (4 × 1.008) +(1 × 12) + (2 × 14)+ ( 1 × 16)
= 4.032 + 12 + 28 + 16 = 60.032 g mol-1

(ii) acetone[CH3COCH3]
Molar mass of Acetone = (6 × Atomic mass of hydrogen) + ( 3 × Atomic mass of carbon) + (1 × Atomic mass of oxygen) = (6 × 1.008) + (3 × 12) + ( 1 × 16)
= 6.048 + 36 + 16 = 52.024 g mol-1.

(iii) boric acid [H3BO3]:
Molar mass of Boric acid = (3 × Atomic mass of hydrogen) + ( 1 × Atomic mass of boron) + ( 3 × Atomic mass of oxygen)
= (3 × 1.008) + (3 × 11) + ( 1 × 16)
= 3.024 + 33 + 16 = 52.024 g mol-1.

(iv) sulphuric acid [H2SO4] = (2 × Atomic mass of hydrogen) + ( 1 × Atomic mass of sulphur) + ( 4 × Atomic mass of oxygen)
= (2 × 1.008) +(1 × 32) + (4 × 16)
= 2.016 + 32 +64 = 98.016 g mol-1.

Question 32.
The density of carbon dioxide is equal to 1.965 kgm-3 at 273 K and 1 atm pressure. Calculate the molar mass of CO2.
Answer:
Given:
The density of C02 at 273 K and 1 atm pressure = 1.965 kgm-3
Molar mass of CO2 =?
At 273 K and 1 atm pressure, 1 mole of CO2 occupies a volume of 22.4 L
Mass of 1 mole of CO2 = \(\frac{1.965 \mathrm{Kg}}{1 \mathrm{~m}^{3}}\) × 22.4 L
= \(\frac{1.965 \times 10^{3} \mathrm{~g} \times 22.4 \times 10^{-3} \mathrm{~m}^{3}}{1 \mathrm{~m}^{3}}\)
= 44.01 g
Molar mass of CO2 = 44 gmol-1

Question 33.
Which contains the greatest number of moles of oxygen atoms.
1 mol of ethanol
1 mol of formic acid
1 mol of H2O
Answer:

Compound

Given no. of moles

No. of oxygen atoms

Ethanol – C2H5OH

1 1 × 6.022 × 1023

Formic acid -HCOOH

1

2 × 6.022 × 1023

Water – H2O 1

1 × 6.022 × 1023

Answer: Formic acid

Question 34.
Calculate the average atomic mass of naturally occurring magnesium using the following data:

Isotope Isotopic atomic mass Abundance (%)
Mg24 23.99 78.99
Mg26 24.99 10.00
Mg25 25.98 11.01

Answer:
Average atomic mass
= \(\frac{(78.9923 .99)(1024.99)(11.0125 .98)}{100}\)
= \(\frac{2430.9}{100}\)
= 24.31 u

Question 35.
In a reaction x + y + xyz. identify the Limiting reagent if any, in the following reaction mixtures.
(a) 200 atoms of x + 200 atoms of y + 50 molecules of z2.
(b) 1 mol of x + 1 mol ofy + 3 mol of z2.
(c) 50 atoms of x + 25 atoms of y + 50 molecules of z2.
(d) 2.5 mol ofx + 5 mol ofy + 5 mol of z2.
Reaction:
x + y + z2 → xyz2

(a) 200 atoms of x + 200 atoms of y + 50 molecules of z2 According to the reaction, 1 atom of x reacts with one atom of y and one molecule of z to give product. In the case (a) 200 atoms of x, 200 atoms of y react with 50 molecules of z2 (4 part) i.e. 50 molecules of z2 react with 50 atoms of x and 50 atoms of y. Hence z is the limiting reagent.

(b) 1 mol of x + 1 mol of y + 3 mol of z2
According to the equation 1 mole of z2 only react with one mole of x and one mole of y. If 3 moles of z2 are there, z is limiting reagent.

(c) 50 atoms of x + 25 atoms of y + 50 molecules of z2
25 atoms of y react with 25 atoms of x and 25 molecules of z2. So y is the limiting reagent.

(d) 2.5 mol of x + 5 mol of y + 5 mol of z2
2.5 mol of x react with 2.5 mole of y and 2.5 mole of z2. So x is the limiting reagent.

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 36.
Mass of one atom of an element is 6.645 × 1023 g How many moles of element are there in 0.320 kg.
Answer:

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 26
there in 0.320 kg
Given:
mass of 1 atom = 6.645 × 10-23 g
∴ mass of 1 mole of atom = 6.645 × 10-2323 g × 6.022 × 1023 = 40 g
∴ Number of moles of element in 0.320 kg = \(\frac{1 \mathrm{~mol}}{40 \mathrm{~g}}\) × 0.320 kg
= \(\frac{1 \mathrm{~mol} \times 320 \mathrm{~g}}{40 \mathrm{~g}}\) = 8 mol

Question 37.
What is the difference between molecular mass and molar mass? Calculate the molecular mass and molar mass for carbon monoxide.
Answer:
Molecular mass:

  • Relative molecular mass is defined as the ratio of the mass of the molecule to the unified atomic mass unit.
  •  It can be calculated by adding the relative atomic masses of its constituent atoms.
  • For carbon monoxide (CO) Molecular mass = Atomic mass of carbon + Atomic mass of oxygen 12 + 16 = 28 u.

Molar mass:

  • It is defined as the mass of one mole of a substance.
  • The molar mass of a compound is equal to the sum of the relative atomic masses of its constituent expressed in g mol-1.
  • For carbon monoxide (CO) 12 + 16 = 28 g mol-1 Both molecular mass and molar mass are numerically the same but the units are different.

Question 38.
What is the empirical formula of the following?
(i) Fructose (C6H12O6) found in honey
(ii) Caffeine (C8H10N4O2 )a substance found in tea and coffee.
Answer:

Compound Molecular formula Empirical formula
Fructose C6H12O6 CH2O
Caffeine C8H10N4O2 C4H5N2O

Question 39.
The reaction between aluminium and ferric oxide can generate temperatures up to 3273 K and is used in welding metals. (Atomic mass of AC = 27 u Atomic mass of O = 16 u)
2Al + Fe2O3 → Al2O3 + 2Fe; If, in this process, 324 g of aluminium is allowed to react with 1.12 kg of ferric oxide
(i) Calculate the mass of Al2O3 formed.
(ii) How much of the excess reagent is left at the end of the reaction?
Answer:
Given:
2Al + Fe2O3 → Al2O3 + 2Fe

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 27
Molar mass of Al2O3 formed = 6mol × 102 g mol-1 = 612 g
[Al2O3 = (2 × 27) + (3 × 16) = 54 + 48 = 102]
Excess reagent = Fe2O3.
Amount of excess reagent left at the end of the reaction = 1 mol × 160 g mol-1 = 160 g
[Fe2O3 = (2 × 56) + (3 × 16) = 112 + 48 = 160 g] = 160 g

Question 40.
How many moles of ethane is requircd to produce 44 g of CO2 (g) after combustion. Balanced equation for the combustion of ethane.
Answer:
C2H6 + \(\frac{7}{2}\)O2 → 2CO2 + 3H2O
⇒ 2C2H6 + 7O2 → 4CO2 + 6H2O
∴ To produce 4 moles of CO2, 2 moles of ethane is required
To produce 1 mole (44g) of CO2 required number of moles of ethane
= \(\frac{1}{2}\) mole of ethane
= 0.5 mole of ethane

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 41.
Hydrogen peroxide is an oxidising agent. It oxidises ferrous ion to ferric ion and reduced itself to water. Write a balanced equation.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 5

Question 42.
Calculate the empirical and molecular formula of a compound containing 76.6% carbon, 6.38 % hydrogen and rest oxygen its vapour density is 47.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 6
Empirical formula = C6H6O
η = Molar mass / Calculated empirical formula mass = \(\frac{2 \times \text { vapour density }}{94}\)
= \(\frac{2 \times 47}{94}\) = 1
∴ Molecular formula(C6H6O) × 1 = C6H6O

Question 43.
A Compound on analysis gave Na = 14.31% S = 9.97% H= 6.22% and 0= 69.5% calculate the molecular formula of the compound if all the hydrogen in the compound is present in combination with oxygen as water of crystallization, (molecular mass of the compound is 322).
Answer:

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 7

∴ Empirical formula = Na2SH20O14
η = Molar mass / Calculated empirical formula mass
= \(\frac{322}{322}\) = 1
[Na2SH20O14 = (2 × 23) + (1 × 32) + (20 × 1) + (14 × 16)
= 46 + 32 + 20 + 224 = 322]
Molecular formula = Na2SH20O14
Since all the hydrogen in the compound present as water
∴ Molecular formula is Na2SH20O14

Question 44.
Balance the following equations by oxidation number method
(i) K2Cr2O7 + KI + H2SO4 → K2SO4 + Cr2(SO4)3 + I2 + H2O
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 8
K2Cr2O7 + 6KI + H2SO4 → K2SO4 + Cr2(SO4)3 + I2 + H2O
K2Cr2O7 + 6KI + H2SO4 → K2SO4 + Cr2(SO4)3 + 3I2 + H2O
K2Cr2O7 + 6KI + 7H2SO4 → 4K2SO4 + Cr2(SO4) + I2 + 7H2O

(ii) KMnO4 + Na2SO3 → MnO2 + Na2SO4 + KOH
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 9
⇒ 2KMnO4 + 3Na2SO3 → MnO2 + Na2SO4 + KOH
⇒ 2KMnO4 + 3Na2SO3 → 2MnO2 + 3Na2SO4 + KOH
⇒ 2KMnO4 + 3Na2SO3 → MnO2 + Na2SO4 + 2KOH

(iii) Cu + HNO3 → Cu(NO3)2 + NO2 + H2O
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 10
Cu + 2HNO3 → Cu(NO3)2 + NO2 + H2O
Cu + 2HNO3 + 2HNO3 → Cu(NO3)2 + 2NO2 + 2H2O
Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O

(iv) KMnO4 + H2C2O4 + H2SO4 → K2SO + MnSO4 + CO2 + H2O
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 11
KMnO4 + 5H2C2O4 + H2SO4 → K2SO + MnSO4 + CO2 + H2O
2KMnO4 + 5H2C2O4 + H2SO4 → 2MnSO4 + 10CO2 + H2O
2KMnO4 + 5H2C2O4 + 3H2SO4 → K2SO + 2MnSO4 + 10CO2 + 8H2O

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 45.
Balance the following equations by ion electron method.
(i) KMnO4 + SnCl2 + HCl → MnCl2 SnCl4 + H2O + KCl
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 12

(ii) C2O42- + Cr2O72-
Cr3+ + CO2 (in acid medium)
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 13

(iii) Na2S2O3 + I2 → Na2S4O6 + NaI 2 (in acid medium)
Answer:
S2O32- → S4O62- ………….(1)
Half reaction ⇒ I2 → I …………….(2)
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 14

(iv) Zn + NO3 → Zn+2 + NO
Half reactions are
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 15

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

11th Chemistry Guide Basic Concepts of Chemistry and Chemical Calculations Additional Questions and Answers

I. Choose the best answer:

Question 1.
_____ consists of more than one chemical entity present without any chemical interactions.
(a) Mixtures
(b) Pure substances
(c) Compounds
(d) Elements
Answer:
(a) Mixtures

Question 2.
_______ are made up of molecules which contain two or more atoms of different elements.
(a) Mixtures
(b) Compounds
(c) Pure substances
(d) Elements
Answer:
(b) Compounds

Question 3.
Match the correct pair:

A. Compound (i) S8
B. Mixture (ii) Glucose
C. Element (iii) Air

(a) A – ii, B – i, C – iii
(b) A – i, B – ii, C – iii
(c) A – ii, B – iii, C – i
(d) A – iii, B – ii, C – i
Answer:
(c) A – ii, B – iii, C – i

Question 4.
_______ atom is considered as standard by the IUPAC for calculating atomic masses.
(a) C – 13
(b) C – 15
(c) C – 14
(d) C – 12
Answer:
(d) C – 12

Question 5.
The value of unified mass is equal to
(a) 1.6605 × 10-27 kg
(b) 1.6736 × 10-27 kg
(c) 1.6605 × 1027 kg
(d) 1.6736 × 10-29 kg
Answer:
(a) 1.6605 × 10-27 kg

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 6.
Chlorine consists of two naturally occurring isotopes 17Cl35 and 17Cl37 in the ratio 77 : 23. The average relative atomic mass of chlorine is
(a) 36.45 u
(b) 35.56 u
(c) 35.46 u
(d) 35.65 u
Answer:
(c) 35.46 u

Question 7.
The relative atomic masses of hydrogen, oxygen, and carbon are 1.008 u, 16 u, and 12 u respectively. The relative molecular mass of glucose (C6 H12 O6 ) is
(a) 170.096 u
(b) 189.096 u
(c) 180.096 u
(d) 190.086 u
Answer:
(c) 180.096 u

Question 8.
The specific amount of a substance is represented in SI unit is
(a) amu
(b) mole
(c) atomic mass
(d) equivalent mass
Answer:
(b) mole

Question 9.
Commonly used medicines for treating heart bum and acidity are called
(a) antipyretics
(b) analgesics
(c) antiseptics
(d) antacids
Answer:
(d) antacids

Question 10.
The typical concentration of hydrochloric acid in gastric acid is
(a) 0.1 M
(b) 0.01 M
(c) 0.082 M
(d) 0.82 M
Answer:
(c) 0.082 M

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 11.
Antacids used to treat acidity contain mostly
(a) Al(OH)3
(b) Mg(OH)2
(c) (a) or (b)
(d) Ca(OH)2
Answer:
(c) (a) or (b)

Question 12.
The value of Avogadro number is
(a) 6.022 × 1023
(b) 6.022 × 1022
(e) 6.022 × 10-25
(d) 6.022 × 10-23
Answer:
(a) 6.022 × 1023

Question 13.
The volume occupied by one mole of any substance in the gaseous state st 273 K and 1 atm pressure is _______ (in litres)
(a) 24.5
(b) 22.4
(e) 22.71
(d) 21.18
Answer:
(b) 22.4

Question 14.
The equivalent mass of KMnO4(Molar mass 158 g mol -1 based on the equation
MnO4 + 8H+ + 5e → Mn2+ + 4H2O
(a) 158
(b) 52.66
(c) 31.6
(d) 40
Answer:
(c) 31.6

Question 15.
The molecular formula of acetic acid is C2H4O2 Its empirical formula is
(a) C2H4O2
(b) C2H2O
(c) CH2O2
(d) CH2O
Answer:
(d) CH2O

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 16.
The number of moles of hydrogen required to produce 20 moles of ammonia is
(a) 10
(b) 15
(c) 30
(d) 20
Answer:
(c) 30

Question 17.
Which of the following is/are redox reactions?
(i) 4Fe + 3O2 → 2Fe2O3
(ii) H2S + Cl2 → 2HCl + S
(iii) CuO + C → Cu + CO
(iv) S +H2 → H2S
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv)
Answer:
(d) (iii) and (iv)

Question 18.
The oxidation number of sulphur on H2SO4 is
(a) -2
(b) -6
(c) +2
(d) +6
Answer:
(d) +6

Question 19.
Choose the correct pair:

Compound Oxidation number
A. Cr in Cr2O7 (i) +4
B. C in CO2 (ii) +2
C. C in CH2F2 (iii) +6
D. Mn in MnSO4 (iv) 0

(a) A – i, B – iv, C- iii, D – ii
(b) A – iii, B – i, C – iv, D – ii
(c) A – i, B – ii, C – iv, D – iii
(d) A – iii, B – iv, C – i, D – ii
Answer:
(b) A – iii, B – i, C – iv, D – ii

Question 20.
The change in the oxidation number of Manganese in the following reaction 2 KMnO4 + 10 FeSO4 + 8 H2SO4 → K2SO4 + 5 Fe2(SO4)3 + 8 H2O
(a) +2 to +7
(b) +2 to +5
(c) +7 to +2
(d) +5 to +2
Answer:
(c) +7 to +2

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 21.
Which of the following statements are correct about oxidation?
(i) Removal of oxygen
(ii) Addition of hydrogen
(iii) Loss of electron
(iv) Gain of electron
(v) Addition of oxygen
(vi) Removal of hydrogen
(a) i, ii, iv
(b) iii, v, vi
(c) i, iii, v
(d) iii, iv, v
Answer:
(b) iii, v, vi

Question 22.
Choose the correct oxidation number of oxygen in the following compounds.

(A) KO2 (i) -2
(B) H2O (ii) +2
(C) H2O2 (iii) -1/2
(D) OF2 (iv) -1

(a) A -iii, B- iv, C – i, D – ii.
(b) A – iv, B – i, C – iii, D – ii
(c) A – iii, B – i, C – iv, D – ii
(d) A – iv, B – iii, C – i, D – ii
Answer:
(c) A – iii, B – i, C – iv, D – ii

Question 23.
The correct order of electron releasing tendency of the following elements is
(a) Zn > Cu > Ag
(b) Cu > Zn > Ag
(c) Ag > Zn > Cu
(d) Ag > Cu > Zn
Answer:
(a) Zn > Cu > Ag

Question 24.
H2O2 → 2 H2O + O2 is a
(a) Displacement reaction
(b) Combinationreaction
(c) Decomposition reaction
(d) Disproportionate reaction
Answer:
(d) Disproportionate reaction

Question 25.
The molar mass and empirical formula mass of a compound are 78 and 13 respectively. The molecular formula of the compound is (Empirical formula is CH)
(a) C2H2O2
(b) C2H4
(c) C6H6
(d) C3H8
Answer:
(c) C6H6

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

II. Very Short Question and Answers (2 Marks):

Question 1.
State Avogadro’s Hypothesis.
Answer:
It states that ‘Equal volume of all gases under the same conditions of temperature and pressure contain the same number of molecules’.

Question 2.
How is matter classified physically?
Answer:
Matter can be classified as solids, liquids, and gases based on their physical state. The physical state of matter can be converted into one another by modifying the temperature and pressure suitably.

Question 3.
How is matter classified chemically?
Answer:
Matter can be classified into mixtures and pure substances based on chemical compositions.

Question 4.
Calculate a number of moles of carbon atoms ¡n three moles of ethane.
Answer:
Ethane – Molecular formula = C2H6
1 mole of ethane contains 2 atoms of carbon (6.023 x 1023 C)
∴ 3 moles of ethane contains 6 atoms of Carbon.
∴ No. of moles of Carbon atoms = 3 x 6.023 x 1023 Carbon atoms.
= 18.069 x 1023 Carbon atoms.

Question 5.
What are pure substances? How are they classified?
Answer:
Pure substances are composed of simple atoms or molecules. They are further classified as elements and compounds.

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 6.
Mass of one atom of an element ¡s 6.66 x 1023 g. How many moles of the element are there in 0.320 kg?
Answer:
Mass of one atom of an element = 6.66 x 1023g
No. of moles = \(\frac {Mass}{Molecular mass}\) 3
Molecular mass = Mass of 1 atom x Avogadro number
6.66 x 1023 x 6.023 x 1023
= 6.66 x 6.023 = 40.11318
Number of moles = \(\frac {Mass}{Molecular mass}\) = \(\frac{0.320 \mathrm{kg} \times 10^{3}}{40}\) = 8 moles.

Question 7.
What are compounds? Give examples.
Answer:
Compounds are made up of molecules which contain two or more atoms of different elements.
Example: Carbon dioxide (CO2), Sodium Chloride (NaCl).

Question 8.
Calculate the weight of 0.2 moles of sodium carbonate.
Answer:
Sodium carbonate = Na2CO3
Molecular mass of Na2CO3 = (23 x 2)+(12 x 1)+(16 x 3)
= 46 + 12 + 48 = 106 g
Mass of 1 mole of Na2CO3 = \(\frac{106 \times 0.2}{1}\) = 21.2 g

Question 9.
Define relative atomic mass.
Answer:
The relative atomic mass is defined as the ratio of the average atomic mass factor to the unified atomic mass unit.
Relative atomic mass
(Ar) = Average mass of the atom / Unified atomic mass

Question 10.
What is the average atomic mass?
Answer:
Average atomic mass is defined as the average of the atomic masses of all atoms in their naturally occurring isotopes.

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 11.
Calculate the equivalent mass of barium hydroxide.
Answer:
Barium hydroxide = Ba(OH)2
Molecular mass of Ba(OH)2 = 137 + (16 x 2) + (1 x 2) = 171.0 g / mol.
Acidity = 2
Equivalent mass of Ba(OH)2 = \(\frac {17 1.0}{2}\) = 85.5

Question 12.
What is a mole?
Answer:
One mole is the amount of substance of a system, which contains as many elementary particles as there are atoms in 12 g of carbon – 12 isotope. The elementary particles can be molecules, atoms, ions, electrons or any other specified particles.

Question 13.
What do you understand by the terms empirical formula and molecular formula?
Answer:
Empirical Formula:

  • It is the simplest formula.
  • It shows the ratio of the number of atoms of different elements in one molecule of the compound.

Molecular Formula:

  • It is the actual formula.
  • It shows the actual number of different types of atoms present in one molecule of the compound.

Question 14.
What is Avogadro’s number?
Answer:
The total number of entities present in one mole of any substance is equal to 6.022 × 1023. This number is called the Avogadro number.

Question 15.
State Avogadro hypothesis.
Answer:
Equal volume of all gases under the same conditions of temperature and pressure contain equal number of molecules.

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 16.
Calculate the mass of sodium (in kg) present in 95 kg of a crude sample of sodium nitrate whose percentage purity is 70%.
Answer:
Sodium Nitrate = NaNO3
Molecular mass of Sodium Nitrate = 23 + 14 + 48 = 85
100% pure 85 g of NaNO3 contains 23 g of Sodium.
100% pure 95 x 103 g of NaNO3 will contains \(\frac {23}{85}\) x 95 x 103
= 25.70 x 103 g of Sodium.
100% pure NaNO3  contains 25.70 x 103 g of Sodium.
∴ 70% pure NaNO3  will contains = 17990 g (or) 17.99 Kg of Na.

Question 17.
Define molar volume.
Answer:
The volume occupied by one mole of any substance in the gaseous state at a given temperature and pressure is called molar volume.

Question 18.
What is gram equivalent mass?
Answer:
Gram equivalent mass of an element, compound or ion is the mass that combines or displaces 1.008 g hydrogen or 8 g oxygen or 35.5 g chlorine.
Gram equivalent mass = (Molar mass(g mol-1)) / Equivalence factor (eq mol-1)

Question 19.
What is meant by Plasma state? Give an example.
Answer:
The gaseous state of matter at a very high temperature containing gaseous ions and free-electron is referred to as the Plasma state. e.g. Lightning.

Question 20.
What is the acidity of a base? Give an example.
Answer:
The acidity of a base is the number of moles of ionizable OH- ions present in 1 mole of the base. The acidity of potassium hydroxide (KOH) is 1.

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 21.
What is empirical formula ola compound?
Answer:
The empirical formula of a compound is the formula written with the simplest ratio of the number of different atoms present in one molecule of the compound as a subscript to the atomic symbol.

Question 22.
Chlorine has a fractional average atomic mass. Justify this statement.
Answer:
Chlorine molecule has two isotopes as in 17Cl35, 17 Cl37 in the ratio of 77 : 23, so when we are calculating the average atomic mass, it becomes fractional.
The average relative atomic mass of Chlorine = \(\frac {(35 x 77) + (37 x 23)}{100}\) = 35.46 amu

Question 23.
What is meant by Stoichiometry?
Answer:
Stoichiometry is the quantitative relationship between reactants and products in a balanced chemical equation in moles. The quantity of reactants and products can be expressed in moles or in terms of mass unit or as volume.

Question 24.
What are limiting and excess reagents?
Answer:
When a reaction is carried out using non-stoichiometric quantifies of the reactants, the product yield will be determined by the reactant that is completely consumed and is called the limiting reagent. It limits the further reaction to take place. The other reagent which is in excess is called the excess reagent.

Question 25.
Define Avogadro Number.
Answer:
Avogadro number is the number of atoms present in one mole of an element or number of molecules present in one mole of a compound. The value of Avogadro number (N) = 6.023 x 1023

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 26.
Define oxidation number.
Answer:
The oxidation number is defined as the imaginary charge left on the atom when all other atoms of the compound have been removed in their usual oxidation states that are assigned according to a set of rules.

Question 27.
Calculate the equivalent mass of Copper. (Atomic mass of copper = 63.5)
Answer:
Equivalent mass = \(\frac {Atomic mass}{Valency}\)
Equivalent mass of Copper = \(\frac {63.5}{2}\) = 31.75 g eq-1.

Question 28.
Mention the types of redox reactions?
Answer:
The types of redox reactions are combination reaction, decomposition reaction, displacement reaction, disproportionate reaction and competitive electron transfer reactions.

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

III. Short Question and Answers (3 Marks):

Question 1.
Describe the chemical classification of matter.
Answer:
Matter can be classified into mixtures and pure substances based on chemical compositions. Mixtures consists of more than one chemical entity present without any chemical interactions. They can be further classified as homogeneous or heterogeneous mixtures based on their physical appearance. Pure substances are composed of simple atoms or molecules. They are further classified as elements and compounds.

Question 2.
Distinguish between element and compound.
Answer:
An element consists of only one type of atom. Element can exists as monoatomic or polyatomic units. The polyatomic elements are called molecules.
Example: Monoatomic unit – Gold (Au), Copper (Cu);
Polyatomic unit: Hydrogen (H2), Phosphorous (P4).

Compounds are made up of molecules which contain two or more atoms of different elements.
Example: Carbon dioxide (CO2), Glucose (C6 H12 O6)

Properties of compounds are different from those of their constituent elements. For example, sodium is a shiny metal, and chlorine is an irritating gas. But the compound formed from these two elements, sodium chloride shows different characteristics as it is crystalline solid, vital for biological functions.

Question 3.
What is average atomic mass? How is average atomic mass of chlorine calculated?
Answer:
Average atomic mass is defined as the average of the atomic masses of all atoms in their naturally occurring isotopes. Chlorine consists of two naturally occurring isotopes 17Cl37 and 17Cl35 in the ratio 77 : 23, the average relative atomic mass of chlorine is
= ((35 × 77) + (37 × 23)) / 100
= 35.46 u

Question 4.
What will be the mass of one 12C atom in g?
Answer:
Molar mass of 12C = 12.00 g mol-1.
∴ Mass of 6.023 x 1023 carbon atom = 12.0 g
∴ Mass of 1 carbon atom = \(\frac{12}{6.023 \times 10^{23}}\) = 1.992 x 10 g.

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 5.
Discuss the role of antacids.
Answer:
Gastric acid is a digestive fluid formed in the stomach and it contains hydrochloric acid. The typical concentration of the acid in gastric acid is 0.082 M. When the concentration exceeds 0.1M it causes heartburn and acidity. Antacids used to treat acidity contain mostly magnesium hydroxide or aluminium hydroxide that neutralizes the excess acid. The chemical reactions are as follows.

3 HCl + Al(OH)3 → AlCl3 + 3H2O
2HCl + Mg(OH)2 → MgCl2 + 2H2O.

From the above reactions, we know that 1 mole of aluminium hydroxide neutralizes 3 moles of HCl while 1 mole of magnesium hydroxide neutralizes 2 moles of HCl.

Question 6.
Explain gram equivalent mass.
Answer:
Gram equivalent mass of an element, compound, or ion is the mass that combines or displaces 1.008 g hydrogen or 8 g oxygen or 35.5 g chlorine.
Consider the following reaction:
Zn + H2 SO4 → ZnSO4 + H2
In this reaction, 1 mole of zinc (65.38 g) displaces one mole of hydrogen molecule (2.016 g). Mass of zinc required to displace 1.008 g hydrogen is
= \(\frac{65.38}{2.016}\) × 1.008
= \(\frac{65.38}{2}\)
= 32.69
The equivalent mass of zinc = 32.69
The gram equivalent mass of zinc = 32.69 g eq-1.
The expression used to calculate gram equivalent mass is
Gram equivalent mass = Molar mass(g mol-1) / Equivalence factor (eq mol-1)

Question 7.
Calculate the gram equivalent mass of sulphuric acid.
Answer:
Basicity of sulphuric acid (H2SO4) = 2eq mol-1
Molar mass of H2SO4 = (2 × 1) +(1 × 32) +(4 × 16) = 56 g mol-1
Gram equivalent mass of H2SO4 = \(\frac{98}{2}\)
= 49 g eq-1

Question 8.
Calculate the gram equivalent mass of potassium hydroxide.
Answer:
Acidity of potassium hydroxide (KOH) = 1 eq mol-1
Molar mass of KOH = (1 × 39) + (1 × 16) + (1 × 1) = 56 g mol-1
Gram equivalent mass of KOH = \(\frac{56}{1}\) = 56 g eq-1.

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 9.
Calculate the gram equivalent mass of Potassium permanganate.
Answer:
Potassium permanganate is an oxidizing agent. Molar mass of
KMnO4 = (1 × 39)+ (1 × 55)+ (4 × 16)
= 158 g mol-1

In an acidic medium, permanganate is reduced during oxidation and is given by the following equation,
MnO4 + 8H+ + 5e → Mn2 + 4H2O
Therefore, n = 5.
Gram equivalent mass of KMnO4 = \(\frac{158}{5}\) = 31.6 g mol-1

Question 10.
An acid found in tamarind on analysis shows the following percentage composition: 32 % Carbon; 4 % Hydrogen; 64 % Oxygen. Find the empirical formula of the compound.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 16

The ratio of C : H : O is 2 : 3 : 3 and hence, the empirical formula of the compound is CH2O

Question 11.
An organic compound present in vinegar has 40 % Carbon, 6.6 % Hydrogen: and 53.4 % Oxygen. Find the empirical formula of the compound.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 17

The ratio of C: H : O is 1 : 2 : 1 and hence, the empirical formula of the compound is CH2O.

Question 12.
How much copper can be obtained from 100 g of anhydrous copper sulphate?
Answer:
Anhydrous copper sulphate = CuSO4
Molecular mass of CuSO4 = 63.5 + 32 + (16 x 4)
= 63.5 + 32 + 64
= 159.5 g
159.5 g of CuSO4 contains 63.5 g of copper.
∴ 100 g of CuSO4 contains \(\frac{6.35}{159.5}\) x 100 = 0.39811 x 100 = 39.81 g of Copper.

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 13.
Explain the classical concept of oxidation and reduction.
Answer:
According to classical concept, the addition of oxygen or removal of hydrogen is called oxidation.
Consider the following reactions,
4Fe + 3O2 → 2Fe2O3
H2S +Cl2 → 2 HCl + S

In the first reaction, which is responsible for the rusting of iron, the oxygen adds on to the metal, iron. In the second reaction, hydrogen is removed from Hydrogen sulphide.
According to classical concept, addition of hydrogen or removal of oxygen is called reduction.
Consider the following reactions,
CuO + C → Cu + CO
S + H2 → H2S
In the first reaction, oxygen is removed from cupric oxide and in the second reaction, hydrogen is added to sulphur.

Question 14.
Describe the electron concept of oxidation and reduction.
Answer:
The reaction involving loss of electron is termed as oxidation and gain of electron is termed as reduction.
Fe2+ → Fe3+ + e (loss of electron – oxidation)
Cu2+ + 2e → Cu ( gain of electron – reduction)

Question 15.
Describe the oxidation number concept of oxidation and reduction.
Answer:
During redox reactions, the oxidation number of elements changes. A reaction in which oxidation number of the element increases is called oxidation whereas the oxidation number of the element 3 decreases is called reduction.
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 18
In this reaction, manganese in potassium permanganate favours the oxidation of ferrous sulphate into ferric sulphate by gets reduced.

Question 16.
Write notes on displacement reaction.
Answer:
Redox reactions in which an ion or an atom in a compound is replaced by an ion or an atom of another element are called displacement reactions. They are further classified into
(i) metal displacement reactions
(ii) non-metal displacement reactions.

(i) Metal displacement reactions:
Place a zinc metal strip in an aqueous copper sulfate solution taken in a beaker. The intensity of blue colour of the solution slowly reduced and finally disappeared. The zinc metal strip became coated with brownish metallic copper. This is due to the following metal displacement reaction.
CuSO4(aq)+ Zn(s) → Cu + ZnSO4.

(ii) Non-metal displacement reaction:
Zn + 2HCl → ZnCl2 + H2

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

IV. Long Question and Answers:

Question 1.
What is a matter? Explain its classification.
Answer:
Matter is defined as anything that has mass and occupies space. All matter is composed of atoms.

Physical Classification:
Matter can be classified as solids, liquids and gases based on their physical state. The physical state of matter can be converted into one another by modifying the temperature and pressure suitably.

Chemical Classification:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 19
Classification of Matter

Matter can be classified into mixtures and pure substances based on chemical compositions. Mixtures consists of more than one chemical entity present without any chemical interactions. They can be further classified as homogeneous or heterogeneous mixtures based on their physical appearance. Pure substances are composed of simple atoms or molecules. They are further classified as elements and compounds.

An element consists of only one type of atom. Element can exists as monoatomic or polyatomic units. The polyatomic elements are called molecules.
Example: Monoatomic unit – Gold (Au), Copper (Cu);
Polyatomic unit: Hydrogen (H2), Phosphorous (P4).

Compounds are made up of molecules which contain two or more atoms of different elements.
Example: Carbon dioxide (CO2), Glucose (C6H12O6)
Properties of compounds are different from those of their constituent elements. For example, sodium is a shiny metal, and chlorine is an irritating gas. But the compound formed from these two elements, sodium chloride shows different characteristics as it is a crystalline solid, vital for biological function.

Question 2.
How is empirical formula of a compound determined from the elemental analysis?
Answer:
The empirical formula of a compound determined from the elemental analysis by the following steps.
(i) Since the composition is expressed in percentage, we can consider the total mass of the compound as 100 g and the percentage values of individual elements as a mass in grams.
(ii) Divide the mass of each element by its atomic mass. This gives the relative number of moles of various elements in the compound.
(iii) Divide the value of a relative number of moles obtained in the step – (ii) by the smallest number of them to get the simplest ratio.
(iv) In case the simplest ratios obtained in step – (iii) are not whole numbers then they may be converted into the whole numbers by multiplying a suitable smallest number.

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 3.
An organic compound present in vinegar has 40% carbon, 6.6 % hydrogen, and 53.4 % oxygen, Find the empirical formula and molecular formula of the compound. (Given, Molar mass: 60 g mol-1).
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 20

The ratio of C: H : O is 1 : 2 : 1 and hence, the empirical formula of the compound is CH2O.
Empirical Formula mass = (1 × 12 + 1 × 2 + 1 × 16) = 12 + 2 + 16 = 30.
Whole number η =  Molar mass / Empirical formula mas = \(\frac{60}{30}\) = 2
Therefore, Molecular formula = (CH2O)2 = C2H4O2

Question 4.
An organic compound present in vinegar has 40% carbon, 6.6 % hydrogen, and 53.4 % oxygen, Find the empirical formula and molecular formula of the compound. (Given, Molar mass: 90 g mol-1).
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 21

The ratio of C : H : O is 1 : 2 : 1 and hence, the empirical formula of the compound is CH2O.
Empirical Formula mass = (1 × 12 + 1 × 2 × 16) = 12 + 2 + 16 = 30.
Whole number, n = \(\frac{\text { Molar mass }}{\text { Empirical formula mass }}\)
= \(\frac{90}{30}\) = 3
Therefore, Molecular formula = (CH2O)3 = C3H6O3

Question 5.
What is a redox reaction? Explain the different concepts of redox reaction.
Answer:
The reaction involving loss of electron is oxidation and gain of electron is reduction. Both these reactions take place simultaneously and are called as redox reactions.

Classical concept of oxidation and reduction:
According to classical concept, addition of oxygen or removal of hydrogen is called oxidation.
Consider the following reactions,

4Fe + 3O2 → 2Fe2O3
H2S + Cl2 → 2HCl + S

In the first reaction, which is responsible for the rusting of iron, the oxygen adds on to the metal, iron. In the second reaction, hydrogen is removed from Hydrogen sulphide.
According to classical concept, addition of hydrogen or removal of oxygen is called reduction.
Consider the following reactions,
CuO + C → Cu + CO
S + H2 → H2S

In the first reaction, oxygen is removed from cupric oxide, and in the second reaction, hydrogen is added to sulphur.

Electron concept of oxidation and reduction.
The reaction involving loss of electron is termed as oxidation and gain of an electron is termed as reduction.
Fe2+ → Fe3+ + e (loss of electron – oxidation)
Cu2+ + 2e → Cu (gain of electron – reduction)

Oxidation number concept of oxidation and reduction:
During redox reactions, the oxidation number of elements changes. A reaction in which oxidation number of the element increases is called oxidation whereas the oxidation number of the element decreases is called reduction.
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 22
In this reaction, manganese in potassium permanganate favours the oxidation of ferrous sulphate into ferric sulphate by gets reduced.

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 6.
What is an oxidation number? State the rules to find the oxidation number.
Answer:
Oxidation number is defined as the imaginary charge left on the atom when all other atoms of the compound have been removed in their usual oxidation states that are assigned according to set of rules. A term that is often used interchangeably with oxidation number is oxidation state.

  1. The oxidation state of a free element (i.e., uncombined state) is zero.
    Example: H2 Cl2, Na, and S8 have the oxidation number of zero.
  2. For a monoatomic ion, the oxidation state is equal to the net charge on the ion.
    Example: The oxidation number of sodium in Na+ is +1.
    The oxidation number of chlorine in Cl is -1.
  3. The algebraic sum of oxidation states of all atoms in a molecule is equal to zero, while in ions, it is equal to the net charge on the ion.
    Example: In H2SO4, 2 × (oxidation number of hydrogen) + 1 × (oxidation number of sulphur) + 4 × (oxidation number of oxygen) = 0.
  4. Hydrogen has an oxidation number of +1 in all its compounds except in metal hydrides where it has -1 value.
    Example: Oxidation number of hydrogen in hydrogen chloride (HCl) is + 1.
    Oxidation number of hydrogen in sodium hydride (NaH) is -1.
  5. Fluorine has an oxidation state of -1 in all its compounds.
  6. The oxidation state of oxygen in most compounds is -2. Exceptions are peroxides, superoxides, and compounds with fluorine.
    Example: Oxidation number of oxygen
    (i) in water is -2,
    (ii) in hydrogen peroxide is -1,
    (iii) in superoxides such as KO2 is 4, and
    (iv) in oxygen difluoride (OF2) is +2.
  7. Alkali metals have an oxidation state of +1 and alkaline earth metals have an oxidation state of +2 in all their compounds.

Question 7.
Balance the following chemical equation by oxidation number method.
KMnO4 + FeSO4 + H2SO4 → K2SO4 + Fe(SO4)3 + 8H2O
Answer:
Using oxidation number concept, the reactants which undergoes oxidation and reduction are as follows:
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 23

The oxidation number of Mn in KMnO4 changes from +7 to +2 by gaining five electrons and the oxidation number of Fe FeSO4 changes from +2 to +3 by loosing one electron.
Since the total number of electrons lost is equal to the total number of electrons gained, the number of electrons, by cross multiplication of the respective formula with suitable integers on reactant side.

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 24

Based on the reactant side, the products are balanced.

2KMnO4 + 10 FeSO4 + H2SO4 → K2SO4 + MnSO4 + Fe( SO4)3 + H2O

Balance the other elements except H and O atoms. K and S are balanced as follows

2KMnO4 + 10 FeSO4 + H2SO4 → K2SO4 + MnSO4 + Fe( SO4)3 + H2O
The difference of 8 – S atoms in reactant side, has to be balanced by multiplying H2SO4 by ‘8’. The equation now becomes,

2KMnO4 + 10 FeSO4 + 8 H2SO4 → K2SO4 + MnSO4 + Fe( SO4)3 + H2O

‘H’ and ‘O’ atoms are balanced by multiplying H2O molecules in the product side by ‘8’.

2KMnO4 + 10 FeSO4 + 8 H2SO4 → K2SO4 + 2 MnSO4 + 5 Fe( SO4)3 + 8 H2O
The above equation is a balanced equation.

Question 8.
How is the following equation ¡s balanced by Ion electron method?
MnO4 + Fe2+ + H+ → Mn2+ + Fe3+ + H2O.
Answer:
Using the oxidation number concept, the reactants which undergoes oxidation and reduction are as follows;
Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations 25
The two half reactions are,
Fe2+ → Fe3+ + 1e ……….(1)
MnO4 + 5e + 8H+ → Mn2+ + 4H2O ……….(2)
Balancing the atoms and charges on both sides of the half reactions.
There is no change in the equation (1) whereas in equation (2), there are four ‘O’ atoms on the reractant side .
Therefore, four H2 is added on the product side, to balance ‘H’ – add, 8 H+ in the reactant side.
MnO4 + 5e + 8H+ → Mn2+ + 4H2O ………..(3)
The two half reactions are equated in such a way that the number of electrons lost is equal to number of electrons gained.
Adding the two half reactions as follows:
(1) × 5 5 Fe2+ → 5 Fe3+ + 5e ……………(4)
(3) × 1 MnO4 + 5e + 8H+ → Mn2+ + 4H2O ………(5)

(4) + (5)
MnO4 + 5Fe2+ + 8H+ → Mn2+ + 5 Fe3+ + 4H2O ………..(6)
The equation (6) is a balanced equation.

Samacheer Kalvi 11th Chemistry Guide Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 9.
Calculate the percentage composition of the elements present in lead nitrate. How many Kg of 02 can be obtained from 50 kg of 70% pure lead nitrate?
Answer:
Lead nitrate = Pb (NO3)2
Molecular mass of lead nitrate = 207 + (14 x 2) + (16 x 6)
= 207 + 28 + 96 = 331 g / mol.
331 g of lead nitrate contains 96 g of oxygen.
∴ 50 x 103 g of lead nitrate will contain \(\frac {96}{331}\) x 50 x 103
= 14501.5 g
= 14.501 Kg of oxygen.
100 % pure lead nitrate contains 14.501 Kg of oxygen.
70 % pure lead nitrate will contain = \(\frac {14.501}{100}\) x 70 = 10.15 Kg of oxygen.
.’. 70 % pure lead nitrate will contain 10.15 Kg of oxygen.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 6 Gaseous State Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

11th Chemistry Guide Gaseous State Text Book Back Questions and Answers

Textual Questions:

I. Choose the best answer:

Question 1.
Gases deviate from ideal behavior at high pressure. Which of the following statement(s) is correct for non-ideality?
(a) at high pressure the collision between the gas molecule become enormous
(b) at high pressure the gas molecules move only in one direction
(c) at high pressure, the volume of gas become insignificant
(d) at high pressure the intermolecular interactions become significant
Answer:
(d) at high pressure the intermolecular interactions become significant

Question 2.
Rate of diffusion of a gas is
(a) directly proportional to its density
(b) directly proportional to its molecular weight
(c) directly proportional to its square root of its molecular weight
(d) inversely proportional to the square root of its molecular weight
Answer:
(d) inversely proportional to the square root of its molecular weight

Question 3.
Which of the following is the correct expression for the equation of state of van der Waals gas?
(a) [P + \(\frac{a}{n^{2} V^{2}}\)](V – nb) = nRT

(b) [P + \(\frac{n a}{n^{2} V^{2}}\)](V – nb) = nRT

(c) [P + \(\frac{a n^{2}}{V^{2}}\)](V – nb) = nRT

(d) [P + \(\frac{n^{2} a^{2}}{V^{2}}\)(V – nb) = nRT]
Answer:
(c) [P + \(\frac{a n^{2}}{V^{2}}\)](V – nb) = nRT

Question 4.
When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules
(a) are above inversion temperature
(b) exert no attractive forces on each other
(c) do work equal to the loss in kinetic energy
(d) collide without loss of energy
Answer:
(b) exert no attractive forces on each other

Question 5.
Equal weights of methane and oxygen are mixed in an empty container at 298 K. The fraction of total pressure exerted by oxygen is
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{3}\) × 273 × 298
Answer:
(a) \(\frac{1}{3}\)

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 6.
The temperatures at which real gases obey the ideal gas laws over a wide range of pressure is called
(a) Critical temperature
(b) Boyle temperature
(c) Inversion temperature
(d) Reduced temperature
Answer:
(b) Boyle temperature

Question 7.
In a closed room of 1000 m3 a perfume bottle is opened up. The room develops a smell. This is due to which property of gases?
(a) Viscosity
(b) Density
(c) Diffusion
(d) None
Answer:
(c) Diffusion

Question 8.
A bottle of ammonia and a bottle of HCl connected through a long tube are opened simultaneously at both ends. The white ammonium chloride ring first formed will be
(a) At the center of the tube
(b) Near the hydrogen chloride bottle
(c) Near the ammonia bottle
(d) Throughout the length of the tube
Answer:
(b) Near the hydrogen chloride bottle

Question 9.
The value of the universal gas constant depends upon
(a) Temperature of the gas
(b) Volume of the gas
(c) Number of moles of the gas
(d) units of pressure and volume
Answer:
(d) units of pressure and volume

Question 10.
The value of the gas constant R is
(a) 0.082 dm3atm
(b) 0.987 cal mol-1K-1
(c) 8.3J mol-1K-1
(d) 8erg mol-1K-1
Answer:
(c) 8.3J mol-1K-1

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 11.
The use of hot air balloon in sports at meteorological observation is an application of
(a) Boyle’s law
(b) Newton’s law
(c) Kelvin’s law
(d) Brown’s law
Answer:
(a) Boyle’s law

Question 12.
The table indicates the value of vanderWaals constant ‘a’ in (dm3)2 atm. mol-2. The gas which can be most easily liquefied is

Gas O2 N2 NH3 CH4
A 1.360 1.390 4.170 2.253

(a) O2
(b) N2
(c) NH3
(d) CH4
Answer:
(c) NH3

Question 13.
Consider the following statements
(i) Atmospheric pressure is less at the top of a mountain than at sea level
(ii) Gases are much more compressible than solids or liquids
(iii) When the atmospheric pressure increases the height of the mercury column rises
Select the correct statement
(a) I and II
(b) II and III
(c) I and III
(d) I, II, and III
Answer:
(b) II and III

Question 14.
The compressibility factor for CO2 at 400 K and 71.0 bar is 0.8697. The molar volume of CO2 under these conditions is
(a) 22.04 dm3
(b) 2.24 dm3
(c) 0.41 dm3
(d) 19.5 dm3
Answer:
(c) 0.41 dm3

Question 15.
If the temperature and volume of an ideal gas is increased to twice its values the initial pressure P becomes
(a) 4P
(b) 2P
(c) P
(d) 3P
Answer:
(b) 2P

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 16.
At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3 times that of a hydrocarbon having molecular formula CnH2n – 2. What is the value of n?
(a) 8
(b) 4
(c) 3
(d) 1
Answer:
(d) 1

Question 17.
Equal moles of hydrogen and oxygen gases are placed in a container, with a pin-hole through which both can escape what fraction of oxygen escapes in the time required for one-half of the hydrogen to escape.
(a) \(\frac{3}{8}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{8}\)
(d) \(\frac{1}{4}\)
Answer:
(c) \(\frac{1}{8}\)

Question 18.
The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion i.e., α = 1\(\left[\frac{\partial V}{\delta T}\right]\)Vp. For an ideal gas, α is equal to
(a) T
(b) 1/T
(c) P
(d) none of these
Answer:
(a) T

Question 19.
Four gases P, Q, R, and S have almost the same values of ‘b’ but their a’ values (a. h are Vander Waals Constants) are in the order Q < R < S < p. At a particular temperature, among the four gases, the most easily liquelìable one is
(a) P
(b) Q
(c) R
(d) S
Answer:
(c) R

Question 20.
Maximum deviation from ideal gas is expected
(a) CH4(g)
(b) NH3(g)
(c) H2 (g)
(d) N2 (g)
Answer:
(b) NH3(g)

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 21.
The units of Vander Waals constants ‘b’ and ‘a’ respectively
(a) mol L-1 and L atm2 mol-1
(b) mol L and L atm mol2
(c) mol-1 L and L2 atm mol-2
(d) none of these
Answer:
(c) mol-1 L and L2 atm mol-2

Question 22.
Assertion:
Critical temperature of CO2 is 304 K. it can be liquefied above 304 K.
Reason:
For a given mass of gas, volume is to directly proportional to pressure at constant temperature
(a) both assertion and reason are true and reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(d) both assertion and reason are false

Question 23.
What is the density of N2 gas at 227°C and 5.00 atm pressure? (R = 0.082 L atm K-1 mol-1)
(a) 1.40 g/L
(b) 2.81 g/L
(c) 3.41 g/L
(d) 0.29 g/L
Answer:
(c) 3.41 g/L

Question 24.
Which of the following diagrams correctly describes the behaviour of a fixed mass of an ideal gas? (T is measured in K)
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 1
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 2

Question 25.
25g of each of the following gases are taken at 27°C and 600 mm Hg pressure. Which of these will have the least volume?
(a) HBr
(b) HCl
(c) HF
(d) HI
Answer:
(d) HI

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

II. Answer these questions briefly:

Question 26.
State Boyle’s law.
Answer:
Boyle’s law states that at a given temperature. the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
V α \(\frac {1}{P}\);
where T and n are fixed or PV = Constant = k

Question 27.
A balloon filled with air at room temperature and cooled to a much lower temperature can be used as a model for Charle’s law.
Answer:
Yes, a balloon filled with air at room temperature and cooled to a much lower temperature can be used as a model for Charle’s law. The volume of balloon decreases when the temperature reduced from room temperature to low temperature. When cooled, the kinetic energy of the gas molecules decreases, so that the volume of the balloon also decreases.

Question 28.
Name two items that can serve as a model for ‘Gay Lusaac’ law and explain.
Answer:
Firing a bullet:
When gunpowder burns, it creates; a significant amount of superheated gas. The high pressure of the hot gas behind the bullet forces it out, of the barrel of the gun.
Heating food in an oven:
When you keep food in an oven for heating, the air inside the oven is heated, thus pressurized.

Question 29.
Give the mathematical expression that relates gas volume and moles. Describe in words what j the mathematical expression means.
Answer:

  1. The mathematical relationship betwêen the volume of a gas and the number of moles is V α n
  2. \(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant
    Where V1 and n1 are the volume and number of moles of a gas and V2 and n2 are the values of volume and number of moles of same gas at a different set of conditions.
  3. If the volume of the gas increase then the number of moles of the gas also increases.
  4. At a certain temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 30.
What are ideal gases? In what way real gases differ from ideal gases?
Answer:
The kinetic theory of gases which is the basis for the gas equation (PV = nRT), assumes that the individual gas molecules occupy negligible volume when compared to the total volume of the gas and there is no attractive force between the gas molecules. Gases whose behaviour is consistent with these assumptions under all conditions are called ideal gases.

But in practice both these assumptions are not valid under all conditions. For example, the fact that gases can be liquefied shows that the attractive force exists among molecules. Hence, there is no gas which behaves ideally under all conditions. The non-ideal gases are called real gases. The real gases f tend to approach the ideal behaviour under certain conditions.

Question 31.
Can a Vander Waals gas with a = 0 be liquefied? Explain.
Answer:

  • a = 0 for a Van der Waals gas i.e. for a real gas. Van der Waals constant a = 0. It cannot be liquefied.
  • If a = 0, there is a very less interaction between the molecules of gas.
  • ‘a’ is the measure of strength of Van der Waals force of attraction between the molecules of the gas.
  • If a is equal to zero, the Van der Waals force of attraction is very less and the gas cannot be liquefied.

Question 32.
Suppose there is a tiny sticky area on the wall of a container of gas. Molecules hitting this area stick there permanently. Is the pressure greater or less than on the ordinary area of walls?
Answer:
Gaseous pressure is developed by the continuous bombardment of the molecules of the gas among themselves and also with the walls of the container. When the molecules hit the sticky area of the container, the number of molecules decreases and hence, the pressure decreases. Therefore, pressure is less than the ordinary area of walls.

Question 33.
Explain the following observations
(a) Aerated water bottles are kept underwater during summer
(b) Liquid ammonia bottle is cooled before opening the seal
(c) The tyre of an automobile is inflated to slightly lesser pressure in summer than in winter
(d) The size of a weather balloon becomes larger and larger as it ascends up to a larger altitude.
Answer:
(a) In aerated water bottles, CO2 gas is passed through the aqueous solution under pressure because the solubility of the gas in water is not very high. In summer, the solubility of the gas in water is likely to decrease because of the rise in temperature. Thus, in summer, more gas will be present above the liquid surface in the glass bottle.

In case, the pressure of the gas becomes too high, the glass will not be able to withstand the pressure and the bottle may explode. To avoid this, the bottles are kept underwater. As a result, the temperature is likely to decrease and the solubility of CO2 is likely to increase in aqueous solution resulting in decreased pressure.

(b) Liquid ammonia bottle contains the gas under very high pressure. If the bottle is opened as such, then the sudden decrease in pressure will lead to a large increase in the volume of the gas. As a result, the gas will come out of the bottle all of a sudden with force. This will lead to the breakage of the bottle and also causes an accident.

However, if the bottle is cooled under tap water for some time, there will be a decrease in the volume of a gas to a large extent. if the seal is opened now, the gas will come out of the bottle at a slower rate, reduces the chances of an accident.

(c) The pressure of air is directly proportional to the temperature. Since the temperature is higher in summer than in higher, the pressure of the air in the tube of the lyre is likely to be quite high as compared to winter. It is quite likely that the tube may burst under high pressure in summer, Therefore, it is advisable to inflate the types to lesser pressure in summer than in winter.

(d) The volume of the gas is inversely proportional to the pressure at a given temperature according to Boyle’s law. As the weather balloon ascends, the pressure tends to decrease. As a result, the volume of the gas inside the balloon or the size of the balloon is likely to increase.

Question 34.
Give a suitable explanation for the following facts about gases.
(a) Gases don’t settle at the bottom of a container.
Answer:
According to kinetic theory, gas molecules are moving continuously at random. Hence, they do not settle at the bottom of a container.

(b) Gases diffuse through all the space available to them.
Answer:
Gases have a tendency to occupy all the available space. The gas molecules migrate from a region of higher concentration to a region of lower concentration. Hence, gases diffuse through all the space available to them.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 35.
Suggest why there is no hydrogen in our atmosphere. Why does the moon have no atmosphere?
Answer:
1. Hydrogen is the lightest element thus when produced in a free state, it rises above all the other gases to the top of the atmosphere, where it is open to cosmic storms and solar flares. There is literally leaks from the atmosphere to the empty space. Hydrogen easily gains the velocity required to escape Earth’s magnetic field. Hydrogen is very reactive in nature. So it would have reacted with O2, in its way to produce H2O. So majority portion of H2 reacts and a very less amount of it present in the upper level of the atmosphere and gains velocity to escape the atmosphere.

2. Moon has no atmosphere because the value of acceleration due to gravity ‘g’ on the surface of the moon is small. Therefore, the value of escape velocity on the surface of the moon is very small. The molecule of the atmospheric gases on the surface of the moon has thermal velocities greater than the escape velocity. That’s why all the molecules of gases have escaped and there is no atmosphere on the moon. The moon has insufficient gravity to retain an atmosphere. So we conclude that the moon has no atmosphere.

Question 36.
Explain whether a gas approaches ideal behavior or deviates from ideal behaviour if
(a) it is compressed to a smaller volume at f constant pressure.
Answer:
When the gas is compressed to a smaller volume, the compressibility factor (Z) decreases. Hence, the gas deviates from ideal behavior

(b) the temperature is raised while keeping the volume constant.
Answer:
When the temperature is increased, the compressibility factor approaches unity. Hence, the gas behaves ideally.

(c) More gas is introduced into the same volume and at the same temperature.
Answer:
When more gas is introduced into a container of the same volume and at the same temperature, the compressibility factor tend to unity. Hence, the gas behaves ideally.

Question 37.
Which of the following gases would you expect to deviate from ideal behavior under conditions of low-temperature F2, Cl2, or Br2? Explain.
Answer:
1. Bromine deviates (Br2) from the ideal gas maximum than Cl2 and F2. Because Br2 has the biggest size (atomic weight 79.9) provides maximum attraction between bromine molecules which is directly proportional to the size of the molecule and the boiling point of the liquid made from those molecules.

2. Br2 deviates from ideal behaviour because it has the largest atomic radii compared to Cl2 and F2. So it contains more electrons than the other two, and the Vander Waals forces are stronger in Br2 than in Cl2 and F2. So Br2 deviates from ideal behaviour.

Question 38.
Distinguish between diffusion and effusion.
Answer:

Diffusion Effusion
1. The property of gas which involves the movement of the gas molecules through another gas is called diffusion. It is the property in which a gas escapes from a container through a very small hole.
2. It is the ability of gases to mix with each other It is the ability of gas to travel through a small hole.
3. The rate of diffusion of a gas depends on how fast the gas molecules are moving The rate that this happens depends on how many gas molecules “collide” with the pore.
4. e.g., Smell of perfume diffuses into the air e.g., Air escaping slowly through the pinhole in a tire.

Question 39.
Aerosol cans carry a clear warning of heating of the can. Why?
Answer:
If aerosol cans are heated, then they will produce more vapour inside the can, which will make the pressure rise very quickly. The rise in temperature can double the pressure inside. Even though the cans are tested, they will burst if the pressure goes up too far. A bursting can could be dangerous.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 40.
When the driver of an automobile applies the brake, the passengers are pushed toward the front of the car but a helium balloon is pushed toward the back of the car. Upon forward acceleration, the passengers are pushed toward the front of the car. Why?
Answer:
1. When the driver of an automobile applies the brake, the passengers are pushed toward the front of the car due to the inertia of the body, but a helium balloon pushed toward the back of the car. A helium balloon responds to the air around it. Helium molecules are lighter than the air of our atmosphere, and so they move toward the back by gravity as a result of the accelerating frame.

2. Upon forwarding acceleration, the passenger’s arc pushed toward the front of the car, because the body in motion tends to stay in motion until acted upon by an outside force. A helium balloon is going to move opposite to this pseudo gravitational force.

Question 41.
Would it be easier to drink water with a straw on the top of Mount Everest?
Answer:
It would be harder on the top of the mountain; because the external pressure pushing on the liquid to force it up the straw is less.

Question 42.
Write the Van der Waals equation for a real gas. Explain the correction term for pressure and Volume.
Answer:
Vander Waals equation for a real gas is given by
(P + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT
Pressure Correction:
The pressure of a gas is directly proportional to the force created by the bombardment of molecules on the walls of the container. The speed of a molecule moving towards the wall of the container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is lower than the ideal pressure of the gas. Hence, van der Waals introduced a correction term to this effect.
Where n is the number of moles of gas and V is the volume of the container
⇒ P ∝ \(\frac{n^{2}}{V^{2}}\)
⇒ P = \(\frac{a n^{2}}{V^{2}}\)
Where a is proportionality constant and depends on the nature of gas
Therefore, P = P + \(\frac{a n^{2}}{V^{2}}\)

Volume Correction:
As every individual molecule of a gas occupies a certain volume, the actual volume is less than the volume of the container, V. Van der Waals introduced a correction factor V to this effect. Let us calculate the correction term by considering gas molecules as spheres.
V = excluded volume
Excluded volume for two molecules = \(\frac{4}{3}\) π(2r)3 = 8Vm
where Vm is a volume of a single molecule.
Excluded volume for single molecule = \(\frac{8 V_{m}}{2}\) = 4Vm
Excluded volume for n molecule
= n(4Vm) = nb
Where b is van der Waals constant which is equal to 4Vm
⇒ V’ = nb
Videal = V – nb
Replacing the corrected pressure and volume in the ideal gas equation PV = nRT we get the van der Waals equation of state for real gases as below,
(P + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT
The constants a and b are van der Waals constants and their values vary with the nature of the gas.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 43.
Derive the values of van der Waals equation constants in terms of critical constants.
Answer:
The Van der walls equation for n moles is
(p + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT ……… (1)
For 1 mole
(P + \(\frac{n^{2}}{V^{2}}\)) (V – b) = RT ……………(2)

From the equation we can derive the values of critical constants Pc, Vc, and Tc in terms of a and b, the van der Waals constants, On expanding the above equation,
PV + \(\frac{a}{V}\) – Pb – \(\frac{a b}{V^{2}}\) – RT = 0 ………….(3)

Multiply equation (3) by \(\frac{V^{2}}{P}\)
\(\frac{V^{2}}{P}\) (PV +\(\frac{a}{V}\) – pb – \(\frac{a b}{V^{2}}\) – RT) = 0

V3 + \(\frac{a}{P}\)V – bV2 – \(\frac{a b}{P}\) – \(\frac{R T V^{2}}{P}\) = 0 ………..(4)
When the above equation is rearranged in powers of V

V3 – [\(\frac{R T}{P}\) + b]V2 + \(\frac{a}{P}\)V – \(\frac{a b}{P}\) = 0 ……………..(5)

The equation (5) is a cubic equation in V. On solving this equation,
we will get three solutions. At the critical point all these three solutions of V are equal to the critical volume Vc. The pressure and temperature becomes Pc and Tc respectively
V = Vc
V – Vc = 0
(V – Vc)3 = 0
V3 – 3VcV2 + 3Vc2V – Vc3 = 0 …………..(6)
As equation (5) is identical with equation (6), we can equate the coefficients of V2, V and constant terms in (5) and (6).
-3VcV2 = –[\(\frac{R T_{c}}{P_{c}}\) + b] V2

3Vc = \(\frac{R T_{c}}{P_{c}}\) + b ………………(7)
3Vc2 = \(\frac{a}{P_{c}}\) ……………(8)
Vc3 = \(\frac{ab}{P_{c}}\) …………….(9)

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 3
The critical constants can be calculated using the values of van der waals constant of a gas and vice versa.
a = 3Vc2 Pc and
b = \(\frac{V_{c}}{3}\)

Question 44.
Why do astronauts have to wear protective suits when they are on the surface of the moon?
Answer:
In space, there is no pressure, if we do wear a pressurized suit, our body will die. In space, we have to wear a pressurized suit, otherwise, our body will continue to push out and blow up like a balloon. It would look cool, but we will be dead. So the astronauts in space must wear a pressurized suit (protective suits).

Question 45.
When ammonia combines with HCl, NH4Cl is formed as white dense fumes. Why do more fumes appear near HCl?
Answer:
HCl and NH4Cl molecules diffuse through the air towards each other. When they meet, they reactto
form a white powder called ammonium chloride, NH4Cl.
HCl(g) + NH3(g) ⇌ NHC1

Hydrogen chloride + ammonia ⇌ ammonium chloride.
The ring of white powder is closer to the HCl than
the NH3. This is because the NH3 molecules are lighter (smaller) and have diffùsed more quickly through the air in the tube.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 46.
A sample of gas at 15°C at 1 atm. has a volume of 2.58 dm3. When the temperature is raised to 38°C at 1 atm does the volume of the gas Increase? If so, calculate the final volume.
Answer:
T1 = 15°C + 273;
T2 = 38 + 273
T1 = 228;
T2 = 311K
V1 = 2.58dm3;
V2 = ?
(P = 1 atom constant)
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)

V2 = \(\left[\frac{V_{1}}{T_{1}}\right]\) × T2

= \(\frac{2.58 d m^{3}}{288 K}\) × 311K
V2 = 2.78 dm3 i.e, volume increased from 2.58 dm3 to 2.78 dm3

Question 47.
Of two samples of nitrogen gas, sample A contains 1.5 moles of nitrogen In a vessel of volume of 37.6 dm3 at 298K, and sample B is in a vessel of volume 16.5 dm3 at 298K. Calculate the number of moles in sample B.
Answer:
nA = 1.5mol; nB = ?
VA = 37.6 dm3; VB = 16.5 dm3
(T = 298K constant)
\(\frac{V_{A}}{n_{A}}=\frac{V_{B}}{n_{B}}\)

nA = \(\left[\frac{n_{A}}{n_{B}}\right] V_{B}\)

Question 48.
Sulphur hexafluoride Is a colourless, odourless gas; calculate the pressure exerted by 1.82 moles of the gas In a steel vessel of volume 5.43 dm3 at 69.5°C, assumIng Ideal gas behaviour.
Answer:
n = 1.82 mole
V = 5.43 dm3
T = 69.5 + 273 = 342.5
P =?
PV = nRT
P = nRT/V
P = Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 4
P = 94.25 atm

Question 49.
Argon is an Inert gas used In light bulbs to retard the vaporization of the tungsten filament. A certain light bulb containing argon at 1.2 atm and 18°C Is heated to 85°C at constant volume. Calculate its final pressure in atm.
Answer:
P1 = 1.2 atm
T1 = 180°C + 273 = 291 K
T2 = 850°C + 273 = 358 K
P2 =?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

P2 = \(\left[\frac{P_{1}}{T_{1}}\right] \times T_{2}\)

P2 = \(\frac{1.2 a t m}{291 K}\) × 358 K
P2 = 1.48 atm

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 50.
A small bubble rises from the bottom of a lake where the temperature and pressure are 6°C and 4 atm. to the water surface, where the temperature is 25°C and pressure Is I arm. Calculate the final volume in (mL) of the bubble, If its initial volume 1.5 mL.
Answer:
T1 = 6°C + 273 = 279 K
P1 = 4 atm; V2 = 1.5 ml
T2 = 25°C + 273 = 298 K
P2 = 1 atm; V2 =?
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)

= \(\frac{4 a t m \times 1.5 m l \times 298 K}{279 K \times 1 a t m}\) = 6.41 mol

Question 51.
Hydrochloric acid Is treated with a metal to produce hydrogen gas. Suppose a student carries out this reaction and collects a volume of 154.4 × 10-3 dm3 of a gas at a pressure of 742 mm of Hg at a temperature of 298 K. What mass of hydrogen gas (in mg) did the student collect?
Answer:
V = 154.4 × 10-3 dm3
P = 742 mm of Hg
T = 298K;
m =?
m = \(\frac{P V}{R T}\)

= \(\frac{742 m m H g \times 154.4 \times 10^{-3} L}{62 m m H g L K^{-1} m o l^{-1} \times 298 K}\)

n = \(\frac{\text { Mass }}{\text { Molar Mass }}\)

Mass = n × Molar Mass
= 0.0006 × 2.016
= 0.0121 g = 12.1 mg

Question 52.
It takes 192 sec for an unknown gas to diffuse through a porous wall and 84 sec for N2 gas to effuse at the same temperature and pressure. What Is the molar mass of the unknown gas?
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 5

A gas’s partial pressure formula is the gas pressure exerted if that gas were alone.

Question 53.
A tank contains a mixture of 52.5 g of oxygen and 65.1 g of CO2 at 300 K the total pressure In the tanks Is 9.21 atm. calculate the partial pressure (in atm.) of each gas in the mixture.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 6
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 7

Question 54.
Combustible gas is stored in a metal tank at a pressure of 2.98 atm at 25°C. The tank can withstand a maximum pressure of 12 atm after which It will explode. The building in which the tank has been stored catches fire. Now predict whether the tank will blow up first or start melting? (Melting point of the metal 1100 K).
Answer:
Pressure of the gas in the tank at its melting point
T = 298 K;
P1 = 2.98 atom;
T2 = 1100 K;
P2 =?
\(\frac{P_{1} P_{2}}{T_{1} T_{2}}\) =????
⇒ P2 = \(\frac{P}{T_{1}}\) × T2
= \(\frac{2.98 \text { atm }}{298 K}\) × 1100 K = 11 atm

At 1100 K the pressure of the gas inside the tank will become 11 atm. Given that tank can withstand a maximum pressure of 12 atm, the tank will start melting first.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

11th Chemistry Guide Gaseous State Additional Questions and Answers

I. Choose the best Answer:

Question 1.
For one mole of a gas, the ideal gas equation is ………..
(a) PV = \(\frac {1}{2}\) RT
(b) PV = RT
(c) PV = \(\frac {3}{2}\)RT
(d) PV = \(\frac {5}{2}\)RT
Answer:
(b) PV= RT

Question 2.
The SI unit of pressure is
(a) pascal
(b) atmosphere
(c) bar
(d) torr
Answer:
(a) pascal

Question 3.
Which of the following is the correct mathematical relation for Charles’ law at constant pressure?
(a) V ∝ T
(b) V ∝ t
(c) V ∝ – \(\frac {1}{T}\)
(d) all of above
Answer:
(a) V ∝ T

Question 4.
“For a fixed mass of a gas at constant pressure, the volume is directly proportional to its temperature”. This statement is
(a) Boyle’s law
(b) Gay-Lussac law
(c) Avogadro’s law
(d) Charle’s law
Answer:
(d) Charle’s law

Question 5.
With rise in temperature, the surface tension of a liquid …………
(a) decreases
(b) increases
(c) remaining the same
(d) none of the above
Answer:
(a) decreases

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 6.
The plot of the volume of the gas against its temperature at a given pressure is called
(a) isotone
(b) isobar
(c) isomer
(d) isotactic
Answer:
(b) isobar

Question 7.
The cleansing action of soaps and detergents is due to …………..
(a) internal friction
(b) high hydrogen bonding
(c) viscosity
(d) surface tensions
Answer:
(d) surface tensions

Question 8.
The precise value of temperature at which the volume of the gas becomes zero is
(a) -273.15 °C
(b) -273 °C
(c) -298.15°C
(d) -298 °C
Answer:
(a) -273.15 °C

Question 9.
The compressibility factor, z for an ideal gas is ………….
(a) zero
(b) less than one
(c) greater than one
(d) equal to one
Answer:
(d) equal to one

Question 10.
The value of gas constant, R, in terms of JK-1 mol-1 is
(a) 8.314
(b) 4.184
(c) 0.0821
(d) 1.987
Answer:
(a) 8.314

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 11.
Which of the following is a monoatomic gas in nature?
(a) Oxygen
(b) Hydrogen
(c) Helium
(d) Ozone
Answer:
(c) Helium

Question 12.
A mixture of gases containing 1 mole of He, 4 moles of Ne, and 5 moles of Xe. The correct order of partial pressure of the gases, if the total pressure is 10 atm is
(a) Xe < Ne < He
(b) He < Ne < Xe
(c) Xe < Ne < He
(d) He < Xe < Ne
Answer:
(b) He < Ne < Xe

Question 13.
Which one of the following is not a monoatomic gas?
(a) Neon
(b) Xenon
(c) Argon
(d) Oxygen
Answer:
(d) Oxygen

Question 14.
The process in which agas escapes from a container through a very small hole is called
(a) diffusion
(b) effusion
(c) occlusion
(d) dilution
Answer:
(a) diffusion

Question 15.
Which of the following is a tri atomic gas at room temperature?
(a) Oxygen
(b) Helium
(c) Ozone
(d) Nitrogen
Answer:
(c) Ozone

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 16.
An unknown gas X diffuses at a rate of 2 times of oxygen at the same temperature and pressure. The molar mass (in g mol-1) of the gas ‘X is (Molar mass of oxygen is 32 g mol-1)
(a) 8
(b) 16
(c) 20
(d)12
Answer:
(a) 8

Question 17.
Among the following, which is deadly poison?
(a) CO2
(b) HCN
(c) HCl
(d) NH3
Answer:
(b) HCN

Question 18.
The gases which deviate from ideal behavior at
(a) low temperature and high pressure
(b) high temperature and low pressure
(c) low temperature and low pressure
(d) high temperature and high pressure
Answer:
(b) high temperature and low pressure

Question 19.
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called temperature
(a) inversion
(b) ideal
(c) Boyle
(d) reversible
Answer:
(c) Boyle

Question 20.
The pressure of a gas is equal to ………..
(a) \(\frac {F}{a}\)
(b) F x a
(c) \(\frac {a}{F}\)
(d) F – a
Answer:
(a) \(\frac {F}{a}\)

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 21.
The pressure correction introduced by Van der Waals is,
Pideal =
(a) P + \(\frac{a V^{2}}{n^{2}}\)

(b) P + \(\frac{a n}{V^{2}}\)

(c) P + \(\frac{a n^{2}}{V^{2}}\)

(d) P + \(\frac{a V}{n^{2}}\)
Answer:
(c) P + \(\frac{a n^{2}}{V^{2}}\)

Question 22.
Statement-I: The pressure cooker takes more time for cooking at high altitude.
Statement-II: Air is subjected to Earth’s gravitational force. The pressure of air gradually decreases from the surface of the Earth to higher altitude.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(e) Statement-I is wrong but Statement-II is correct
(ð) Statement-I is correct but Statement-II is wrong
Answer:
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I

Question 23.
Van der waals equation for one mole of a gas is
(a) (P + \(\frac{a}{V}\))(V-b) = RT
(b) (P – \(\frac{a}{V^{2}}\))(V + -b) = RT
(c) (P + \(\frac{a}{V^{2}}\))(V – b) = RT
(d) (P + \(\frac{a}{V}\))(V + b) = RT
Answer:
(c) (P + \(\frac{a}{V^{2}}\))(V – b) = RT

Question 24.
The standard atmospheric pressure at sea level at 0°C is equal to ……………..
(a) 1 mm Hg
(b) 76 mm Hg
(c) 760 mm Hg
(d) 680 mm Hg
Answer:
(c) 760 mm Hg

Question 25.
The unit of Van der waals constant ‘b’ is
(a) lit mol-1
(b) lit mol
(c) atm lit mol-1
(d) atm lif-1 mol-2
Answer:
(a) lit mol-1

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 26.
The temperature above which a gas cannot be liquefied even at high pressure is called is ________ temperature.
(a) ideal
(b) inversion
(c) critical
(d) real
Answer:
(c) critical

Question 27.
Which one of the following represents Charles’ law?
(a) PV = Constant
(b) \(\frac {V}{T}\) = Constant
(c) VT Constant
(d) \(\frac {T}{V}\) = R
Answer:
(b) \(\frac {V}{T}\) = Constant

Question 28.
Which of the following gas has the highest critical temperature?
(a) NH3
(b) CO2
(c) N2
(d) CH4
Answer:
(a) NH3

Question 29.
\(\frac {P}{T}\) = Constant is known as …………..
(a) Boyle’s law
(b) Charles’ law
(c) Gay Lussac’s law
(d) Dalton’s law
Answer:
(c) Gay Lussac’s law

Question 30.
The temperature below which a gas obeys the Joule-Thomson effect is called ______ temperature.
(a) critical
(b) Inversion
(c) ideal
(d) real
Answer:
(b) Inversion

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 31.
The value of the Universal gas constant in an ideal gas equation is equal to ………….
(a) 8.3 14 KJ
(b) 0.082057 dm3 atm mol-1 K-1
(c) 1 Pascal
(d) 8.314 x 10-2Pascal
Answer:
(b) 0.082057 dm3 atm mol-1 K-1

Question 32.
A sample of a given mass of gas at constant temperature occupies a volume of 95 cm3 under pressure of 10.13 × 104 Nm-2. At the same temperature, its volume at a pressure of 10.13 × 104 Vm-2 is
(a) 190 cm3
(b) 93 cm3
(c) 46.5 cm3
(d) 4.75 cm3
Answer:
(b) 93 cm3

Question 33.
Which law is used in the isotopic separation of deuterium and protium?
(a) Boyle’s law
(b) Charles’ law
(c) Graham’s law
(d) Gay Lussac’s law
Answer:
(c) Graham’s law

Question 34.
The use of hot air balloons in sports and meteorological observations is an application of
(a) Boyle’s law
(b) Newton’s law
(c) Kelvin’s law
(d) Charle’s law
Answer:
(d) Charle’s law

Question 35.
The value of critical volume is equal in terms of Vander Waals constant is ……….
(a) 3b
(b) \(\frac{8a}{27 Rb}\)
(c) \(\frac{a}{27 b^{2}}\)
(d) \(\frac{2a}{Rb}\)
Answer:
(a) 3b

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 36.
7.0 g of a gas at 300K and 1 atm occupies a volume of 4.1 litre. What is the molecular mass of the gas?
(a) 42
(b) 38.24
(c) 14.5
(d) 46.5
Answer:
(a) 42

Question 37.
If most probable velocity is represented by a and fraction possessing it by / then with increase in temperature which one of the following is correct?
(a) α increases, f decreases
(b) α decreases, f increases
(c) Both α and f decrease
(d) Both α and f increase
Answer:
(a) α increases, f decreases

Question 38.
The value of critical pressure of CO2 is ……………….
(a) 173 atm
(b) 73 atm
(c) 1 atm
(d) 22.4 atm
Answer:
(b) 73 atm

Question 39.
To which of the following gaseous mixtures Dalton’s law is not applicable?
(a) Ne + He + SO2
(b) NH3 + HCl + HBr
(c) O2 + N2 +CO2
(d) N2 + H2 + O2
Answer:
(b) NH3 + HCl + HBr

Question 40.
The substance used in the adiabatic process of liquefaction is ……………
(a) liquid helium
(b) gadolinium sulphate
(c) iron sulphate
(d) liquid ammonia
Answer:
(b) Gadolinium sulphate

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 41.
50 mL of gas A effuses through a pinhole in 146 seconds. The same volume of CO2 under identical condition effuses in 115 seconds. The molar mass of A is
(a) 44
(b) 35.5
(c) 71
(d) None of these
Answer:
(c) 71

Question 42.
The molecules of gas A travel four times faster than the molecules of gas B at the same temperature. The ratio of molecular weight MA/ MB is ………….
(a) 1/16
(b) 4
(c) 1/4
(d) 16
Answer:
(a) 1/16

Question 43.
Which of the following gases is expected to have the largest value of van der Waals constant ‘a’?
(a) He
(b) H2
(c) NH3
(d) O2
Answer:
(c) NH3

Question 44.
In van der Waals equation of state for a real gas, the term that accounts for intermolecular forces is
(a) Vm – b
(b) P + \(\frac{a}{V m^{2}}\)
(c) RT
(d) \(\frac{1}{R T}\)
Answer:
(b) P + \(\frac{a}{V m^{2}}\)

Question 45.
The value of Vander Waals constant “a” is maximum for ……………….
(a) helium
(b) nitrogen
(c) methane
(d) ammonia
Answer:
(d) Ammonia

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 46.
Which of the following has a non-linear relationship?
(a) P vs V
(b) P vs \(\frac{1}{V}\)
(c) both (a) and (b)
(d) none of these
Answer:
(a) P vs V

Question 47.
Why is that the gases show ideal behavior when the volume occupied is large?
(a) So that the volume of the molecules can be neglected in comparison to it.
(b) So that the pressure is very high
(c) So that the Boyle temperature of the gas is constant
(d) all of these
Answer:
(a) So that the volume of the molecules can be neglected in comparison to it.

Question 48.
At a given temperature, the pressure of a gas obeying the Van der Waals equation is
(a) less than that of an ideal gas
(b) more than that of an ideal gas
(c) more or less depending on the nature of gas
(d) equal to that of an ideal gas
Answer:
(a) less than that of an ideal gas

Question 49.
The rate of diffusion of a gas is ………….
(a) directly proportional to its density
(b) directly proportional to its molecular mass
(c) directly proportional to the square root of its molecular mass
(d) inversely proportional to the square root of its molecular mass
Answer:
(d) inversely proportional to the square root of its molecular mass

Question 50.
Maximum deviation from ideal gas is expected from
(a) CH4(g)
(b) NH3(g)
(c) H2(g)
(d) N2(g)
Answer:
(b) NH3(g)

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

II. Very short question and answers (2 Marks):

Question 1.
Identify the elements that are in the gaseous state under normal atmospheric conditions.
Answer:

  • Hydrogen, nitrogen, oxygen, fluorine, and chlorine exist as gaseous diatomic molecules.
  • Another form of oxygen namely the ozone triatomic molecule exists as a gas at room temperature.
  • Noble gases namely helium, neon, argon, krypton, xenon, and radon are monoatomic gases.

Question 2.
State Gay Lussac ‘s law.
Answer:
At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature.
P ∝ T or \(\frac{P}{T}\) = constant K
If P1 and P2 are the pressures at temperatures T1 and T2, respectively, then from Gay Lussac’s law
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

Question 3.
Define pressure. Give its units.
Answer:

  • Pressure is defined as the force exerted by a gas on unit area of the wall. Force F
  • Pressure = \(\frac {Force}{Area}\) = \(\frac {F}{a}\)
  • The SI unit of pressure is Pascal (Pa)

Question 4.
What is the Compressibility factor?
Answer:
The deviation of real gases from ideal behaviour is measured in terms of a ratio of PV to nRT. This is termed as compressibility factor. Mathematically,
Z = \(\frac{P V}{n R T}\)

Question 5.
Deep-sea divers ascend slowly and breathe continuously by the time they reach the surface. Give reason.
Answer:

  • For every 10 m of depth, a diver experiences an additional 1 atm of pressure due to the weight of water surrounding him.
  • At 20 m, the diver experiences a total pressure of 3 atm. So the most important rule in diving is never to hold a breath.
  • Divers must ascend slowly and breathe continuously allowing the regulator to bring the air pressure in their lungs to 1 atm by the time they reach the surface.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 6.
How does cooling is produced in the Adiabatic Process?
Answer:
In the Adiabatic process, cooling is produced by removing the magnetic property of magnetic material such as gadolinium sulphate. By this method, a temperature of 10-4 K i.e., as low as 0 K can be achieved.

Question 7.
What is the reason behind the cause of ear pain while climbing a mountain? How it can be rectified?
Answer:

  • When one ascends a mountain in a plain, the external pressure drops while the pressure within the air cavities remains the same. This creates an imbalance.
  • The greater internal pressure forces the eardrum to bulge outward causing pain.
  • With time and with the help of a yawn or two, the excess air within your ear’s cavities escapes thereby equalizing the internal and external pressure and relieving the pain.

Question 8.
Write a note on the Consequence of Boyle’s law.
Answer:
The pressure-density relationship can be derived from the Boyle’s law as shown below.
P1V1 = P2V2
P1 \(\frac{m}{d_{1}}\) = P2 \(\frac{m}{d_{2}}\)
where “m” is the mass, d1 and d2 are the densities of gases at pressure P1 and P2.
\(\frac{P_{1}}{d_{1}}=\frac{P_{2}}{d_{2}}\)
In other words, the density of a gas is directly proportional to pressure.

Question 9.
What are the applications of Charles’ law?
Answer:

  • The hot air inside the balloon rises because of its decreased density and causes the balloon to float inside the balloon rises because of its decreased density and causes the balloon to float.
  • If you take a helium balloon outside on a chilly day, the balloon will crumble. Once you get back into a warm area, the balloon will return to its original shape. This is because, in accordance with Charles’ law, a gas like helium takes up more space when it is warm.

Question 10.
Write a note on the application of Dalton’s law.
Answer:
In a reaction involving the collection of gas by downward displacement of water, the pressure of dry vapor collected can be calculated using Dalton’s law.
Pdry gas collected = Ptotal – Pwatervapour
Pwatervapour has generally referred to as aqueous tension and its values are available for air at various temperatures.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 11.
Define Dalton’s law of partial pressure.
Answer:
Dalton’s law of partial pressure:
It states that the total pressure of a mixture of gases is the sum of partial pressures of the gases present.
Ptotal = P1 + P2 + P3

Question 12.
What happens when a balloon is moved from an ice cold water bath to a boiling water bath?
Answer:
If a balloon is moved from an ice-cold water bath to a boiling water bath, the temperature of the gas increases. As a result, the gas molecules inside the balloon move faster and gas expands. Hence, the volume increases.

Question 13.
Write notes on the coefficient of expansion(α).
Answer:
The relative increase in volume per °C (α) is equal to \(\frac{V}{V_{0} T}\)
Therefore, \(\frac{V}{V_{0} T}\)
TV = V0(αT + 1)
Charles found that the coefficient of expansion is approximately equal to 1/273. It means that at constant temperature for a given mass, for each degree rise in temperature, all gases expand by 1/273 of their volume at 0°C.

Question 14.
Define Graham’s law of diffusion.
Answer:
Graham’s law of diffusion:
The rate of diffusion or effusion is inversely proportional to the square root of the molecular mass of a gas through an orifice.
\(\frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{A}}}}\)
rArB = rate of diffusion of gases A, B
MA, MB = Molecular mass of gases A, B

Question 15.
State Avogadro’s hypothesis.
Answer:
Equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 16.
Explain the applications of Graham’s law of diffusion.
Answer:

  • Graham’s law of diffusion is useful to determine the molecular mass of the gas if the rate of diffusion is known.
  • Graham’s law forms the basis of the process of enriching the isotopes of U235 from other isotopes and also useful in the isotopic separation of deuterium and protium.

Question 17.
State Graham’s law of diffusion.
Answer:
The rate of diffusion or effusion is inversely proportional to the square root of molar mass. This statement is called Graham’s law of diffusion/effusion.
Mathematically, rate of diffusion ∝ \(\frac{1}{M}\)

Question 18.
What is Boyle temperature?
Answer:
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called Boyle temperature or Boyle point.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

III. Short Question and Answers (3 Marks):

Question 1.
Write notes on Boyle’s point.
Answer:
The temperature at which a real gas obeys the ideal gas law over an appreciable range of pressure is called Boyle temperature or Boyle point. The Boyle point varies with the nature of the gas. Above the Boyle point, for real gases, Z > 1, i.e., the real gases show a positive deviation. Below the Boyle point, the real gases first show a decrease for Z, reaches a minimum, and then increase with the increase in pressure.

Question 2.
What are the consequences of Boyle’s law?
Answer:
1. if the volume of a fixed mass of a gas is reduced to half at constant temperature the gas pressure doubles.
2. Boyle’s law also helps to relate the pressure to density.
P1V1 = P3V3 (Boyle’s law)
\(P_{1} \frac{m}{d_{1}}=P_{2} \frac{m}{d_{2}}\)
Where ‘m’ is the mass, d1 and d2 are the densities of gases at pressure P1 and P2. The density of the gas is directly proportional to pressure.

Question 3.
48 liter of dry N2 is passed through 36g of H2O at 27°C and this results In a loss of 1.20 g of water. Find the vapour pressure of water?
Answer:
Water loss is observed because of the escape of water molecules with N2 gas. These water vapour occupy the volume of N2 gas i.e., 48 litres.
Using,
PV = \(\frac{m}{M}\)RT;
V = 48L
m = 1.2g;
M = 18u;
R = 0.082 L atm K-1mol-1
T = 27 + 273 = 300 K
P = \(\frac{1.2}{18} \times \frac{0.0821 \times 300}{48}\) = 0.034 atm

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 4.
A flask of capacity one litre is heated from 25°C to 35°C. What volume of air will escape from the flask?
Answer:
Applying, \(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)
V1 = 1L;
T1 = 25 + 273 = 298 K
V2 = ?
T2 = 35 + 273 = 308 K
V2 = \(\frac{1}{298}\) × 308 = 1.033 L
Capacity of Flask = 1 L
So, volume of air escaped = 1.033 – 1 = 0.033 L = 33 mL

Question 5.
Discuss the graphical representation of Boyle’s law.
Answer:
Boyle’s law is applicable to all gases regardless of their chemical identity (provided the pressure is low). Therefore, for a given mass of a gas under two different sets of conditions at constant temperature we can write,
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 10
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 11

Question 6.
A certain gas takes three times as long to effuse out as helium. Find its molecular mass.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 12

Question 7.
Derive ideal gas equation.
Answer:
The gaseous state is described completely using the following four variables T, P, V and n. Each gas law relates one variable of a gaseous sample to another while the other two variables are held constant. Therefore, combining all equations into a single equation will enable to account for the change in any or all of the variables.
Boyle’s law: V ∝ \(\frac{1}{P}\)
Charles’ law: V ∝ T
Avogadro’s law: V ∝ n
We can combine these equations into the following general equation that describes the physical behaviour of all gases.
V ∝ \(\frac{nT}{P}\)
V = \(\frac{nRT}{P}\)
where R = Proportionately constant.
The above equation can be rearranged to give PV = nRT – Ideal gas equation. Where R is also known as Universal gas constant.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 8.
Write notes on the compressibility factor for real gases.
Answer:
The compressibility factor Z for real gases can be rewritten,
Z = PVreal / nRT …………..(1)
Videal = \(\frac{n R T}{P}\) ……………(2)
substituting (2) in (1)
where Vreal is the molar volume of the real gas and Videal is the molar volume of it when it behaves ideally.

Question 9.
Derive ideal gas equation.
Answer:
The gaseous state is described completely using the following four variables T, P, V, and n and their relationships were governed by the gas laws studied so far.
Boyle’s law V ∝ P1
Charles law V ∝ T
Avogadro’s law V ∝ n
We can combine these equations into the following general equation that describes the physical behaviour of all gases.
V ∝ \(\frac{n T}{P}\)

V = \(\frac{n R T}{P}\)
where, R is the proportionality constant called universal gas constant.
The above equation can be rearranged to give the ideal gas equation
PV = nRT

Question 10.
Explain the different methods used for the liquefaction of gases.
Answer:

  • Linde’s method: Joule-Thomson effect is used to get liquid air or any other gas.
  • Claude’s process: In addition to the Joule-Thomson effect, the gas is allowed to perform mechanical work so that more cooling is produced.
  • Adiabatic process: This method of cooling is produced by removing the magnetic property of magnetic material e.g. Gadolinium sulphate. By this method, a temperature of 10-4 K i.e. as low as zero Kelvin can be achieved.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

IV. Long Question and Answers (5 Marks):

Question 1.
Derive Van der Waals equation of state.
Answer:
J.D.Van der Waals made the first mathematical analysis of real gases. His treatment provides us an interpretation of real gas behaviour at the molecular level. He modified the ideal gas equation PV = nRT by introducing two correction factors, namely, pressure correction and volume correction.

Pressure Correction:
The pressure of a gas is directly proportional to the force created by the bombardment of molecules on the walls of the container. The speed of a molecule moving towards the wall of the container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is lower than the ideal pressure of the gas. Hence, van der Waals introduced a correction term to this effect.

Van der Waals found out the forces of attraction experienced by a molecule near the wall are directly proportional to the square of the density of the gas.
P’ ∝ p2
p = \(\frac{n}{v}\)
where n is the number of moles of gas and V is the volume of the container
⇒ P’ = \(\frac{n^{2}}{V^{2}}\)
⇒ P’ = a\(\frac{n^{2}}{V^{2}}\)

where a is the proportionality constant and depends on the nature of gas.
Therefore, ideal P = P + a\(\frac{n^{2}}{V^{2}}\)

Volume Correction:
As every individual molecule of a gas occupies a certain volume, the actual volume is less than the volume of the container, V. Van der Waals introduced a correction factor V’ to this effect. Let us calculate the correction term by considering gas molecules as spheres.
V = excluded volume
Excluded volume for two molecules = \(\frac{4}{3}\)π(2r)3
= \(\left|\frac{-8}{(3 \pi)}\right|\) = 8Vm

where Vm is a volume of a single molecule
Excluded volume for single-molecule = \(\frac{8 V_{m}}{2}\) = 4Vm

Excluded volume for n molecule = n(4Vm) = nb
Where b is van der Waals constant which is equal to 4 Vm
=> V’ = nb
Videal = V nb

Replacing the corrected pressure and volume in the ideal gas equation PV = nRT we get the van der Waals equation of state for real gases as below,
(P + \(\frac{a n^{2}}{V}\))(V – nb) = nRT
The constants a and b are van der Waals constants and their values vary with the nature of the gas. It is an approximate formula for the non-ideal gas.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 2.
Derive the relationship between Van der Waals constants and critical constants.
Answer:
The van der Waals equation for n moles is
(P + \(\frac{a n^{2}}{V}\))(V – nb) = nRT
For 1 mole
(p + \(\frac{a}{V^{2}}\))(V – b) = RT
From the equation, we can derive the values of critical constants Pc, Vc, and Tc in terms of a and b, the van der Waals constants, On expanding the above equation
PV + \(\frac{a}{V}\) – Pb – \(\frac{a b}{V^{2}}\) – RT = 0

Multiply the above equation by
\(\frac{V_{2}}{P}\)(PV + \(\frac{a}{V}\) – Pb – \(\frac{a b}{V^{2}}\) – RT) = 0

V3 + \(\frac{a V}{P}\) + -bV2 – \(\frac{a b}{p}\) – \(\frac{R T V^{2}}{P}\) = 0

when the above equation is rearranged in powers of V

V3 \(\frac{R T}{P}\) + bV2 \(\frac{a}{P}\) V – \(\frac{a b}{P}\) = 0.

The above equation is a cubic equation in V. On solving this equation,
we will get three solutions. At the critical point, all these three solutions of Vc are equal to the critical volume. The pressure and temperature becomes Pc and Tc respectively
i.e., V = Vc
V – Vc = 0
V – (Vc)3 = 0
V3 – 3VcV2 + 3VVc2 – Vc3 = 0.
As equation identical with the equation above, we can equate the coefficients of Vc, V, and constant terms.

-3VcV2 = –\(\frac{R T_{C}}{P}\) + bV2

3Vc = \(\frac{R T_{C}}{P_{C}}\) + b ……….(1)

3Vc2 = \(\frac{a}{P_{C}}\) …………..(2)

Vc3 = \(\frac{a b}{P_{C}}\) …………..(3)

Divide equation (3) by equation (2)

\(\frac{V_{c}^{3}}{3 V_{C}^{2}}=\frac{a b / P_{C}}{a / P_{C}}\)

\(\frac{V_{C}}{3}\) = b
i.e., Vc = 3b ……………(4)

when equation (4) is substituted in (2)
Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 13
The critical constants can be calculated using the values of Vander walls constant of a gas and vice versa.

a = 3Vc2 Pc and b = \(\frac{V_{C}}{3}\)

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State

Question 3.
Explain Andrew’s isotherm for Carbon dioxide.
Answer:
Thomas Andrew gave the first complete data on the pressure-volume temperature of a substance in the gaseous and liquid states. He plotted isotherms of carbon dioxide at different temperatures which are shown in Figure. From the plots we can infer the following.

At low-temperature isotherms, for example, at 130°C as the pressure increases, the volume decreases along with AB and is a gas until point B is reached. At B, a liquid separates along the line BC, both the liquid and gas co-exist and the pressure remains constant. At C, the gas is completely converted into liquid. If the pressure is higher than at C, only the liquid is compressed so, there is no significant change in the volume. The successive isotherms show a similar trend with the shorter flat region. i.e., The volume range in which the liquid and gas coexist becomes shorter.

At the temperature of 31.1°C the length of the shorter portion is reduced to zero at point P. In other words, the CO2 gas is liquefied completely at this point. This temperature is known as the liquefaction temperature or critical temperature of CO3. At this point, the pressure is 73 atm. Above this temperature, CO3 remains as a gas at all pressure values. It is then proved that many real gases behave in a similar manner to carbon dioxide.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 14

Question 4.
Explain Boyle’s law experiment:
Answer:
Robert Boyle performed a series of experiments to j study the relation between the pressure and volume of gases. The schematic diagram of the apparatus j used Boyle is shown in the figure.

Samacheer Kalvi 11th Chemistry Guide Chapter 6 Gaseous State 15

Mercury was added through the open end of the apparatus such that the mercury level on both ends is equal as shown in figure (a). Add more amount of mercury until the volume of the trapped air is reduced to half of its original volume as shown in figure (b). The pressure exerted on the gas by the addition of excess mercury is given by the difference in mercury levels of the tube.

Initially, the pressure exerted by the gas is equal to 1 atm as the difference in height of the mercury levels is zero. When the volume is reduced to half, the difference in mercury levels increases to 760 mm. Now the pressure exerted by the gas is equal to 2 atm. It led him to conclude that at a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
Mathematically, Boyle’s law can be written as
V ∝ \(\frac{1}{P}\) ……….(1)
(T and n are fixed, T-temperature, n- number of moles)
V = k × \(\frac{1}{P}\) ……….(2)
k – proportionality constant When we rearrange equation (2)
PV = k at constant temperature and mass.