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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Ex 8.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Ex 8.2

Question 1.
Mention two branches of statistical inference?
Solution:
The two branches of statistical inference are estimation and testing of hypothesis.

Question 2.
What is an estimator?
Solution:
Any sample statistic which is used to estimate an unknown population parameter is called an estimator ie., an estimator is a sample statistic used to estimate a population parameter.

Question 3.
What is an estimate?
Solution:
When we observe a specific numerical value of our estimator, we call that value is an estimate. In other words, an estimate is a specific value of a statistic.

Question 4.
What is point estimation?
Solution:
Point estimation involves the use of sample data to calculate a single value which is to serve as a best estimate of an unknown population parameter. For example the mean height of 145 cm from a sample of 15 students is‘a point estimate for the mean height of the class of 100 students.

Question 5.
What is interval estimation?
Solution:
Generally, there are situation where point estimation is not desirable and we are interested in finding limits within which the parameter would be expected to lie is called an interval estimation.

Question 6.
What is confidence interval?
Solution:
Let us choose a small value of a which is known as level of significance (1% or 5%) and determine two constants says c₁ and c₂ such that p(c₁ < θ < c₂|t) = 1 – α

The quantities c₁ and c₂ so determined are known as the confidence Limits and the interval [c₁, c₂] within which the unknown value of the population parameter is expected to lie is known as a confidence interval.(1 – α)is called as confidence coefficient.

Question 7.
What is null hypothesis? Give an example.
Solution:
According to prof R. A. fisher, “Null hypothesis is the hypothesis which is tested for possible rejection under the assumption that it is true”, and it is denoted by H₀.

For example: If we want to find the population mean has a specified value µ₀, then the null hypothesis H₀ is set as follows H₀ : µ = µ₀

Question 8.
Define alternative hypothesis.
Solution:
A null hypothesis is a type of hypothesis, that proposes that no statistical significance exists in a set of given observations. For example, let the average time to cook a specific dish is 15 minutes. The null hypothesis would be stated as “The population mean is equal to 15 minutes”, (i.e) H₀ : µ = 15

Any hypothesis which is complementary to the null hypothesis is called as the alternative hypothesis and is usually denoted by H₁.

For example: If we want to test the null hypothesis that the population has specified mean µ i.e., H₀ : µ = µ₀ then the alternative hypothesis could be anyone among the following:
(i) H₁ : µ ≠ µ₀ (µ > or µ < µ₀)
(ii) H₁ : µ > µ₀
(iii) H₁ : µ < µ₀

Question 9.
Define critical region.
Solution:
A region corresponding to a test statistic in the sample space which tends to rejection of H₀ is called critical region or region of rejection.

Question 10.
Define critical value.
Solution:
The value of test statistic which separates the critical (or rejection) region and the acceptance region is called the critical value or significant value. It depends upon.
(i) The level of significance
(ii) The alternative hypothesis whether it is two-tailed or single-tailed

Question 11.
Define the level of significance.
Solution:
The level of significance is defined as the probability of rejecting a null hypothesis by the test when it is really true, which is denoted as α. That is P(Type 1 error) = α. For example, the level of significance 0.1 is related to the 90% confidence level.

Question 12.
What is a type I error
Solution:
There is every chance that a decision regarding a null hypothesis may be correct or may not be correct. The error of rejecting H₀ when it is true is called type I error.

Question 13.
What is single tailed test.
Solution:
When the hypothesis about the population parameter is rejected only for the value of sample statistic falling into one of the tails of the sampling distribution, then it is known as a one-tailed test. Here H₁: µ > µ₀ and H₁: µ < µ₀ is known as a one-tailed alternative.

Question 14.
A sample of 100 items, draw from a universe with mean value 64 and S.D 3, has a mean value 63.5. Is the difference in the mean significant?
Solution:
sample size n = 100 ; sample mean \(\bar { x}\) = 63.5
sample SD S = 3;
population mean µ = 64 population SD σ = 3
Null Hypothesis H₀ : µ = 64 (the sample has been drawn from the population mean µ = 64 and SD σ = 3)
Alternative Hypothesis H₁ : µ ≠ 64 (two tail) i.e the sample has not been drawn from the population mean µ = 64 and SD σ = 3
The level of significance α = 5% = 0.05
Test statistic
z = \(\frac { 63.5-64}{\frac{3}{\sqrt{100}}}\) = \(\frac { -0.5 }{(\frac{3}{10})}\) = \(\frac { -0.5 }{0.3}\) = -1.667
|z| = 1.667
∴ calculated z = 1.667
critical value at 5% level of
significance is z_{\(\frac { α }{2}\)} = 1.96
Inference:
At 5% level of significance Z < Z_{\(\frac { α }{2}\)} since the calculated value is less than the table value the null hypothesis is accepted.

Question 15.
A sample of 400 individuals is found to have a mean height of 67.47 inches. Can it be reasonably regarded as a sample from a large population with mean height of 67.39 inches and standard deviation 1.30 inches?
Solution:
sample size n = 400; sample mean \(\bar { x}\) = 67.47 inches
sample SD S = 1.30 inches population mean
µ = 67.39 inches
population SD σ = 1.30 inches
Null Hypothesis H₀ : µ = 67.39 inches (the sample has been drawn from the population mean µ = 67.39 inches; population SD σ = 1.30 inches)
Alternative Hypothesis H₁ = µ ≠ 67.39 inches(two tail)
i.e the sample has not been drawn from the population mean µ = 67.39 inches and SD σ = 1.30 inches
The level of significance α = 5% = 0.05
Test static:

Thus the calculated and the significant value or Z_{\(\frac { α }{2}\)} = 1.96
table value comparing the calculated and table values
Z_{\(\frac { α }{2}\)} (i.e.,) 1.2308 < 1.96
Inference: since the calculated value is less than value i.e Z > Z_{\(\frac { α }{2}\)} at 5% level of significance, the null hypothesis is accepted Hence we conclude that the data doesn’t provide us any evidence against the null hypothesis. Therefore, the sample has been drawn from the population mean µ = 67.39 inches and SD σ = 1.30 inches.

Question 16.
The average score on a nationally administered aptitude test was 76 and the corresponding standard deviation was 8. In order to evaluate a state’s education system, the scores of 100 of the state’s students were randomly selected. These students had an average score of 72. Test at a significance level of 0.05 if there is a significant difference between the state scores and the national scores.
Solution:
sample size n = 100
sample mean \(\bar { x}\) = 72
sample SD S = 8
population mean µ = 76
under the Null hypothesis H₀ : p = 76
Against the alternative hypothesis H₀ : µ ≠ 76 (two mail)
Level of significance µ = 0.05
Test statistic:

since alternative hypothesis is of two tailed test we can take |Z| = 5.
∴ critical value 5% level of significance is z > z_{\(\frac { α }{2}\)} = 1.96
Inference:
Since the calculated value is less than table value i.e z > z_{\(\frac { α }{2}\)} at 5% level of significance the null hypothesis H0 is rejected Therefore, we conclude that there is significant difference between the sample mean and population mean µ = 76 and SD σ = 8.

Question 17.
The mean breaking strength of cables supplied by a manufacturer is 1,800 with a standard deviation 100. By a new technique in the manufacturing process it is claimed that the breaking strength of the cables has increased. In order to test this claim a sample of 50 cables is tested. It is found that the mean breaking strength is 1,850. Can you support the claim at 0.01 level of significance.
Solution:
Sample size n = 50
Sample mean \(\bar { x}\) = 1800
Sample SD S = 100
population mean µ = 1850
Null hypothesis H₀ : µ = 1850
Level of significance µ = 0.01

= -3.5355
∴ z = -3.536
calculated value |z| = 3.536
Critical value at 1% level of significance is z_{\(\frac { α }{2}\)} = 2.58
Inference:
Since the calculated value is greater than table (ie) z > z_{\(\frac { α }{2}\)} at 1% level of significance, the null hypothesis is rejected, therefore we conclude that we is rejected, Therefore we conclude that we can not support we conclude that we can support the claim of 0.01 of significance.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Ex 9.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Ex 9.1

Question 1.
Define Time series.
Solution:
When quantitative data are arranged in the order of their occurrence, the resulting series is called the Time Series.

Question 2.
What is the need for studying time series?
Solution:
Time series analysis is one of the statistical methods used to determine the patterns in data collected for a period of time. Generally, each of us should know about the past data to observe and understand the changes that have taken place in the past and current time. One can also identify the regular or irregular occurrence of any specific feature over a time period in a time series data.

Question 3.
State the uses of time series.
Solution:

Question 4.
Mention the components of the time series.
Solution:
There are four types of components in a time series. They are as follows
(i) Secular Trend
(ii) Seasonal variations
(iii) Cyclic variations
(iv) Irregular variations

Question 5.
Define secular trend.
Solution:
Secular Trend: It is a general tendency, of time series to increase or decrease, or stagnates during a long period of time. An upward tendency is usually observed in the population of a country, production, sales, prices in industries, the income of individuals, etc., A downward tendency is observed in deaths, epidemics, prices of electronic gadgets, water sources, mortality rate, etc…

Question 6.
Write a brief note on seasonal variations
Solution:
As the name suggests, tendency movements are due to nature which repeats themselves periodically in every season. These variations repeat themselves in less than one year time. It is measured in an interval of time. Seasonal variations may be influenced by natural force, social customs and traditions. These variations are the results of such factors which uniformly and regularly rise and fall in the magnitude. For example, selling of umbrellas’ and raincoat in the rainy season, sales of cool drinks in the summer season, crackers in the Deepawali season, purchase of dresses in a festival season, sugarcane in Pongal season.

Question 7.
Explain cyclic variations.
Solution:
Cyclic Variations: These variations are not necessarily uniformly periodic in nature. That is, they may or may not follow exactly similar patterns after equal intervals of time. Generally, one cyclic period ranges from 7 to 9 years and there is no hard and fast rule in the fixation of years for a cyclic period. For example, every business cycle has a Start- Boom-Depression- Recover, maintenance during booms and depressions, changes in government monetary policies, changes in interest rates.

Question 8.
Explain the stratified random sampling with a suitable example.
Solution:
These variations do not have a particular pattern and there is no regular period of time of their occurrences. These are accidental changes which are purely random or unpredictable. Normally they are short-term variations, but their occurrence sometimes has its effect so intense that they may give rise to new cyclic or other movements of variations. For example floods, wars, earthquakes, Tsunami, strikes, lockouts, etc.

Question 9.
Define seasonal index.
Solution:
Seasonal Index for every season (i.e) months, quarters, or year is given by
Seasonal Index (S.I) = \(\frac{\text { Seasonal Average }}{\text { Grand average }}\) × 100
Where seasonal average is calculated for month, (or) quarter depending on the problem and Grand Average (G) is the average of averages.

Question 10.
Explain the method of fitting a straight line.
Solution:
(i) The straight-line trend is represented by the equation Y = a + bX …………. (1)
Where Y is the actual value, X is time, a, b are constants.
(ii) The constants ‘a and ‘b’ are estimated by solving the following two normal Equations
ΣY = n a + b ΣX ………… (2)
ΣXY = a ΣX + b ΣX² …………… (3)
Where ‘n’ = a number of years given in the data.
(iii) By taking the mid-point of the time as the ori-gin, we get ΣX = 0
(iv) When ΣX = 0, the two normal equations reduces to
ΣY = na + b(0) ; a = \(\frac { ΣY }{n}\) = \(\bar { Y}\)
ΣXY = a(0) + bΣX² ; b = \(\frac { ΣXY }{ΣX^2}\)
The constant ‘a’ gives the mean of Y and ‘b gives the rate of change (slope),
(v) By substituting the values of ‘a and ‘b’ in the trend equation (1), we get the Line of Best Fit.

Question 11.
State the two normal equations used in fitting a straight line.
Solution:
The normal equations used in fitting a straight line are
ΣY = na + b ΣX and ΣXY = a ΣX + b ΣX²
Where n = number of years given in the data,
X = time
Y = actual value
a, b = constants

Question 12.
State the different methods of measuring trends.
Solution:
Following are the methods by which we can measure the trend.
(i) Freehand or Graphic Method
(ii) Method of Semi-Averages
(iii) Method of Moving Averages

Question 13.
Compute the average seasonal movement for the following series

Solution:
Computation of seasonal. Index by the method of simple averages.

Question 14.
The following figures relates to the profits of a commercial concern for 8 years

Find the trend of profits by the method of three yearly moving averages.
Solution:

Question 15.
Find the trend of production by the method of a five-yearly period of moving average for the following data:

Solution:
I Computation of five – yearly moving averages

Question 16.
The following table gives the number of small- scale units registered with the Directorate of Industries between 1985 and 1991. Show the growth on a trend line by the free hand method.

Solution:

Question 17.
The annual production of a commodity is given as follows:

Fit a straight line trend by the method of least squares
Solution:

Therefore the required equation of the straight line
Y = a + bx
Y = 169.428 + 3.285 X
⇒ Y = 169.428 + 3.285 (x – 1998)

Question 18.
Determine the equation of a straight line which best fits the following data

Compute the trend values for all years from 2000 to 2004
Solution:
Computation of trend values by the method of least squares (ODD Years)

Therefore, the required equation of the straight line trend is given by
Y = a + bX
Y = 54 + 5.4X
Y = 54 + 5.4 (x – 2002)
The trend value can be obtained by when x = 2000
Yt = 54 + 5.4 (2000 – 2002)
Y = 54+ 5.4 (-2)
= 54 – 10.8
= 43.2
When x = 2001
Yt = 54 + 5.4 (2001 – 2002)
Y = 54 + 5.4 (-1)
= 54 – 5.4
= 48.6
When x = 2002
Yt = 54 + 5.4 (2002 – 2002)
V = 54 + 5.4 (0)
= 54
When x = 2003
Yt = 54 + 5.4 (2003 – 2002)
Y = 54 + 5.4 (1)
= 54 + 5.4
= 59.4
When x = 2004
Yt = 54 + 5.4 (2004 – 2002)
Y = 54 + 5.4(2)
= 54 + 10.8
= 64.8

Question 19.
The sales of a commodity in tones varied from January 2010 to December 2010 as follows: in year 2010 Sales (in tones)

Fit a trend line by the method of semi-average.
Solution:
Since the number of years is even (twelve). We can equally divide the given data it two equal parts and obtain averages of the first six months and the last six.

Question 20.
Use the method of monthly averages to find the monthly indices for the following data of production of a commodity for the years 2002, 2003, and 2004.

Solution:

Question 21.
Use the method of monthly averages to find T the monthly indices for the following data of production of a commodity for the years 2002, 2003, and 2004.

Solution:
Computation of Seasonal Index by the method of simple averages

Question 22.
The following table shows the number of salesmen working for a certain concern:

Use the method of least squares to fit a straight line and estimate the number of salesmen in 1997
Solution:
Computation of trend values by the method of least squares (ODD Years)

Therefore, the required equation of the straight line trend is given by
Y = a + bX
Y = 48.8 + 2X
(i.e) Y = 48.8 + 2 (x – 1994)
When x = 1992
Yt = 48.8 + 2 (1992 – 1994)
Y = 48.8 + 2 (-2)
= 48.8 – 4
= 44.8
When x = 1993
Yt = 48.8 + 2 (1993 – 1994)
Y = 48.8 + 2 (-1)
= 48.8 – 2
= 46.8
When x = 1994
Yt = 48.8 + 2 (1994 – 1994)
Y = 48.8 + 2(0)
= 48.8
When x = 1995
Yt = 48.8 + 2 (1995 – 1994)
Y = 48.8 + 2(1)
= 50.8
When x = 1996
Yt = 48.8 + 2 (1996 – 1994)
Y = 48.8 + 2 (2)
= 48.8 + 4
= 52.8
When x = 1997
Yt = 48.8 + 2 (1997 – 1994)
Y = 48.8 + 2 (3)
= 48.8 + 6
= 54.8

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Accountancy Guide Pdf Chapter 13 Final Accounts of Sole Proprietors – II Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

11th Accountancy Guide Final Accounts of Sole Proprietors – II Text Book Back Questions and Answers

I. Multiple Choice Questions

Choose the correct answer.

Question 1.
A prepayment of insurance premium will appear in __________.
a) The trading account on the debit side
b) The profit and loss account on the credit side
c) The balance sheet on the assets side
d) The balance sheet on the liabilities side
Answer:
c) The balance sheet on the assets side

Question 2.
Net profit is __________.
a) Debited to capital account
b) Credited to capital account
c) Debited to drawings account
d) Credited to drawings account
Answer:
b) Credited to capital account

Question 3.
Closing stock is valued at __________.
a) Cost price
b) Market price
c) Cost price or market price whichever is higher
d) Cost price or net realisable value whichever is lower
Answer:
d) Cost price or net realisable value whichever is lower

Question 4.
Accrued interest on investment will be shown __________.
a) On the credit side of profit and loss account
b) On the assets side of balance sheet
c) Both (a) and (b)
d) None of these
Answer:
c) Both (a) and (b)

Question 5.
If there is no existing provision for doubtful debts, provision created for doubtful debts is __________.
a) Debited to bad debts account
b) Debited to sundry debtors account
c) Credited to bad debts account
d) Debited to profit and loss account
Answer:
d) Debited to profit and loss account

II. Very Short Answer Questions

Question 1.
What are adjusting entries?
Answer:
Adjusting entries are the journal entries made at the end of the accounting period to account for items which are omitted in the trial balance and to make adjustments for outstanding and prepaid expenses and revenues accrued and received in advance.

Question 2.
What is an outstanding expense?
Answer:

1. Expenses which have been incurred in the accounting period but not paid till the end of the accounting period are called outstanding expenses.
2. In other words, if certain benefits or services are received during the year but payment is not made for the services received and utilized, these are termed outstanding expenses.
3. An outstanding expense account is a representative personal account and an expense account is a nominal account.

Question 3.
What is a prepaid expense?
Answer:

1. Prepaid expenses refer to any expense or portion of expense paid in the current accounting year but the benefit of services of which will be received in the next accounting period.
2. They are also called unexpired expenses.
3. Though these expenses are paid in the accounting period, they are not incurred during the accounting period.
4. A prepaid expense account is a representative personal account.
5. An expense account is a nominal account.

Question 4.
What are accrued incomes?
Answer:

1. Accrued income is income or portion of income which has been earned during the current accounting year but not received till the end of that accounting year.
2. It generally happens in the case of the amount to be received on account of commission, interest, dividend, etc.

Question 5.
What is the provision for discounts on debtors?
Answer:

1. A cash discount is allowed by the suppliers to customers for prompt payment of the amount due either on or before the due date.
2. Provision for discount on debtors is made on the basis of past experience at an estimated rate on sundry debtors.

III. Short Answer Questions

Question 1.
What is the need for preparing final accounts?
Answer:

1. To record omissions in trial balance such as closing stock, interest on capital, interest on drawings, etc.
2. To bring into account outstanding and prepaid expenses.
3. To bring into account income accrued and received in advance.
4. To create reserves and provisions.

Question 2.
What is meant by provision for doubtful debts? Why is it created?
Answer:
Provision for bad and doubtful debts refers to the amount set aside as a charge against profit to meet any loss arising due to bad debt in the future. The amount of doubtful debts is calculated on the basis of some percentage on debtors at the end of the accounting period after deducting further bad debts (if any). A provision for doubtful debts is created and is charged to the profit and loss account.

Question 3.
Explain how dosing stock is treated in final accounts.
Answer:
The unsold goods in the business at the end of the accounting period are termed as closing stock. As per AS-2 (Revised), the stock is valued at cost price or net realizable value, whichever is lower.
Adjusting entry

Presentation in final accounts

Question 4.
Give the adjusting entries for interest on capital and interest on drawings.
Answer:
Adjusting entry

Interest on Drawings

Question 5.
Explain the accounting treatment of bad debts, provision for doubtful debts, and provision for discount on debtors.
Answer:
Bad debts: In other words, debts which cannot be recovered or irrecoverable debts are called bad debts. It is a loss for the business and should be charged against profit.

Provision for bad and doubtful debts:

• Provision for bad and doubtful debts refers to the amount set aside as a charge against profit to meet any loss arising due to bad debt in the future.
• At the end of the accounting period, there may be certain debts which are doubtful, i.e., the amount to be received from debtors may or may not be received.
• The reason may be the incapacity to pay the amount of deceit.
• In general, based on past experience, the amount of doubtful debts is calculated on the basis of some percentage on debtors at the end of the accounting period after deducting further bad debts (if any).
• Since the amount of loss is impossible to ascertain until it is proved bad, doubtful debts are charged against profit and loss accounts in the form of provision.
• A provision for doubtful debts is created and is charged to the profit and loss account. When bad debts occur, it is transferred to provision for doubtful debts account and not to profit and loss account.

Provision for discount on debtors:

• A cash discount is allowed by the suppliers to customers for prompt payment of the amount due either on or before the due date.
• A provision created on sundry debtors for allowing such discount is called a provision for discount on debtors.

IV. Exercises

Question 1.
Pass adjusting entries for the following
a) The closing stock was valued at ₹ 5,000
b) Outstanding salaries ₹ 150
c) Insurance prepaid ₹ 450
d) ₹ 20,000 was received in advance for commission.
e) Accrued interest on investments is ₹ 1,000.
Solution::
Adjusting Entries

Question 2.
For the following adjustments, pass adjusting entries
a) Outstanding wages ₹ 5,000.
b) Depreciate machinery by ₹ 1,000.
c) Interest on capital @ 5% (Capital ₹ 20,000)
d) Interest on drawings ₹ 50
e) Write off bad debts ₹ 500
Solution:
Adjusting Entries

Question 3.
On preparing final accounts of Suresh, bad debt account has a balance of ₹ 800 and sundry debtors account has a balance of ₹ 18,000 of which ₹ 1,200 Is to be written off as further bad debts. Pass adjusting entry for bad debts. And also show how it would appear in profit and loss account and balance sheet.
Solution:
Adjusting Entries

Profit & loss account for the year ended

Balance sheet

Question 4.
The trial balance on March 31, 2016, shows the following:
Sundry debtors ₹ 30,000; Bad debts ₹ 1,200
It is found that 3% of sundry debtors is doubtful of recovery and is to be provided for. Pass journal entry for the amount of provision and also show how it would appear in the profit and loss account and balance sheet.
Solution:
Adjusting Entries

Profit & loss account for the year ended 31^st March 2016

Balance sheet as of 31^st March 2016

Question 5.
The trial balance of a trader on 31^st December 2016 shows sundry debtors as ₹ 50,000.
Adjustments:
a) Write off ₹ 1,000 as bad debts
b) Provide 5% for doubtful debts
c) Provide 2% discount on debtors
Show how these items will appear in the profit and loss A/c and balance sheet of the trader.
Solution:
Profits loss account for the year ended 31^st December 2016

Balance Sheet as of 31^st December 2016

Question 6.
On 1st January 2016, the provision for doubtful debts accounts had a balance of ₹ 3,000. On December 31, 2016, sundry debtors amounted to ₹ 80,000. During the year, bad debts to be written off were ₹ 2,000. A provision for 5% was required for next year. Pass journal entries and show how these items would appear in the final accounts.
Solution:
Adjusting Entries

Profit & loss account for the year ended 31^st December 2016

Balance sheet as of 31^st December 2016

Question 7.
The following are the extracts from the trial balance.
Sundry debtors ₹ 30,000; Bad debts ₹ 5,000
Additional information
a) Write off further bad debts ₹ 3,000.
b) Create 10% provision for bad and doubtful debts.
You are required to pass necessary adjusting entries and show how these items will appear in profit and loss account and balance sheet.
Adjusting Entries

Profit and Loss A/c

Balance sheet

Question 8.
The following are the extracts from the trial balance

Additional information
a) Additional bad debts ₹ 3,000.
b) Keep a provision for bad and doubtful debts @ 10% on sundry debtors.
You are required to pass necessary adjusting entries and show how these items will appear in the profit and loss account and balance sheet.
Solution:
Adjusting Entries

Profit and Loss A/c

Balance sheet

Question 9.
The accounts of Lakshmi traders showed the following balance on 31^st March 2016.
Sundry debtors ₹ 60,000;
Bad debts ₹ 2,000
Provision for doubtful debts ₹ 4,200
At the time of preparation of final accounts on 31^st March, it was found that out of sundry debtors, ₹ 1,000 will be irrecoverable. It was decided to create a provision of 2% on debtors to meet any future possible bad debts.
Pass necessary journal entries and show how these items would appear in the final accounts.
Solution:
Adjusting Entries

Profit and Loss A/c

Balance sheet as on 31^st Mar 2016

Question 10.
The following are the extracts from the trial balance.

Additional information:
a) Create a provision for doubtful debts @ 10% on sundry debtors.
b) Create a provision for discount on debtors @ 5% on sundry debtors.
You are required to pass necessary adjusting entries and show how these items will appear in the final accounts.
Solution:
Adjusting Entries

Profit and Loss A/c

Balance sheet

Question 11.
Following are the extracts from the trial balance.

Additional information:
a) Additional bad debts ₹ 1,000
b) Create a provision for doubtful debts @ 5% on sundry debtors,
c) Create a provision for discount on debtors @ 2% on sundry debtors.
You are required to pass necessary journal entries and show how these items will appear in the final accounts.
Solution:
Adjusting Entries

Profit and Loss A/c

Balance sheet as on 31^st December 2016

Question 12.
The following are the extracts from the trial balance.

You are required to show how these items will appear in the profit and ioss account and balance sheet.
Solution:
Profit and Loss A/c
Balance sheet

Question 13.
Prepare trading account of Archana for the year ending 31^st December, 2016 from the following information

Adjustments:
a) Closing stock ₹ 1,00,000
b) Wages outstanding ₹ 12,000
c) Freight inwards paid in advance ₹ 5,000
Solution:
Trading A/c for the year ending 31^st Dec 2016

Question 14.
Prepare profit and loss account of Manoj for the year ending on 31^st March, 2016

Adjustments:
i) Salary outstanding ₹ 400
ii) Rent paid in advance ₹ 50
iii) Commission receivable ₹ 100
Solution:
Profit and loss A/c for the year ending 31^st Mar 2016

Question 15.
From the trial balance of Sumathi and the adjustments prepare the trading and profit and loss account for the year ended 31^st March 2016, and a balance sheet as on that date.

Adjustments:
a) Six months interest on the loan is outstanding.
b) Two months rent is due from the tenant, the monthly rent being ₹ 25.
c) Salary for the month of March 2016, ₹ 75 is unpaid.
d) Stock in hand on March 31, 2016, was valued at ₹ 1,030.
Solution:
Profit and loss A/c of Sumathi as 31^st Dec 2016

Balance sheet as on 31^st Dec 2016

Question 16.
The following trial balance was extracted from the books of Arun Traders as of 31^st March 2018.

Prepare trading and profit and loss account for the year ending 31^st March 2018 and balance sheet as of that date after considering the following:
a) Depreciate Plant and machinery @ 20%
b) Wages outstanding amounts to ₹ 750.
c) Half of the repairs and maintenance paid is for the next year.
d) Closing stock was valued at ₹ 15,000.
Solution:
Trading and loss account for the year ended 31^st March 2018

Balance sheet for the year ending 31^st March 2016.

Question 17.
The following is the trial balance of Babu as of 31^st December 2016.

Prepare trading and profit and loss accounts for the year ended 31^st December 2016 and a balance sheet as on that date after the following adjustments.
a) Salaries outstanding ₹ 500
b) Interest on investments receivable at 10%.
c) Provision required for bad debts is 5%.
d) Closing stock is valued at ₹ 9,900.
Solution:
Profit and loss A/c

Balance sheet for the year ended 31^st December 2016.

Question 18.
From the following trial balance of Ramesh as of 31^st March 2017, prepare the trading and profit and loss account and the balance sheet as of that date.

a) Closing stock was valued at ₹ 35,000
b) Unexpired advertising ₹ 250
c) Provision for bad and doubtful debts is to be increased to ₹ 3,000
d) Provide 2% for discount on debtors :
Solution:
Trading profit and loss account of Mrs. Ramesh for the year ended 31^st March, 2017

Balance sheet for the year ended 31^st March 2016

Question 19.
Following are the ledger balances of Devi as on 31^st December, 20:6

Prepare trading and profit and loss accounts for the year ended 31^st December 2016 and balance sheet as on that date.
a) Stock on 31^st December 2016 ₹ 5,800.
b) Write off bad debts ₹ 500.
c) Make a provision for bad debts @ 5%.
d) Provide for discount on debtors @ 2%.
Solution:
Trading and profit and loss account of Mrs. Devi for the year ended 31^st Dec 2016

The balance sheet of Mrs. Devi for the year ended 31^st December 2016

Question 20.
From the following trial balance of Mohan for the year ended 31^st March 2017 and additional information, prepare trading and profit and loss account and balance sheet.

Additional information:
a) Closing stock is valued at ₹ 15,500
b) Write off t 500 as bad debts and create a provision for bad debts @ 10% on debtors.
c) Depreciation @ 10% required
Solution:
Trading and profit and loss account of Mr. Mohan for the year ended 31^st March 2017

Balance sheet for the year ended 31^st March 2017

Question 21.
From the following t I balance of Subramaniam, prepare his trading and profit and loss account and balance sheet as on 31^st December, 2016.

Take into account the following adjustments:
a) Charge interest on drawings at 8%.
b) Outstanding salaries ₹ 3,000
c) Closing stock was valued at ₹ 48,000
d) Provide for 5% interest on capital.
Solution:
Trading and profit and loss A/c of Mrs. Subramanian for the year ended 31^st Dec.2016

Balance sheet for the ended 31^st December 2016.

Question 22.
Prepare trading and profit and loss account and balance sheet from the following trial balance of Madan as of 31^st March, 2018

Adjustments
a) The closing stock was ₹ 80,000
b) Provide depreciation on plant and machinery @ 20%
c) Write off ₹ 800 as further bad debts
d) Provide the doubtful debts @ 5% on sundry debtors
Solution:
Trading profit and loss account of Mrs.Madan for the year ended 31^st March 2018.

Balance sheet for the year ended 31^st March 2018.

Question 23.
From the following information prepare trading and profit and loss account and balance sheet of Kumar for the year ending 31^st December 2017.

Adjustments
a) The closing stock on 31^st December 2017 was valued at ₹ 3,900.
b) Carriage inwards prepaid ₹ 250
c) Rent received in advance ₹ 100
d) Manager is entitled to receive commission @ 5% of net profit after providing such commission
Solution:
Trading profit and loss a/c of Mrs.Kumar for the ended 31^st December 2017

Balance sheet for the year ended 31^st December 2017.

Question 24.
From the following information, prepare a trading and profit and loss account and balance sheet in the books of Sangeetha for the year ending 31^st March 2018.

Adjustments
a) Stock on 31^st March 2018 ₹ 14,200
b) Income tax of Sangeetha paid ₹ 800
c) Charge interest on drawings @12% p.a.
d) Provide managerial remuneration @ 10% of net profit before charging such commission.
Solution:
Trading and loss A/c of Mrs. Sangeetha for the ended 31^st March 2018

Balance sheet for the year ended 31^st March 2018.

11th Accountancy Guide Final Accounts of Sole Proprietors – II Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
If closing stock is already adjusted, adjusted purchase account and ………………. stock will appear in trial balance.
(a) Opening
(b) Closing
(c) Average
(d) None of these
Answer:
(b) Closing

Question 2.
The unsold goods in the business at the end of the accounting period are termed as __________.
a) Opening stock
b) Closing stock
c) Average stock
d) None of these
Answer:
b) Closing stock

Question 3.
When bad debts already appear in the trial balance, it is taken only to the debit side of ………………. account.
(a) Profit and Loss
(b) Balance sheet
(c) Asset side
(d) None of these
Answer:
(a) Profit and Loss

Question 4.
Unexpired expenses are also called as __________.
a) Outstanding expenses
b) Prepaid expenses
c) Accrued income
d) Income received in advance
Answer:
b) Prepaid expenses

Question 5.
All the items given in the adjustment will appear at __________ in the final account.
a) Four place
b) Three place
c) Two place
d) One place.
Answer:
c) Two place

Question 6.
The Trial balance shows bank loan ₹ 5,00,000 at 12% on 01-01-2017. Interest paid ₹ 40,000. Interest outstanding is ₹ __________ as on 31-12-2018.
a) 40,000
b) 5,40,000
c) 4,60,000
d) 20,000
Answer:
d) 20,000

Question 7.
As per Trial balance, capital as on 31.03,2018 is ₹ 5,00,000. Provide 6% Interest on capital ₹ __________.
a) 10,000
b) 20,000
c) 30,000
d) 40,000
Answer:
c) 30,000

Question 8.
Trial balance as of 31.12.2017. Shows sundry debtors t 55,000. As per given adjustments, if ₹ 5,000 is to be written off as bad debts, the provision for bad and doubtful debts at 2% will be ₹ __________.
a) 1,000
b) 1,100
c) 1,200
d) 1,300
Answer:
a) 1,000

Question 9.
The Trial balance shows Investment ₹ 3,00,000 at 10% on 61-01-2017. Income received ₹ 20,000. Accrued Interest on investment, yet to be received is ₹ __________ as on 31-12-2018.
a) 30,000
b) 10,000
c) 50,000
d) 25,000
Answer:
b) 10,000

Question 10.
__________ contributed by proprietor is a liability to the business.
a) Drawings
b) Profit
c) Loss
d) Capita
Answer:
d) Capita

Question 11.
__________ represents the number of goods withdrawn by the proprietor front the business for his personal use.
a) Drawings
b) Purchases
c) Sales
d) Capital
Answer:
a) Drawings

Question 12.
The decrease in book value of fixed assets due to usage or passage of time is called __________.
a) Bad debts
b) Depreciation
c) Closing stock
d) Capital
Answer:
b) Depreciation

Question 13.
Debts Which cannot be recovered or irrecoverable debts are called __________.
a) Bad debts
b) Depreciation
c) Closing stock
d) Capita
Answer:
a) Bad debts

Question 14.
Prepaid insurance is __________.
a) an asset
b) a liability
c) an income
d) an expense
Answer:
a) an asset

Question 15.
Wages outstanding is __________.
a) an asset
b) a liability
c) an income
d) an expense
Answer:
b) a liability

II. Short Answer Questions

Question 1.
What is closing stock?
Answer:
The unsold goods in the business at the end of the accounting period are termed dosing stock. As per AS-2 (Revised), the stock is valued at cost price or net realizable value, whichever is the tower.

Question 2.
What is Income received in advance?
Answer:
Income received in advance refers to income or portion of income received in an accounting year which is not earned in the accounting period. It is also known as unearned income or unexpired income. Though the amount is received in the current accounting year, the benefit is yet to be offered to the concerned person in the next accounting year.

Question 3.
What is Interest in the capital?
Answer:
According to the separate entity concept, business and proprietor are two separate entities. Capital contributed by the proprietor is a liability to the business. Hence, interest may be provided on capita contributed by the proprietor. It is treated as a business expense. The purpose is to know the true profit of the business.

Question 4.
What is Interest in drawings?
Answer:
Drawings represent the number of goods withdrawn by the proprietor from the business for his personal use. As business is separate from owner, interest charged on drawings, if any, is to be treated as business income.

Question 5.
What is Interest on a loan?
Answer:
Business entities may have loans borrowed from banks and other financial institutions, private money lenders, etc. If any interest is payable on the loan and not yet provided at the time of preparation of the trial balance, it is necessary to provide for outstanding interest on the loan. It is an outstanding expense.

Question 6.
What is Interest in investment?
Answer:
Business entitles may have investments in outside securities carrying a specified rate of interest. If interest is due but not yet received, adjustment is to be made for the same in the accounting records before preparation of final accounts. Interest receivable on any Investments in the form of shares, deposits, etc made outside the business is called accrued interest It is an accrued income.

Question 7.
What is Depreciation?
Answer:
The decrease in book value of fixed assets due to usage or passage of time is called depreciation. It is a loss to the business. Therefore, it must be written off from the value of asset. Generally, a certain percentage on the value of the asset is calculated as the amount of depreciation.

Question 8.
What is bad debt?
Answer:
In other words, debts which cannot be recovered or irrecoverable debts are called bad debts. It is a loss for the business and should be charged against profit.

Question 9.
Write notes on Provision for bad and doubtful debts?
Answer:
1. Provision for bad and doubtful debts refers to the amount set aside as a charge against profit to meet any loss arising due to bad debt in the future. At the end of the accounting period, there may be certain debts which are doubtful, i.e., the amount to be received from debtors may or may not be received.

2. In general, based on past experience, the amount of doubtful debts is calculated on the basis of some percentage on debtors at the end of the accounting period after deducting further bad debts.

Question 10.
Write notes on Provision for discount on creditors?
Answer:
Similar to the cash discount allowed to debtors, the firm may have a chance to receive the cash discount from the creditors for prompt payment. Provision for discount on Creditors is calculated at a certain percentage on Sundry Creditors.

III. Additional Sums

Question 1.
From the following Trial Balance prepare the Trading and Profit and Loss Account for the year ended 31^st December 2017 and a Balance Sheet as of that date

Adjustments
i. Closing Stock ₹ 9,800
ii. Salaries outstanding ₹ 600
iii. Rent paid in advance ₹ 400
iv. Depreciate Machinery at 10%
v. Bad debts written off ₹ 500
Solution:
Trading, Profit, and Loss Account for the year ended 3 .12.2017

Balance Sheet as on 31.12.2017

Question 2.
From the following trial balance of Shri, prepare Trading and Profit and Loss account for the year ended December 31^st, 2017, and a balance sheet as on that date.
Adjustments
1. Stock on 31.12.2017 ₹ 4,900
2. Salaries unpaid ₹ 300
3. Rent paid in advance ₹ 200
4. Insurance prepaid ₹ 90

Solution:
Trading, Profit and loss Account for the year ended 31.12.2017

Balance Sheet as on 31.12.2017

Question 3.
From the following trial balance prepare trading and profit and loss account for the year ended 2016 and balance sheet as on that data

Adjustments
1. Closing Stock ₹ 9,800
2. Salaries Outstanding ₹ 600
3. Rent paid in advance ₹ 400
4. Depreciate Machinery at 10%
5. Bad debts are written off ₹ 500
Solution:
Trading, Profit, and Loss Account for the year ended 31.12.2017

Balance Sheet as on 31.12.2017

Question 4.
Pass necessary adjusting entries for the following adjustments.

Solution:
Adjustment Entries

Question 5.
Pass necessary adjusting entries for the following adjustments.

Solution:
Journal Entries

Question 6.
The Trial balance as of 31^st March 2004 shows the following.
Sundry debtors – 40800
Bad debts are written off – 1400
Adjustment: Written off ₹ 800/- bad debts.
Solution:
Journal Entries

Profit and loss account for the year ending 31^st March 2004

Balance sheet as on 31^st March 2004.

Question 7.
The following items are found in the Trial Balance of Mr. Vivekanandan as of 31^st March 2004.
Sundry debtors – 64000
Bad debts – 1200
Provision for bad and doubtful debts – 2800
Adjustments:
Provide for bad and doubtful debts at 5% on sundry debtors give necessary entries and show how these items will appear in the final accounts.
Solution:
Transfer of bad debts

Adjusting Entry

Profit and Loss account for the period ended 31^st March 2004

Balance sheet as on 31^st March 2004

Question 8.
Following are the balances extracted from the Trial Balance Mr. Mohan as of 31^st March 2002.

Adjustments
Create provision for bad and doubtful debts @ 5% on sundry debtors pass adjusting entry and show how these items will appear final accounts.
Solution:
Adjusting Entry

Profit and loss Account for the Year ending 31^st March 2002

Balance sheet as of 31^st March 2002

Question 9.
Prepare trading account from the following ledger balance presented by P. Sen as on 31^st March 2016

Additional Information
a) Stock on 31^st March 2016, ₹ 20,000
b) Outstanding wages amounted to ₹ 4,000
c) Gas and Fuel paid in advanced ₹ 1000
Solution:
Trading account for the year ended 31^st March 2016

Question 10.
From the following particular presented by Thilak for the year ended 31^st March 2017 Prepare profit and loss A/c

Adjustments
1. Outstanding salaries amounted to ₹ 4,000
2. Rent paid for 11 month
3. Interest due but not received amounted to ₹ 2,000
4. Prepaid insurance amounted to ₹ 2,000
5. Depreciate buildings by 10%
6. Further bad debts amounted to ₹ 3000 and make a provision for bad debts @ 5% on sundry debtors
7. Commission received in advance amounted to ₹ 2,000
Solution:
Profit and loss account for the year ended 31^st Mar 2017

Question 11.
From the following balance as of 31^st December 2017, prepare a profit and loss account

Adjustments
a) Rent accrued by not yet received ₹ 500
b) Fire insurance premium prepaid to the extent of ₹ 1,500
c) Provide manages commission at 10% on profits before charging such commission.
Solution:
Profit and loss account for the year ended 31^st December 2017

Working Note
Manager’s commission = Net profit before charaing commission x \(\frac{\text { Rate of commission }}{100}\)
Net profit = 55,500 – (18,000 + 12,000 + 8,000 + 2,500 + 5,000 ) = ₹ 10,000
Manager’s commission = 10,000 x 10/100 = ₹ 1,000

Question 12.
From the following particulars, prepare the balance sheet of Madhu, for the year ended 31^st March 2018.

Adjustment
a. Outstanding liabilities: Salaries ₹ 10,000, Wages ₹ 20,000, Interest on bank overdraft ₹ 3000 Interest on bank loan ₹ 6,000
b. Provide interest on capital @ 10% P.a
c. Bad debts amounted to ₹ 10,000 and make a provision for bad and doubtful debts 10% on sundry debtors
d. Closing stock amounted to ₹ 1,20,000
e. Depreciate vehicles @10% P.a
Net profit for the year amounted to ₹ 96,000 after considering all the above adjustments.
Solution:
Balance Sheet of Madhu as on 31^st Mar, 2018

Question 13.
The trial balance shows on 31-03-2005 sundry debtors ₹ 56,000
Adjustments
a) Bad debts to be written off ₹ 6,000
b) Provision for bad and doubtful debts by 5%
c) Provision discount on debtors at 2% pass adjusting entries and show how these items will appear in the final accounts.
Solution:
Adjusting Entry

Profit and loss A/c

Balance sheet as on 31-03-2005

Question 14.
The trial balance shows the following as on 31-03-1998
Capital ₹ 5,00,000
Drawings ₹ 50,000
Charge interest on drawings at 5% pass adjusting and transfer entry his will appear in final accounts.
Solution:
Adjusting Entry

Transfer Entry

Balance sheet as on 31, March,1998

Question 15.
The following balance has been extracted from Mrs. Padma as of 31^st March 2002.

Adjustments
1) Closing stock was valued at ₹ 80,000
2) Provide for accrued on investment ₹ 500
3) Commission received in advance ₹ 400
4) A provision for bad and doubtful debts is to be created to the extent of 5% on sundry debtors
5) Sundry creditors is to the created an extent of 2% on sundry creditors
Solution:
Trading and profit loss A/c of Mrs. Padma for the year ended 31^st March 2002

Balance sheet as on Mrs. Padma 31^st March 2002.

Question 16.
The following are the balance extracted from the books of Mrs. Nandhini as 31-03-2002

Adjustments:
1) Closing stock ₹ 40,000
2) Interest on capital 6% to be provided
3) Interest on drawings 5% to be provided
4) Wages yet to be paid ₹ 1,000
5) Rent prepaid ₹ 900
Solution:
Trading, profit loss account of Mrs. Nandhini for the year ended 31^st March 2002

The balance sheet of Mrs. Nandhini as of 31^st March 2002.

Question 17.
From the following trial balance of Thiru. Rehman as of 31^st March 2016, prepare trading profit and loss A/c Balance sheet trading to the adjustments.

Adjustments
a) Stock on 31.3.1995 was ₹ 6,800
b) Salary outstanding ₹ 1,500
c) Insurance prepaid ₹ 150
d) Depreciate machinery @ 10% and patent 20%
e) Create a provision of 2% on debtors bad debts
Solution:
Trading, profit, and loss A/c of Thiru. Rehman for the year ended 31 March 2016

Balance sheet as of Thiru Rahaman 31^st March 2016

Students can download 11th Business Maths Chapter 5 Differential Calculus Ex 5.2 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 5 Differential Calculus Ex 5.2

Samacheer Kalvi 11th Business Maths Differential Calculus Ex 5.2 Text Book Back Questions and Answers

Question 1.
Evaluate the following:

Solution:
(i) \(\lim _{x \rightarrow 2} \frac{x^{3}+2}{x+1}\)

(ii) \(\lim _{x \rightarrow \infty} \frac{2 x+5}{x^{2}+3 x+9}\)

[Takeout x from numerator and take x² from the denominator]

(iii) \(\lim _{x \rightarrow \infty} \frac{\Sigma n}{n^{2}}\)

(iv) \(\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{5 x}\)

(v) \(\lim _{x \rightarrow a} \frac{x^{\frac{5}{8}}-a^{\frac{5}{8}}}{x^{\frac{2}{3}}-a^{\frac{2}{3}}}\)

[Divide both numerator and denominator by x – a; \(\lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n}\)]

(vi) \(\lim _{x \rightarrow 0} \frac{\sin ^{2} 3 x}{x^{2}}\)

Question 2.
If \(\lim _{x \rightarrow a} \frac{x^{9}-a^{9}}{x-a}=\lim _{x \rightarrow 3}(x+6)\), find the value of a.
Solution:
\(\lim _{x \rightarrow a} \frac{x^{9}-a^{9}}{x-a}=\lim _{x \rightarrow 3}(x+6)\)
9 . a^9-1 = 3 + 6
9 . a⁸ = 9
a⁸ = 1
Taking squareroot on bothsides, we get
\(\left(a^{8}\right)^{\frac{1}{2}}\) = ±1
a⁴ = ±1
But a⁴ = -1 is imposssible.
∴ a⁴ = 1
Again taking squareroot, we get
\(\left(a^{4}\right)^{\frac{1}{2}}\) = ±1
a² = ±1
a² = -1 is imposssible
∴ a² = 1
Again taking positive squareroot, a = ±1

Question 3.
If \(\lim _{x \rightarrow 2} \frac{x^{n}-2^{n}}{x-2}=448\), then find the least positive integer n.
Solution:

Question 4.
If f(x) = \(\frac{x^{7}-128}{x^{5}-32}\), then find \(\lim _{x \rightarrow 2} f(x)\)
Solution:
\(\lim _{x \rightarrow 2} f(x)\)

Question 5.
Let f(x) = \(\frac{a x+b}{x+1}\), if \(\lim _{x \rightarrow 0} f(x)=2\) and \(\lim _{x \rightarrow \infty} f(x)=1\), then show that f(-2) = 0
Solution:


Hence Proved.

Read More:

DEEPAKNTR Pivot Point Calculator

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 15 Environmental Chemistry Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 15 Environmental Chemistry

11th Chemistry Guide Environmental Chemistry Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
The gaseous envelope around the earth is known as atmosphere. The region lying between an altitudes of 11 – 50 km is _______.
(a) Troposphere
(b) Mesosphere
(c) Thermosphere
(d) Stratosphere
Answer:
(d) Stratosphere

Question 2.
Which of the following belongs to secondary air pollutant?
(a) Hydrocarbon
(b) Peroxy acetyl nitrate
(c) Carbon monoxide
(d) Nitric oxide
Answer:
(a) Hydrocarbon

Question 3.
Which of the following is natural and human disturbance in ecology?
(a) Forest fire
(b) Floods
(c) Acid rain
(d) Green house effect
Answer:
(b) Floods

Question 4.
Bhopal Gas Tragedy is a case of _______.
(a) thermal pollution
(b) air pollution
(c) nuclear pollution
(d) land pollution
Answer:
(c) nuclear pollution

Question 5.
Haemoglobin of the blood forms carboxy haemoglobin with
(a) Carbon dioxide
(b) Carbon tetra chloride
(c) Carbon monoxide
(d) Carbonic acid
Answer:
(c) Carbon monoxide

Question 6.
Which sequence for green house gases is based on GWP?
(a) CFC > N₂O > CO₂ > CH₄
(b) CFC > CO₂ > N₂O > CH₄
(c) CFC > N₂O > CH₄ > CO₂
(d) CFC > CH₄ > N₂O > CO₂
Answer:
(b) CFC > CO₂ > N₂O > CH₄

Question 7.
Photo chemical smog formed in congested metropolitan cities mainly consists of
(a) Ozone, SO₂ and hydrocarbons
(b) Ozone, PAN and NO₂
(c) PAN, smoke and SO₂
(d) Hydrocarbons, SO₂ and CO₂
Answer:
(d) Hydrocarbons, SO₂ and CO₂

Question 8.
The pH of normal rain water is
(a) 6.5
(b) 7.5
(c) 5.6
(d) 4.6
Answer:
(c) 5.6

Question 9.
Ozone depletion will cause
(a) forest fires
(b) eutrophication
(c) bio magnification
(d) global warming
Answer:
(a) forest fires

Question 10.
_______ is considered to be ozone friendly substitute for CFC’S
(a) HFC (Hydro Fluro Carbon)
(b) Halons
(c) PAN (Peroxy acetyl nitrate)
(d) PAH (Polycyclic aromatic hydrocarbon)
Answer:
(d) PAH (Polycyclic aromatic hydrocarbon)

Question 11.
Identify the wrong statement in the following.
(a) The clean water would have a BOD value of less than 5 ppm
(b) Greenhouse effect is also called Global warming
(c) Minute solid particles in air are known as particulate pollutants
(d) Biosphere is the protective blanket of gases surrounding the earth
Answer:
(c) Minute solid particles in air are known as particulate pollutants

Question 12.
Living in the atmosphere of CO is dangerous because it
(a) combines with O₂ present inside to form CO₂
(b) Reduces organic matter of tissues
(c) Combines with hemoglobin and makes it incapable to absorb oxygen
(d) Dries up the blood
Answer:
(c) Combines with hemoglobin and makes it incapable to absorb oxygen

Question 13.
World Ozone layer protection Day is celebrated in ________.
(a) June 5
(b) Nov – 19
(c) Sep – 16
(d) Jan – 26
Answer:
(c) Sep – 16

Question 14.
Release of oxides of nitrogen and hydrocarbons into the atmosphere by motor vehicles is prevented by using
(a) grit chamber
(b) scrubbers
(c) trickling filters
(d) catalytic converters
Answer:
(b) scrubbers

Question 15.
Biochemical oxygen Demand value less than 5 ppm indicates a water sample to be
(a) highly polluted
(b) poor in dissolved oxygen
(c) rich in dissolved oxygen
(d) low COD
Answer:
(d) low COD

Question 16.
Match the List I with List-II and select the correct answer using the code given below the lists:

List I	List II
A. Depletion of the ozone layer	1. CO2
B. Acid rain	2. NO
C. Photochemical smog	3. SO2
D. Greenhouse effect	4. CFC

Code:

Answer:
(a)

Question 17.
Match List I with List-II and select the correct answer using the code given below the lists.

List I	List II
A. Stone leprosy	1. CO
B. Biological magnification	2. Greenhouse gases
C. Global warming	3. Acid rain
D. Combination with hemoglobin	4. DDT

Code:

Answer:
(d)

Question 18.
Assertion (A):
If the BOD level of water in a reservoir is more than 5 pm it is highly polluted.
Reason (R):
High biological oxygen demand means the high activity of bacteria in water.
(i) Both (A) and (R) are correct and (R) is the correct explanation of (A)
(ii) Both (A) and (R) are correct and (R) is not the correct explanation of (A)
(iii) Both (A) and (R) are not correct
(iv) (A) is correct but (R) is not correct
(a) i
(b) ii
(c) iii
(d) iv
Answer:
(a) i

Question 19.
Assertion (A):
Excessive use of chlorinated pesticides causes soil and water pollution.
Reason (R):
Such pesticides are non – biodegradable.
(i) Both (A) and (R) are correct and (R) is the correct explanation of (A)
(ii) Both (A) and (R) are correct and (R) is not the correct explanation of (A)
(iii) Both (A) and (R) are not correct
(iv) (A) is correct but (R) is not correct
(a) i
(b) ii
(c) iii
(d) iv
Answer:
(a) i

Question 20.
Assertion (A):
Oxygen plays a key role in the troposphere
Reason (R):
The troposphere is not responsible for all biological activities
(i) Both (A) and (R) are correct and (R) is the correct explanation of (A)
(ii) Both (A) and (R) are correct and (R) is not the correct explanation of (A)
(iii) Both (A) and (R) are not correct
(iv) (A) is correct but (R) is not correct
(a) i
(b) ii
(c) iii
(d) iv
Answer:
(b) ii

II. Write brief answer to the following questions:

Question 21.
Dissolved oxygen in water is responsible for aquatic life. What processes are responsible for the reduction in dissolved oxygen in water?
Answer:

• Organic matter such as leaves, grass, trash can pollute water. Microorganisms present in water can decompose this organic matter and consume dissolved oxygen in water.
• Eutrophication is a process by which water bodies receive excess nutrients that stimulates excessive plant growth. This enhanced plant growth in water bodies is called algal bloom.
• The growth of algae in extreme abundance covers the water surface and reduces the oxygen concentration in water. Thus, bloom-infeded water inhibits the growth of others. living organisms in the water body.
• This process in which the nutrient-rich water support a dense plant population kills animal life by depriving it of oxygen and results in loss of biodiversity is known as eutrophication.

Question 22.
What would happen, if the greenhouse gases were totally missing in the earth’s atmosphere?
Answer:
The solar energy radiated back from the earth’s surface is absorbed by the greenhouse gases. (CO₂, CH₄, O₃, CFCs) are present near the earth’s surface. They heat up the atmosphere near the earth’s surface and keep it warm.

As a result of these, there is the growth of vegetation that supports life. In the absence of this effect, there will be no life of both plant and animal on the surface of the earth.

Question 23.
Define smog.
Answer:

• Smog is a combination of smoke and fog which form droplets that remain suspended in the air.
• Smog is a chemical mixture of gases that forms a brownish-yellow haze. It mainly consists of ground-level ozone, oxides of nitrogen, volatile organic compounds, SO₂, acidic aerosols and some other gases.

Question 24.
Which is considered to be the earth’s protective umbrella? Why?
Answer:
At high altitudes to the atmosphere consists of a layer of ozone (O₃) which acts as an umbrella or shield for harmful UV radiations. It protects us from harmful effects such as skin cancer. UV radiation can convert molecular oxygen into ozone as shown in the following reaction.
O₂(g) O(g) + O(g)
O(g) + O₂(g) O₃(g)

Question 25.
What are degradable and non – degradable pollutants?
Answer:
The pollutants are classified as bio-degradable and non-biodegradable pollutants.

Bio-degradable pollutants:
The pollutants which can be easily decomposed by the natural biological processes are called bio-degradable pollutants.
Example:
plant wastes, animal wastes, etc.

Non-bio-degradable pollutants:
The pollutants which cannot be decomposed by the natural biological processes are called Non-bio-degradable
pollutants.
Examples:
Metal wastes (mainly Hg and Pb), D.D.T, plastics, nuclear wastes, etc.,
These pollutants are harmful to living organisms even in low concentrations. As they are not degraded naturally, it is difficult to eliminate them from our environment.

Question 26.
From where does ozone come in the photochemical smog?
Answer:
NO₂ NO + (O)
O₃ are strong oxidizing agent and can react with unburnt hydrocarbons in polluted air to form formaldehyde, acrolein and peroxy acetyl nitrate (PAN).

Question 27.
A person was using water supplied by corporation. Due to shortage of water he started using underground water. He felt laxative effect. What could be the cause?
Answer:
Drinking water containing moderate level of sulphatcs is harmless. But excessive concentration (>500 ppm) of suiphates in drinking water causes laxative effect.

Question 28.
What is green chemistry?
Answer:
Efforts to control environmental pollution resulted in development of science for synthesis of chemicals favorable to environment which is called green chemistry. Green chemistry means science of environmentally favorable chemical synthesis.

Question 29.
Explain how does greenhouse effect cause global warming.
Answer:
Greenhouse effect may be defined as the heating up of the earth surface due to trapping of infrared radiations reflected by earth’s surface by CO₂ layer in the atmosphere”. The heating up of earth through the greenhouse effect is called global warming.

Without the heating caused by the greenhouse effect, Earth’s average surface temperature would be only about -18 °C (CPF). Although the greenhouse effect is a naturally occurring phenomenon, it is intensified by the continuous emission of greenhouse gases into the atmosphere.

During the past 100 years, the amount of carbon dioxide in the atmosphere increased by roughly 30 percent and the amount of methane more than doubled. If these trends continue, the average global temperature will increase which can lead to melting of polar ice caps and flooding of low lying areas. This will increase incidence of infectious diseases like dengue, malaria etc.

Question 30.
Mention the standards prescribed by BIS for quality of drinking water.
Answer:
Standard characteristics prescribed for deciding the quality of drinking water by BIS, in 1991 are shown in Table.

Question 31.
How does classical smog differ from photochemical smog?
Answer:
Classical smog:

• Classical smog is caused by coal smoke and fog.
• It occurs in cold humid climate.
• The chemical composition is the mixture of SO₂, SO₃ gases and humidity.
• Chemically it is reducing in nature because of high concentration of SO, and so it is also called reducing smog.
• It is primarily responsible for acid rain.
• It also causes bronchial irritation.

Photochemical smog:

• Photochemical smog is cause by photochemical oxidants.
• It occurs in warm, dry and sunny climate.
• The chemical composition is the mixture of NO₂ and O₃ gases.
• Chemically it is oxidising in nature because of high concentration of oxidising agents such as NO₂ and O₃
• and so it is also called oxidising smog.
• It causes irritation to eyes, skin and lungs and increase the chances of asthma.
• It causes corrosion of metals, stones

Question 32.
What are particulate pollutants? Explain any three.
Answer:
1. Particulate pollutants are small solid particles and liquid droplets suspended in air. Many of particulate pollutants are hazardous.
Examples: dust, pollen, smoke, soot and liquid droplets (aerosols) etc,

2. Smoke particulate consists of solid particles (or) mixture of solid and liquid particles formed by the combustion of organic matter.
For example, cigarette smoke, oil smoke, smokes from burning fossil fuel, garbage and dry leaves.

3. Dust composed of fine solid particles produced during crushing and grinding of solid materials.
For example, sand from sandblasting, sawdust from woodworks, cement dust from cement factories and fly ash from power generating units.

Question 33.
Even though the use of pesticides increases crop production, they adversely affect the living organisms. Explain the function and the adverse effects of the pesticides.
Answer:
Pesticides are the chemicals that are used to kill or stop the growth of unwanted organisms. But these pesticides can affect the health of human beings. Pesticides are classified as
(a) insecticides,
(b) Fungicides and
(c) Herbicides.

(a) Insecticides:
Insecticides like DDT, BHC, Aidrin can stay in soil for a long period of time and are absorbed by soil. They contaminate root crops like carrot, radish.

(b) Fungicides:
Organomercury compounds dissociate in soil to produce mercury which is highly toxic.

(c) Herbicides:
They are used to control unwanted plants and are also known as weed killers. Eg, Sodium chlorate, sodium nitrate. They are toxic to mammals.

Question 34.
Ethane burns completely in air to give CO₂, while in a limited supply of air gives CO. The same gases are found in automobile exhaust. Both CO and CO₂ are atmospheric pollutants.
Answer:
The major pollutants of oxides of carbon are carbon monoxide and carbon dioxide.

(i) Carbon Monoxide:
Carbon monoxide is a poisonous gas produced as a result of incomplete combustion of coal are firewood. It is released into the air mainly by. automobile exhaust. It binds with haemoglobin and form carboxy haemoglobin which impairs normal oxygen transport by blood and hence the oxygen carrying capacity of blood is reduced. This oxygen deficiency results in headache, dizziness, tension, Loss of consciousness, blurring of eye sight and cardiac arrest.

(ii) Carbon dioxide:
Carbon dioxide is released into the atmosphere mainly by the process of respiration, burning of fossil fuels, forest fire, decomposition of limestone in cement industry etc.

Green plants can convert CO₂ gas in the atmosphere into carbohydrate and oxygen through a process called photosynthesis. The increased CO₂ level in the atmosphere is responsible for global warming. It causes headache and nausea.

Question 35.
On the basis of chemical reactions involved, explain how do CFC’s cause depletion of ozone layer in stratosphere?
Answer:
In the presence of uv radiation, CFC’s break up into chlorine free radical
CF₂Cl₂ CF₂Cl + Cl
CFCl₃ CFCl₂ + Cl
Cl + O₃ → ClO + O₂
ClO + O → Cl + O₂

Chlorine radical is regenerated in the course of reaction. Due to this continuous attack of Cl thinning of Ozone layer takes place which leads to formation of the ozone hole.

It is estimated that for every reactive chlorine atom generated in the stratosphere 1,00,000 molecules of ozone are depleted.

Question 36.
How is acid rain formed? Explain its effect.
Answer:
1. Rainwater has a pli of 5.6 due to the dissolution of CO., into it. Oxides of sulphur and nitrogen in the atmosphere may be absorbed by droplets of water that make up the clouds and get chemically converted into sulphuric acid and nitric acid. Due to this, the pH of rainwater drops below the level of 5.6. Hence it is called acid rain.

2. Acid rain is a by-product of sulphur and Nitrogen oxides in the atmosphere. Burning of fossil fuels in power stations, furnaces and petrol, diesel in motor engines produce SO₂ and NO₂ gases. They are converted into H₂SO₄ and HNO₃ by the reaction with oxygen and water.

3. 2SO₂ + O₃ + 2H₂O → 2H₂SO₄
4NO₂ + O₂+ 2H₂O → 4HNO₃

Harmful effects of acid rain:
1. Acid rain causes damage to buildings made us of marbles. This attack on marble is termed as stone leprosy.
CaCO₃ + H₂SO₂ CaSO₄ + H₂O + CO₂↑

2. Acid rain affects plant and animal life in aquatic ecosystem.

3. It is harmful For agriculture, as it dissolves in the earth and removes the nutrients needed for the growth of plants.

4. It corrodes water pipes resulting in the leaching of heavy metals such as iron, lead and copper into drinking water which has toxic effects.

5. it causes respiratory ailments in humans and animals.

Question 37.
What is sewage? What are the major steps involved in the treatment of sewage waste?
Answer:
Objectives of wastewater treatment:

1. To convert harmful compounds into harmless compounds.
2. To eliminate the offensive smell.
3. To remove the solid content of the sewage.
4. To destroy the disease-producing microorganisms.

Treatment process:
The sewage (or) wastewater treatment process involves the following steps.

I. Preliminary Treatment:
In this treatment, coarse solids and suspended impurities are removed by passing the wastewater through bar and mesh screens.

II. Primary treatment (or) Settling process:
In this treatment, greater proportion of the suspended inorganic and organic solids are removed from the liquid sewage by settling. In order to facilitate quick settling coagulants like alum, ferrous sulphate are added. These produce large gelatinous precipitates, which entrap finely divided organic matter and settle rapidly.
Al₂(SO₄)₃ + 6H₂O → 2Al(OH)₃ ↓ + 3H₂SO₄

III. Secondary (or) biological treatment:
In this treatment, biodegradable organic impurities are removed by aerobic bacteria. It removes upto 90 % of the oxygen-demanding wastes. This is done by trickling filter or activated sludge process.

(a) Trickling filter process:
It is a circular tank and is filled with either coarse or crushed rock. Sewage is sprayed over this bed by means of slowly rotating arms.

When sewage starts percolating downwards, microorganisms present in the sewage grow on the surface of filtering media using organic material of the sewage as food. After completion of aerobic oxidation the treated sewage is taken to the settling tank and the sludge is removed. This process removes about 80 – 85 % of BOD.

(b) Activated sludge process:

Activated sludge is biologically active sewage and it has a large number of aerobic bacterias, which can easily oxidize the organic impurities.

The sewage effluent from primary treatment is mixed with the required amount of activated sludge. Then the mixture is aerated in the aeration tank. Under this condition, organic impurities of the sewage get oxidized rapidly by the microorganisms.

After aeration, the sewage is taken to the sedimentation tank. Sludges settle down in this tank, called activated sludge, a portion of which is used for seeding fresh batch of the sewage. This process removes about 90-95 % of BOD.

IV. Tertiary treatment:
After the secondary treatment, the sewage effluent has a lower BOD (25 ppm), which can be removed by the tertiary treatment process.

In the tertiary treatment, the effluent is introduced into a flocculation tank, where lime is added to remove phosphates. From the flocculation tank the effluent is led to ammonia stripping tower, where pH is maintained to 11 and the NH₄⁺ is converted to gaseous NH₃. Then the effluent is allowed to pass through activated charcoal column, where minute organic wastes are absorbed by charcoal. Finally the effluent water is treated with disinfectant (chlorine).

V. Disposal of sludge:
This is the last stage in the sewage treatment. Sludge formed from different steps can be disposed by

1. dumping into low – lying areas,
2. burning of sludge (incineration),
3. dumping into the sea,
4. using it as low grade fertilizers.

Question 38.
Differentiate the following:
(i) BOD and COD
(ii) Viable and non-viable particulate pollutants.
Answer:
(i) BOD and COD Biochemical oxygen demand (BOD):
The total amount of oxygen in milligrams consumed by microorganisms in decomposing the waste in one litre of water at 200°C for a period of 5 days is called biochemical oxygen demand (BOD) and its value is expressed in ppm. BOD is used as a measure of degree of water pollution. Clean water would have BOD value less than 5 ppm whereas highly polluted water has BOD value of 17 ppm or more.

Chemical Oxygen Demand (COD):
BOD measurement takes 5 days so another parameter called the Chemical Oxygen Demand (COD) is measured. Chemical oxygen demand (COD) is defined as the amount of oxygen required by the organic matter in a sample of water for its oxidation by a strong oxidising agent like K₂Cr₂O₇ in acid medium for a period of 2 hrs.

(ii) Viable and non – viable particulate pollutants:

Viable particulates:
The viable particulates are the small size living organisms such as bacteria, fungi, moulds, algae, etc. which are dispersed in air. Some of the fungi cause allergy in human beings and diseases in plants.

Non-viable particulates:
The non- viable particulates are small solid particles and liquid droplets suspended in air. They help in the transportation of viable particles. There are four types of non-viable particulates in the atmosphere.
Example:
Smoke, Dust, Mists, Fumes.

Question 39.
Explain how oxygen deficiency is cause by carbon monoxide in our blood? Give its effect.
Answer:

1. Carbon monoxide binds with haemoglobin and form carboxyhemoglobin which impairs normal oxygen transport by blood and hence the oxygen-carrying capacity of blood is reduced.
2. This oxygen deficiency results in headache, dizziness, tension, loss of consciousness, blurring of eyesight and cardiac arrest.

Question 40.
What are the various methods you suggest to protect our environment from pollution?
Answer:
Methods to control environmental pollution:

• Waste management Environmental pollution can be controlled by proper disposal of wastes.
• Recycling A large number of disposed waste materials can be reuse by recycling the waste, thus it reduces the landfill.
• By substitution of less toxic solvents for highly toxic ones are used in industrial processes.
• By growing more trees.
• By using fuels with lower sulphur content.
• By control measures in vehicle emissions which are adequate.

11th Chemistry Guide Environmental Chemistry Additional Questions and Answers

I. Choose the best answer:

Question 1.
Which one of the following is a bio-degradable pollutant?
(a) DDT
(b) Plastics
(c) Mercury
(d) Wood
Answer:
(d) Wood

Question 2.
The greenhouse effect is caused by
(a) CO₂
(b) NO₂
(c) NO
(d) CO
Answer:
(a) CO₂

Question 3.
Which of the following pair of oxides is responsible for acid rain?
(a) SO₃ + NO₂
(b) CO₂ + CO
(c) N₂O + CH₄
(d) O₂ + H₂
Answer:
(a) SO₃ + NO₂

Question 4.
In Antarctica ozone depletion is due to the formation of the following compound
(a) acrolein
(b) peroxyacetyl nitrate
(c) SO₂ and NO₂
(d) chlorine nitrate
Answer:
(a) acrolein

Question 5.
Which of the following is not a greenhouse gas’?
(a) CO
(b) O₃
(c) CH₄
(d) Water vapours
Answer:
(a) CO

Question 6.
Classical smog occurs in places of
(a) excess SO₂
(b) low temperature
(c) high temperature
(d) excess NH₃
Answer:
(b) low temperature

Question 7.
Which gas is responsible for ‘Bhopal Gas Tragedy in 1984?
(a) CO
(b) Methyl isocynate
(c) SO₂ and NO₂
(d) Ethyl isocynate
Answer:
(b) Methyl isocynate

Question 8.
Which gas is a main reason behind air pollution, is produced by
(a) sewage pollutant
(b) aerosols
(c) industrial remains
(d) Above all
Answer:
(b) aerosols

Question 9.
Which is a dangerous radiological pollutant?
(a) C¹⁴
(b) S³⁵
(c) Sr⁹⁰
(d) P³²
Answer:
(c) Sr⁹⁰

Question 10.
Which is related to the ‘Green House Effect’?
(a) Farming of Green Plants
(b) Farming of Vegetables in Houses
(c) Global Warming
(d) Biodegradable pollutant
Answer:
(c) Global Warming

Question 11.
The uppermost region of the atmosphere is called
(a) Ionosphere
(b) Mesosphere
(c) Troposphere
(d) Stratosphere
Answer:
(d) Stratosphere

Question 12.
Which of the following is the coldest region of the atmosphere?
(a) Thermosphere
(b) Mesosphere
(c) Troposphere
(d) Stratosphere
Answer:
(b) Mesosphere

Question 13.
The region which is greatly affected by air pollution is
(a) Thermosphere
(b) Stratosphere
(c) Troposphere
(d) Mesosphere
Answer:
(c) Troposphere

Question 14.
The substance which is a primary pollutant?
(a) H₂SO₄
(b) CO
(c) PAN
(d) Aldehydes
Answer:
(b) CO

Question 15.
Depletion of ozone layer causes
(a) breast cancer
(b) blood cancer
(c) lung cancer
(d) skin cancer
Answer:
(d) skin cancer

Question 16.
Formation of London smog takes place in
(a) Winter during day time
(b) summer during day time
(c) summer during morning time
(d) winter during morning time
Answer:
(d) winter during morning time

Question 17.
The substance which is not regarded as a pollutant?
(a) NO₂
(b) CO₂
(c) O₃
(d) Hydrocarbons
Answer:
(b) CO₂

Question 18.
Green house gases
(a) allow shorter wavelength to enter earth’s atmosphere while doesn’t allow longer wavelength to leave the earth’s atmosphere.
(b) allow longer wavelength to enter earth atmosphere while doesn’t allow shorter wavelength to leave the surface.
(c) don’t have wavelength-specific character.
(d) she wavelength-specific behaviour near the earth while far from earth these have wavelength-independent behavior.
Answer:
(a) allow shorter wavelength to enter the earth’s atmosphere while doesn’t allow longer wavelength to leave the earth’s atmosphere.

Question 19.
Carbon monoxide (CO) is harmful to man because
(a) it forms carbolic acid
(b) it generates excess CO₂
(c) it is carcinogenic
(d) it competes with O₂ for haemoglobin
Answer:
(d) it competes with O₂ for haemoglobin

Question 20.
Today the concentration of greenhouse gases is very high because of
(a) use of refrigerator
(b) increased combustion of oils and coal
(c) deforestation
(d) All of the above
Answer:
(d) All of the above

Question 21.
The quantity of CO₂ in atmosphere is
(a) 3.34 %
(b) 6.5 %
(c) 0.034 %
(d) 0.34 %
Answer:
(c) 0.034 %

Question 22.
BOD of pond is connected with
(a) microbes & organic matter
(b) organic matter
(c) microbes
(d) None of these
Answer:
(a) microbes & organic matter

Question 23.
When rain is accompanied by a thunderstorm, the collected rain water will have a pH value
(a) slightly lower than that of rain water without thunderstorm
(b) slightly higher than that when the thunderstorm is not there
(c) uninfluenced by occurrence of thunderstorm
(d) which depends upon the amount of dust in air
Answer:
(a) slightly lower than that of rain water without thunderstorm

Question 24.
Water pollution is caused by
(a) pesticides
(b) SO₂
(c) O₂
(d) CO₂
Answer:
(a) pesticides

Question 25.
Minamata disease of Japan is due to pollution of
(a) Aresenic
(b) Lead
(c) Cynide
(d) Mercury
Answer:
(d) Mercury

Question 26.
Which causes death of fish in water bodies polluted by sewage?
(a) Foul smell
(b) Pathogens
(c) Herbicides
(d) Decrease in D.O.
Answer:
(d) Decrease in D.O.

Question 27.
Sewage water is purified by
(a) aquatic plants
(b) microorganisms
(c) light
(d) fishes
Answer:
(b) microorganisms

Question 28.
Which pollutant is harmful for ‘Tajmahal’?
(a) Hydrogen
(b) O₂
(c) SO₂
(d) Chlorine
Answer:
(c) SO₂

Question 29.
Negative soil pollution is
(a) reduction in soil productivity due to erosion and overuse
(b) reduction in soil productivity due to the addition of pesticides and industrial wastes
(c) converting fertile land into barren land by dumping ash, sludge, and garbage
(d) None of the above
Answer:
(a) reduction in soil productivity due to erosion and overuse

Question 30.
The quantity of DDT in the food chain
(a) decreases
(b) remains same
(c) increases
(d) changes
Answer:
(c) increases

Question 31.
Which is known as “Third poison of environment” and also creates ‘Blue baby syndrome’
(a) Nitrate present in water
(b) Phosphate and detergents found in water
(c) Cyanide
(d) Pesticides
Answer:
(b) Phosphate and detergents found in water

Question 32.
Is the substance having the largest concentration in acid rain?
(a) H₂CO₃
(b) HNO₃
(c) HCl
(d) H₂SO₄
Answer:
(d) H₂SO₄

Question 33.
Water is often treated with chlorine to
(a) remove hardness
(b) increase oxygen content
(c) kill germs
(d) remove suspended particles
Answer:
(c) kill germs

Question 34.
Thermal pollution affects mainly
(a) vegetation
(b) aquatic creature
(c) rocks
(d) air
Answer:
(b) aquatic creature

Question 35.
B.O.D test or biochemical oxygen demand test is made for measuring
(a) air pollution
(b) water pollution
(c) noise pollution
(d) soil pollution
Answer:
(b) water pollution

Question 36.
Brewery and sugar factory water alters the quality of a water body by increasing
(a) temperature
(b) turbidity
(C) pH
(d) COD and BOD
Answer:
(d) COD and BOD

Question 37.
A dental disease characterized by mottling of teeth is due to the presence of a certain chemical element in drinking water. Which is that element?
(a) Boron
(b) Chlorine
(c) Fluorine
(d) Mercury
Answer:
(c) Fluorine

Question 38.
The high amount of E.coli in water is an indicator of
(a) hardness of water
(b) industrial pollution
(c) sewage pollution
(d) presence of chlorine in the water
Answer:
(c) sewage pollution

Question 39.
A lake with an inflow of domestic sewage rich in organic waste may result in
(a) drying of the lake very soon due to algal bloom
(b) an increased production of fish due to a lot of nutrients
(c) death of fish due to lack of oxygen
(d) increased population of aquatic food web organisms
Answer:
(c) death of fish due to lack of oxygen

Question 40.
In which one of the following the BOD (Biochemical Oxygen Demand) of sewage(S), distillery effluent (DE), paper mill effluent (PE) and sugar mill effluent (SE) have been arranged in ascending order
(a) SE < S < PE < DE
(b) SE < PE < S < DE
(c) PE < S < SE < DE
(d) S < DE < PE < SE
Answer:
(c) PE < S < SE < DE

Question 41.
The greenhouse effect is because of the
(a) presence of gases, which in general are strong infrared absorbers, in the atmosphere
(b) presence of CO₂ only in the atmosphere
(c) presence of O₃ and CH₄ in the atmosphere
(d) N₂O and chlorofluoro hydrocarbons in the atmosphere
Answer:
(a) presence of gases, which in general are strong infrared absorbers, in the atmosphere

Question 42.
Which of the following is/are the hazardous pollutant(s) present in automobile exhaust gases?
(a) N₂
(b) CO
(c) CH₄
(d) Oxides of nitrogen
Answer:
(c) CH₄

Question 43.
Green chemistry means such reactions which:
(a) produce colour during reactions
(b) reduce the use and production of hazardous chemicals
(c) are related to the depletion of ozone layer
(d) study the reactions in plants
Answer:
(b) reduce the use and production of hazardous chemicals

Question 44.
Which one of the following statement is not true?
(a) pH of drinking water should be between 5.5 – 9.5.
(b) Concentration of DO below 6 ppm is good for the growth of fish.
(c) Clean water would have a BOD value of less than 5 ppm.
(d) Oxides of sulphur, nitrogen and carbon are the most widespread air pollutant.
Answer:
(b) Concentration of DO below 6 ppm is good for the growth of fish.

Question 45.
Which one of the following statements regarding photochemical smog is not correct?
(a) Carbon monoxide does not play any role in photochemical smog formation.
(b) photochemical smog is an oxidizing agent in character.
(c) photochemical smog is formed through a photochemical reaction involving solar energy.
(d) Photochemical smog does not cause irritation in eyes and throat.
Answer:
(d) Photochemical smog does not cause irritation in eyes and throat.

Question 46.
Frequent occurrence of water blooms in a lake indicates
(a) nutrient deficiency
(b) oxygen deficiency
(c) excessive nutrient availability
(d) absence of herbivores in the lake
Answer:
(b) oxygen deficiency

Question 47.
The smog is essentially caused by the presence of
(a) Oxides of sulphur and nitrogen
(b) O₂ and N₂
(c) O₂ and O₃
(d) O₂ and N₂
Answer:
(a) Oxides of sulphur and nitrogen

Question 48.
Identify the wrong statement in the following.
(a) Chlorofluorocarbons are responsible for ozone layer depletion.
(b) Greenhouse effect is responsible for global warming.
(c) Ozone layer does not permit infrared radiation from the sun to reach the earth.
(d) Acid rain is mostly because of oxides of nitrogen and sulphur.
Answer:
(c) Ozone layer does not permit infrared radiation from the sun to reach the earth.

Question 49.
Identify the incorrect statement from the following.
(a) Ozone absorbs the intense ultraviolet radiation of the sun.
(b) Depletion of ozone layer is because of its chemical reactions with chlorofluoro alkanes.
(c) Ozone absorbs infrared radiation.
(d) Oxides of nitrogen in the atmosphere can cause the depletion of ozone layer.
Answer:
(c) Ozone absorbs infrared radiation

Question 50.
What is DDT among the following?
(a) Greenhouse gas
(b) A fertilizer
(c) Biodegradable pollutant
(d) Non – biodegradable pollutant
Answer:
(d) Non – biodegradable pollutant

Question 51.
The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was:
(a) Methyl isocyanate
(b) Methylamine
(c) Ammonia
(d) Phosgene
Answer:
(a) Methyl isocyanate

Question 52.
Black – foot disease is caused due to groundwater contaminated with an excess of
(a) Nitrate
(b) Fluoride
(c) Arsenic
(d) Sulphur
Answer:
(c) Arsenic

Question 53.
Exposure of an organism to UV system causes
(a) photodynamic action
(b) formation of thymidine
(c) splitting of H – bonds of DNA
(d) splitting of phosphodiester bonds
Answer:
(c) splitting of H – bonds of DNA

Question 54.
Under column – I, a list of gases that are known to have a greenhouse effect is given. Relate them to their main source selecting from the given under Column – II:

Column – I	Column – II
A. Nitrous oxide	1. Secondary pollutant from car exhausts
B. Chlorofluorocarbon (CFCs)	2. Combustion of fossil fuels, wood, etc.
C. Methane	3. Denitrification
D. Ozone (O3)	4. refrigerators, aerosol, sprays
E. Carbon dioxide	5. Cattle, rice fields, toilets

(a) A – 3, B – 4, C – 5, D – 1, E – 2
(b) A – 5, B – 1, C – 3, D – 4, E – 2
(c) A – 4, B – 5 , C – 1, D – 2, E – 3
(d) A – 1, B – 3, C – 4, D – 5, E – 2
Answer:
(a) A – 3, B – 4, C – 5, D – 1, E – 2

Question 55.
Minamata disease is a pollution-related disease results form
(a) oil spills into the sea
(b) accumulation of arsenic into the atmosphere
(c) release of industrial waste mercury into bodies of water
(d) release human organic waste into drinking water
Answer:
(c) release of industrial waste mercury into bodies of water

Question 56.
Air pollution causing photochemical oxidants production include
(a) Carbon monoxide, sulphur dioxide
(b) Nitrous oxide, nitric acid fumes, nitric oxide
(c) Ozone, peroxyacetyl nitrate, aldehydes
(d) Oxygen, chlorine, fuming nitric acid
Answer:
(c) Ozone, peroxyacetyl nitrate, aldehydes

Question 57.
Photochemical smog formed in congested metropolitan cities mainly consists of
(a) ozone, peroxyacetyl nitrate, and NO_x
(b) smoke, peroxyacetyl nitrate and SO₂
(c) hydrocarbons, SO₂ and CO₂
(d) hydrocarbons, ozone, and SO_x
Answer:
(a) ozone, peroxyacetyl nitrate, and NO_x

Question 58.
Which, one of the following statements is correct?
(a) Extensive use of chemical fertilizers may lead to eutrophication of nearby water bodies
(b) Both Azotobacter and Rhizobium fix atmospheric nitrogen in root nodules of plants
(c) Cyanobacteria such as Anabaena and Nostoc are important mobilizers of phosphates and potassium for plant nutrition in the soil
(d) At present it is not possible to grow maize without chemical fertilizers
Answer:
(a) Extensive use of chemical fertilizers may lead to eutrophication of nearby water bodies

Question 59.
Reducing the use of non-biodegradable things will contribute of
(a) Increase in O₂
(b) Cyanophycean blooms occur
(c) Depletion of O₂ layers
(d) Eutrophication
Answer:
(a) Increase in O₂

Question 60.
Which of the following metal is a water pollutant and causes sterility in human. being?
(a) As
(b) Mn
(c) Mg
(d) Hg
Answer:
(b) Mn

Question 61.
Lichens do not like to grow in cities
(a) because of absence of the right type of algae and fungi
(b) because of lack of moisture
(c) because of SO₂ pollution
(d) because natural habitat is missing
Answer:
(c) because of SO₂ pollution

Question 62.
Limit of BOD prescribed by Central pollution Control Board for the discharge of industrial and municipal wastewaters into natural surface waters is
(a) < 100 ppm
(b) < 30 ppm
(c) < 3.0 ppm
(d) < 10 ppm
Answer:
(b) < 30 ppm

Question 63.
Which one of the following pairs is mismatched
(a) Fossil fuel burning – release of CO₂
(b) Nuclear power – radioactive wastes
(c) Solar energy – The greenhouse effect
(d) Biomass burning – release of CO₂
Answer:
(c) Solar energy – The greenhouse effect

Question 64.
In a coal-fired power plant, electrostatic precipitators are installed to control the emission of
(a) SO₂
(b) NO_x
(c) SPM
(d) CO
Answer:
(c) SPM

Question 65.
The term “Biomagnification” refers to the
(a) growth of organism due to food consumption
(b) increase in population size
(c) blowing up of environmental issues by man
(d) increase in the concentration of non-degradable pollutants as they pass through the food chain
Answer:
(d) increase in the concentration of non-degradable pollutants as they pass through the food chain

Question 66.
In almost all Indian metropolitan cities like Delhi, the major atmospheric pollutant(s) is/are
(a) suspended particulate matter (SPM)
(b) oxides of sulphur
(c) carbon dioxide and carbon monoxide
(d) oxides of nitrogen
Answer:
(a) suspended particulate matter (SPM)

Question 67.
In coming years, skin-related disorders will be more common due to
(a) pollutants in air
(b) use of detergents
(c) water pollution
(d) depletion of ozone layer
Answer:
(d) depletion of ozone layer

Question 68.
Statement 1:
Inhabitants close to very busy airports are likely to experience health hazards.
Statement 2:
Sound level of jet aeroplanes usually exceeds 160 dB.
(a) Statement – 1 is True, Statement – 2 is True, Statement – 2 is a correct explanation for Statement – 1.
(b) Statement – 1 is True, Statement – 2 is True, Statement – 2 is not a correct explanation for statement – 1
(c) Statement – 1 is True, Statement – 2 is False
(d) Statement – 1 is False, Statement – 2 is True
Answer:
(a) Statement – 1 is True, Statement – 2 is True, Statement – 2 is a correct explanation for Statement – 1.

Question 69.
Statement 1:
Suspended particulate matter (SPM) is an important pollutant released by diesel vehicles.
Statement 2:
Catalytic converters greatly reduce pollution caused by automobiles.
(a) Statement – 1 is True, Statement – 2 is True, Statement -2 is a correct explanation for Statement – 1.
(b) Statement – 1 is True, Statement – 2 is True, Statement – 2 is not a correct explanation for statement – 1
(c) Statement – 1 is True, Statement – 2 is False
(d) Statement – 1 is False, Statement – 2 is True
Answer:
(b) Statement – 1 is True, Statement – 2 is True, Statement – 2 is not a correct explanation for statement – 1

Question 70.
Statement 1:
Eutrophication shows increase in productivity in water.
Statement 2:
With increasing eutrophication, the diversity of the phytoplankton increases.
(a) Statement – 1 is True, Statement – 2 is True, Statement -2 is a correct explanation for Statement – 1.
(b) Statement – 1 is True, Statement – 2 is True, Statement – 2 is not a correct explanation for statement – 1
(c) Statement – 1 is True, Statement – 2 is False
(d) Statement – 1 is False, Statement – 2 is True
Answer:
(b) Statement – 1 is True, Statement – 2 is True, Statement – 2 is not a correct explanation for statement – 1

Question 71.
Statement 1:
The main cause of the Bhopal gas tragedy was phosgene.
Statement 2:
Phosgene is a volatile liquid.
(a) Statement – 1 is True, Statement – 2 is True, Statement -2 is a correct explanation for Statement – 1.
(b) Statement – 1 is True, Statement – 2 is True, Statement – 2 is not a correct explanation for statement – 1
(c) Statement – 1 is True, Statement – 2 is False
(d) Statement – 1 is False, Statement – 2 is True
Answer:
(d) Statement – 1 is False, Statement – 2 is True

Question 72.
Statement 1:
CO₂ causes the greenhouse effect.
Statement 2:
Other gases do not show such an effect.
(a) Statement – 1 is True, Statement – 2 is True, Statement -2 is a correct explanation for Statement – 1
(b) Statement – 1 is True, Statement – 2 is True, Statement – 2 is not a correct explanation for statement – 1
(c) Statement – 1 is True, Statement – 2 is False
(d) Statement – 1 is False, Statement – 2 is True
Answer:
(c) Statement – 1 is True, Statement – 2 is False

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

Question 1.
What is called environmental pollution?
Answer:
Any undesirable change in our environment that have harmful effects on plants. animals and human beings is called environmental pollution.

Question 2.
What are pollutants?
Answer:
The substances which cause pollution of the environment are called pollutants.

Question 3.
Write the different types of atmospheric pollution.
Answer:

1. The lowest layer of the atmosphere is called the troposphere and il extends from 0 10 km from the earth surface.
2. About 80% of the mass of the atmosphere is in this layer. 3 layers are present in it. They are
   • Hydrosphere
   • Lithosphere
   • Biosphere

Question 4.
What is Air pollution?
Answer:
Any undesirable change in the air which adversely affects living organisms is called air pollution. Air pollution is limited to the troposphere and stratosphere. Air pollution is mainly due to the excessive discharge of undesirable foreign matter into the atmospheric air.

Question 5.
What are the techniques adopted to reduce particulate pollutants?
Answer:
The particulates from the air can be removed by using electrostatic precipitators, gravity settling chambers, and wet scrubbers or by cyclone collectors. These techniques are based on washing away or settling of the particulates.

Question 6.
Define soil pollution.
Answer:
Soil pollution is defined as the buildup of persistent toxic compounds, radioactive materials, chemical salts, and disease-causing agents in soils which have harmful effects on plant growth and animal health.

Question 7.
Write the effects that were caused by classical smog.
Answer:

1. Smog is primarily responsible for acid rain.
2. The smog results in poor visibility and it affects air and road transport.
3. It also causes bronchial irritation.

III. Short question and answers (3 Marks):

Question 1.
How the oxides of nitrogen pollute the atmospheric air?
Answer:
Oxides of nitrogen are produced during high temperature combustion processes, oxidation of nitrogen in air and from the combustion of fuels (coal, diesel, petrol etc.).
N₂ + O₂ 2NO
2NO + O₂ 2NO₂
NO + O₃ → NO₂ + O₂

The oxides of nitrogen are converted into nitric acid which comes, down in the form of acid rain. They also form a reddish-brown haze in heavy traffic. Nitrogen dioxide potentially damages plant leaves and retards photosynthesis. NO₂ is a respiratory irritant and it can cause asthma and lung injury. Nitrogen dioxide is also harmful to various textile fibres and metals.

Question 2.
How the hydrocarbon compounds make harmful effects on living things?
Answer:
The compounds composed of carbon and hydrogen only are called hydrocarbons. They are mainly produced naturally (marsh gas) and also by incomplete combustion of automobile fuel.

They are potential cancer-causing (carcinogenic) agents. For example, polynuclear aromatic hydrocarbons (PAH) are carcinogenic, they cause irritation in eyes and mucous membranes.

Question 3.
Explain the environmental impact of ozone depletion.
Answer:
The formation and destruction of ozone is a regular natural process, which never disturbs the equilibrium level of ozone in the stratosphere. Any change in the equilibrium level of the ozone in the atmosphere will adversely affect life in the biosphere in the following ways.

Depletion of ozone layer will allow more UV rays to reach the earth surface and layer would cause skin cancer and also decrease the immunity level in human beings. UV radiation affects plant proteins which leads to harmful mutation of cells. UV radiation affects the growth of phytoplankton, as a result ocean food chain is disturbed and even damages the fish productivity.

Question 4.
Write the causes of water pollution.
Answer:

• The compounds composed of carbon and Hydrogen only are called hydrocarbons. They are mainly produced naturally and also by incomplete combustion of automobile fuels.
• They are potential cancer-causing (carcinogenic) agents.
• For example, polynuclear aromatic hydrocarbons (PAH) are carcinogenic, they cause irritation in eyes and mucous membranes.

Question 5.
(a) Define eutrophication and pneumoconiosis.
(b) Write differences between photochemical smog and classical smog.
Answer:
(a) Eutrophication:
When the growth of algae increases in the surface of the water, dissolved oxygen in water is greatly reduced. This phenomenon is known as eutrophication. Due to this growth of fishes gets inhibited.

(b) Pueumoconiosis:
It is a disease that irritates the lungs. It causes scarring or fibrosis of the lungs.

Photochemical smog:

• It is formed as a result of the photochemical decomposition of nitrogen dioxide and chemical reactions involving hydrocarbons.
• It takes place during the dry warm season in presence of sunlight.
• It is oxidizing in nature.

Classical smog:

• It’s formed due to the condensation of SO₂ vapours on particles of carbon in cold climates.
• it is generally formed during winter when there is severe cold.
• It is reducing in nature.

Question 6.
Write the harmful effects caused by chemical water pollutants.
Answer:

1. Cadmium and mercury can cause kidney damage.
2. Lead poisoning can lead to severe damage of kidneys, liver, brain, etc. it also affects the central nervous system
3. Polychlorinated biphenyls (PCBs) cause skin diseases and are carcinogenic in nature.

Question 7.
Distinguish between BOD and COD.
Answer:

BOD	COD
1. BOD is the amount of oxygen required for the biological decomposition of organic matter present in the water.	COD is the amount of oxygen required for chemical oxidation of organic matter using some oxidizing agent like K2Cr207 and KMn04.
2. It is an important indication of the amount of organic matter present in the river water.	It is carried out to determine the pollution strength of river water.
3. Since complete oxidation occurs in an indefinite period, the reaction period is taken as 5 days at 20°C.	It is a rapid process and takes only 8 hours.

IV. Long question and answer (5 Marks):

Question 1.
Explain the different layers of the earth’s atmosphere.
Answer:
Troposphere:
The lowest layer of the atmosphere is called the troposphere and it extends from o – 10 km from the earth’s surface. About 80% of the mass of the atmosphere is in this layer. This troposphere is further divided as follows.

i) Hydrosphere:
The hydrosphere includes all types of water sources like oceans, seas, rivers, lakes, streams, underground water, polar icecaps, clouds etc. It covers about 75% of the earth’s surface. Hence the earth is called a blue planet.

ii) Lithosphere:
The lithosphere includes soil, rocks and mountains which are solid components of earth.

iii) Biosphere:
It includes the lithosphere, hydrosphere, and atmosphere integrating the living organism present in the lithosphere, hydrosphere and atmosphere.

Question 2.
How the oxides of sulphur pollute the atmospheric air?
Answer:
Sulphur dioxide and sulphur trioxide are produced by burning sulfur-containing fossil fuels arid roasting sulphide ores. Sulphur dioxide is a poisonous gas to both animals and plants. Sulphur dioxide causes eye irritation, coughing and respiratory asthma, bronchitis, etc.

Sulphur dioxide is oxidised into more harmful sulphur trioxide in the presence of particulate matter present in polluted air.
2SO₂ + O₂ 2SO₃

SO₃ combines with atmospheric water vapour to form H₂SO₄, which comes down in the form of acid rain.
SO₃ + H₂O → H₂SO₄

Question 3.
Explain the health effects of particulate pollutants for human health.
Answer:

1. Dust, mist, fumes,etc., are air-borne particles which are dangerous for human health. Particulate pollutants bigger than 5 microns are likely to settle in the nasal passage whereas particles of about 10 micron enters the lungs easily and causes scarring or fibrosis of lung lining.
2. They irritate the lungs and causes cancer and a,sthma. This disease is also called pneumoconiosis. Coal miners may suffer from black lung disease. Textile workers may suffer from white lung disease.
3. Lead particulates affect children’s brain, interferes maturation of RBCs and even cause cancer.
4. Particulates in the atmosphere reduce visibility by scattering and absorption of sunlight. It is dangerous for aircraft and motor vehicles
5. Particulates provide nuclei for cloud formation and increase fog and rain.
6. Particulates deposit on plant leaves and hinder the intake of CO₂ from the air and affect photosynthesis.

Question 4.
Explain the effects of photochemical smog and its control.
Answer:
The three main components of photochemical smog are nitrogen oxide, ozone and oxidised hydrocarbon like formaldehyde(HCHO), Acrolein (CH₂ = CH – CHO), peroxyacetyl nitrate (PAN). Photochemical smog causes irritation to eyes, skin and lungs, increase in chances of asthma.

High concentrations of ozone and NO can cause nose and throat irritation, chest pain, uncomfortable in breathing, etc. PAN is toxic to plants, attacks younger leaves and cause bronzing and glazing of their surfaces. It causes corrosion of metals stones, building materials and painted surfaces.

Control of Photochemical smog:
The formation of photochemical smog can be suppressed by preventing the release of nitrogen oxides and hydrocarbons into the atmosphere from motor vehicles by using catalytic convertors in engines. Plantation of certain trees like Pinus, Pyrus, Quercus Vitus, and Juniperus can metabolise nitrogen oxide.

Question 5.
List out the major water pollutants and their sources.
Answer:

Pollutant	Sources
1.Microorganisms	Domestic sewage, domestic wastewater, dung heap.
2. Organic wastes	Domestic sewage, animal excreta, food processing factory waste, detergents, and decayed animals and plants.
3. Plant nutrients	Chemical fertilizers.
4. Heavy metals	Heavy metal producing factories.
5. Sediments	Soil erosion by agriculture and strip-mining.
6. Pesticides	Chemicals used for killing insects, fungi, and weeds.
7. Radioactive	Mining of uranium-containing minerals substances.
8. Heat	Water used for cooling in industries.

Question 6.
Describe the causes of water pollution.
Answer:
Causes of water pollution –
1. Microbiological pollutants:
(a) Disease causing microorganisms like bacteria, viruses and protozoa are most senous water pollutants. They come from domestic sewage and animal excreta.
(b) Fish and shellfish can become contaminated from them and people who eat them will also become ill.
(c) Dysentery and cholera are water borne diseases.
(d) Human excrcta contain bacteria such as Escherichia coll and Streptococcus farcical- -is which causes gastrointestinal diseases.

2. Organic wastes:
Organic matter such as leaves, grass, trash can also pollute water. Water pollution is cause by excessive phytoplankton growth within water.

3. Chemical wastes:
A whole variety of chemicals from industries such as metals and solvents are poisonous to fish and other aquatic life. Detergents and oils float spoils the water bodies. Acids from mine drainage and salts form various sources can also contaminate water sources.

Question 7.
Explain the various contribution of green chemistry in our day-to-day life.
Answer:
(1) Dry cleaning of clothes:
Solvents like tetrachloroethylene used in dry cleaning of clothes, pollute the groundwater and are carcinogenic. In the place of tetrachloroethylene, liquefied CO₂ with suitable detergent is an Alternate solvent used. Liquified CO₂ is not harmful to the groundwater. Now a days H₂O₂ used for bleaching clothes in laundry gives better results and utilizes less water.

(2) Bleaching of paper:
The conventional method of bleaching was done with chlorine. Nowadays H₂O₂ can be used for bleaching paper in presence of a catalyst.

(3) Synthesis of chemicals:
Acetaldehyde is now commercially prepared by one-step oxidation of ethene in the presence of anionic catalyst in an aqueous medium with a 90% yield.

CH₂ = CH₂ + O CH₃CHO
Ethylene Acetaldehyde

(4) Instead of petrol, methanol is used as a fuel in automobiles:
Methanol is considered to be less expensive than other commercial fuel and gasoline. During the process of combustion, it provides a higher thermal efficiency and power output because of its high octane rating and high heat vaporization.

(5) Neem based pesticides have been synthesised, which are safer than the chlorinated hydrocarbons:
Every individual has an important role in preventing pollution and improving our environment. We Eire responsible for environmental protection. Let us begin to save our environment and provide clean earth for our future generations.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Ex 9.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Ex 9.3

Question 1.
Define Statistical Quality Control.
Solution:
Statistical quality control (SQC) refers to the use of statistical methods in the monitoring and maintaining of the quality of products and services. This method is used to determine the tolerance limits for accepting a production process.

Question 2.
Mention the types of causes for variation in a production process.
Solution:
There are two causes of variation which affects the quality of a product, namely
1. Chance Causes (or) Random causes
2. Assignable Causes

Question 3.
Define Chance Cause.
Solution:
These are small variations which are natural and inherent in the manufacturing process. The variation occurring due to these causes is beyond human control and cannot be prevented or eliminated under any circumstances, the minor causes which do not affect the quality of the products to an extent are called as Chance Causes (or) Random causes. For example Rain, floods, power cuts, etc.,

Question 4.
Define Assignable cause.
Solution:
The R chart is used to show the variability or dispersion of the samples taken from the given process. The average range is given by \(\overline{\mathrm{R}}=\frac{\sum R}{n}\), where R = x_max – x_min for each ‘n’ samples. For samples of size less than 20, the range provides a good estimate of σ. Hence to measure the variance in the variable, range chart is used.

Question 5.
What do you mean by product control?
Solution:
Product Control means that controlling the quality of the product by critical examination through sampling inspection plans. Product Control aims at a certain quality level to be guaranteed to the customers. It attempts to ensure that the product sold does not contain a large number of defective items. Thus it is concerned with classification of raw materials, semi-finished goods or finished goods into acceptable on rejectable products.

Question 6.
What do you mean by process control?
Solution:
In-Process Control the proportion of defective items in the production process is to be minimized and it is achieved through the technique of control charts.

Question 7.
Define a control chart.
Solution:
A control chart is essentially a graphic device for presenting data so as to directly reveal the frequency and extent of variations from established standards or goals. Control charts are simple to construct and easy to interpret and they tell the manager at a glance whether or not the process is in control, i.e within the tolerance limits.

Question 8.
Name the control charts for variables.
Solution:
(i) Charts for mean (\(\bar { X}\))
(ii) Charts for Range (R)

Question 9.
Define mean chart.
Solution:
The mean chart (\(\bar { X}\) chart) is used to show, the quality average of the samples taken from the given process. The \(\bar { X}\) charts are usually required for decision making to accept or reject the process.
Procedure for \(\bar { X}\)
i. Let X₁, X₂, X₃, etc. be the samples selected each containing “n” observations usually (n = 4.5 or 6)

ii. Calculate mean for each samples \(\bar { X}\)₁, \(\bar { X}\)₂, \(\bar { X}\)₃, ……… by using \(\bar { X}\)_i = \(\frac { ΣX_i }{n}\), i = 1, 2, 3, 4, ….
where Σx_i = total f “n” values included in the sample X₁.

iii. Find the mean (\(\bar { \bar X}\)) of the sample means
\(\bar { \bar X}\) = \(\frac { Σ \bar X }{number of samples}\)
where Σ\(\bar { X}\) = total of all the sample means.

Question 10.
Define R chart.
Solution:
The R chart is used to show the variability or dispersion of the samples taken from the given process. R charts are also required for decision making to accept of reject the process.
Procedure for R-Charts
Calculate R = Xmax – Xmin
Let R₁, R₂, R₃ ………….. be the ranges of the “n” samples. The average range is given by
\(\bar { X}\) = \(\frac { ΣR }{n}\)

Question 11.
What are the uses of statistical quality control?
Solution:
(i) The role of statistical quality control is to collect and analyse relevant data for the purpose of detecting whether the process is under control or not.
(ii) The value of quality control lies in the fact that assignable causes in a process can be quickly detected. Infact the variations are often discovered before the product becomes defective.
(iii) Statistical quality control is only diagnostic. It tells us whether the standard is being maintained or not.
(iv) This technique is used in almost all production industries such as automobile textile, electrical equipment, biscuits, both soaps, chemicals, Petroleum products, etc.
(v) The purpose for which SQC are used in two fold namely (a) process control (b) product control.
The main purpose of SQC is to device statistical techniques which would help in elimination of assignable causes and bring the production process under control.

Question 12.
Write the control limits for the mean chart.
Solution:
The calculation of control limits for x chart in two different cases is

Question 13.
Write the control limits for the R chart.
Solution:
The calculation of control limits for R chart in two different cases are

The values of A₂, D₂ and D₄ are given in the table.

Question 14.
A machine is set to deliver packets of a given j weight. Ten samples of size five each were recorded. Below are given relevant data:

Solution:
Calculate the control limits for mean chart and the range chart and then comment on the state of control, (conversion factors for n = 5, A₂ = 0.58, D₃ = 0 and D₄ = 2.115)

UCL = \(\bar { \bar X}\) + A₂ \(\bar { R}\)
= 16.2 + (0.58)(7.4)
= 16.2 + 4.292
= 20.492
= 20.49
CL = \(\bar { \bar X}\) = 16.2
LCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 16.2 – (0.58) (7.4)
= 16.2 – 4.292
= 11.908
= 11.91
The control limits for Range chart is
UCL = D₄ \(\bar { R}\)
= (2.115)(7.4)
= 15.651
= 15.65
CL = \(\bar { R}\) = 7.4
LCL = D₃ \(\bar { R}\) = (0)(7.4) = 0

Question 15.
Ten samples each of size five are difawn at regular intervals from a manufacturing process. The sample means (X) and their ranges (R) are given below:

Calculate the control limits in respect of \(\bar { X}\) chart. (Given A₂ = 0.58, D₃ = 0 and D₄ = 2.115 ) Comment on the state of control.
Solution:

UCL = \(\bar { \bar X}\) + A₂ \(\bar { R}\)
= 46.2 + (0.58)(6.8)
= 46.2 + 3.944
= 50.144
= 50.14
CL = \(\bar { \bar X}\) = 46.2
LCL = \(\bar { \bar X}\) + A₂ \(\bar { R}\)
= 46.2 – (0.58)(6.8)
= 46.2 – 3.944
= 42.256
= 42.26
The control limits for Range chart is
UCL = D₄ \(\bar { R}\)
= (2.115)(6.8)
= 14.382
= 14.38
CL = \(\bar { R}\) = 6.8
LCL = D₃\(\bar { R}\) = (0)(6.8) = 0

Question 16.
Construct X and R charts for the following data:

Solution:


\(\bar { R}\) = \(\frac { 144 }{8}\) = 18
UCL = \(\bar { \bar x}\) + A₂ \(\bar { R}\)
= 37.71 + (0.58)(18)
= 37.71 + 10.44
= 48.15
CL = \(\bar { \bar x}\) = 37.71
LCL = \(\bar { \bar x}\) – A₂ \(\bar { R}\)
= 37.71 – (0.58)(18)
= 37.71 – 10.44
= 27.27
The control limits for Range chart is
UCL = D₄ \(\bar { R}\) = 2.115 (18) = 38.07
CL = \(\bar { R}\) = 18
LCL = D₃ \(\bar { R}\) = 0(18) = 0

Question 17.
The following data show the values of sample mean (\(\bar {x}\)) and its range (R) for the samples of size five each. Calculate the values for control limits for mean, range chart and determine whether the process is in control.

Solution:

UCL = \(\bar { \bar x}\) – A₂ \(\bar { R}\)
= 10.66 + (0.58)(6.3)
= 10.66 + 3.654 = 14.314
= 14.31
CL = \(\bar { \bar x}\) = 10.66
LCL = \(\bar { \bar x}\) – A₂ \(\bar { R}\)
= 10.66 – (0.58)(6.3)
= 10.66 – 3.654
= 7.006
The control limits for Range chart is
UCL = D₄ \(\bar { R}\) = 2.115 (6.3)
= 13.3245
= 13.32
CL = \(\bar { R}\) = 6.3
LCL = D₃ \(\bar { R}\) = 0(6.3) = 0
Conclusion: Since all the points of sample range is within UCL of R chart, the process is in control.

Question 18.
A quality control inspector has taken ten ” samples of size four packets each from a potato chips company. The contents of the sample are given below, Calculate the control limits for mean and range chart.

(Given for n = 5, A₂ = 0.58, D₃ = 0 and D₄ = 2.115)
Solution:

UCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 12.5 + (0.58)(0.37)
= 12.5 + 0.2146 = 12.7146
= 12.71
CL = \(\bar { \bar X}\) = 12.5
LCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\) = 12.5 – (0.58) (0.37)
= 12.5 – 0.2146 = 12.2854
= 12.29
The control limits for Range chart is
UCL = D₄ \(\bar { R}\) = (2.115)(0.37) = 0.78255
= 0.78
CL = \(\bar { R}\) = 0.37
LCL = D₃\(\bar { R}\) = (0)(0.37) = 0

Question 19.
The following data show the values of sample means and the ranges for ten samples of size 4 each. Construct the control chart for mean and range chart and determine whether the process is in control.

Solution:

UCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 30.1 +(0.73)(20.1)
= 30.1 + 14.673 = 44.773
= 44.77
CL = \(\bar { \bar X}\) = 30.1
LCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 30.1 – (0.73) (20.1)
= 30.1 – 14.673 = 15.427
= 15.43
The control limits for Range chart is
UCL = D₄ \(\bar { R}\)
= 2.28(20.1) = 45.828
= 45.83
CL = \(\bar { R}\) = 20.1
LCL = D₃ \(\bar { R}\) = 0(20.1) = 0

Question 20.
In a production process, eight samples of size 4 are collected and their means and ranges are given below. Construct mean chart and range chart with control limits.

Solution:

UCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 13.25 + (0.73) (3.12)
= 13.25 + 2.2776 = 15.5276
= 15.53
CL = \(\bar { \bar X}\) = 13.25
LCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 13.25 – (0.73)(3.12)
= 13.25 – 2.2776 = 10.972
= 10.97
The control limits for Range chart is
UCL = D₄ \(\bar { R}\)
= 2.28(3.12) = 7.11984
= 7.12
CL = \(\bar { R}\) = 3.12
LCL = D₃\(\bar { R}\) = 0(3.12) = 0

Question 21.
In a certain bottling industry the quality control inspector recorded the weight of each of the 5 bottles selected at random during each hour of four hours in the morning.

Solution:

UCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 41 + (0.58)(4)
41 + 2.32 = 43.32
CL = \(\bar { \bar X}\) = 41
LCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 41 – (0.58)(4)
= 41 – 2.32
= 38.68
The control limits for range chart is
UCL = D₄ \(\bar { R}\) = 2.115(4)
= 8.46
CL = \(\bar { R}\) = 4
LCL = D₂ \(\bar { R}\) = 0(4) = 0
Conclusion: Since all the points of the sample mean and Range are within the control limits, the process is in control.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 14 Haloalkanes and Haloarenes Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 14 Haloalkanes and Haloarenes

11th Chemistry Guide Haloalkanes and Haloarenes Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
The IUPAC name of is
a) 2 – Bromo pent – 3 – ene
b) 4 – Bromo pent – 2 – ene
c) 2 – Bromo pent – 4 – ene
d) 4 – Bromo pent – 1 – ene
Answer:
b) 4 – Bromo pent – 2 – ene

Question 2.
Of the following compounds, which has the highest boiling point?
a) n – Butyl chloride
b) Isobutyl chloride
c) t – Butyl chloride
d) n – Propyl chloride
Answer:
a) n – Butyl chloride

Question 3.
Arrange the following compounds in increasing order of their density
A) CCl₄
B) CHCl₃
C) CH₂Cl₂
D) CH₃Cl
a) D < C < B < A
b) C > B > A > D
c) A < B < C < D
d) C > A > B > D
Answer:
a) D < C < B < A

Question 4.
With respect to the position of – Cl in the compound CH3 – CH = CH – CH₂ – Cl, it is classified as
a) Vinyl
b) Allyl
c) Secondary
d) Aralkyl
Answer:
b) Allyl

Question 5.
What should be the correct IUPAC name of diethyl chloromethane?
a) 3 – Chloro pentane
b) 1 – Chloropentane
c) 1 – Chloro – 1, 1 – diethyl methane
d) 1- Chloro-1-ethyl propane
Answer:
a) 3 – Chloro pentane

Question 6.
C – X bond is strongest in
a) Chloromethane
b) Iodomethane
c) Bromomethane
d) Fluoromethane
Answer:
d) Fluoromethane

Question 7.
In the reaction X is ______.

Answer:
b)

Question 8.
Which of the following compounds will give racemic mixture on nucleophilic substitution by OH^– ion?
i)

ii)

iii)
a) (i)
b) (ii) and (iii)
c) (iii)
d) (i) and (ii)
Answer:
c) (iii)

Question 9.
The treatment of ethyl formate with excess of RMgX gives
a)

b)
c) R – CHO
d) R – O – R
Answer:
c) R – CHO

Question 10.
Benzene reacts with Cl₂ in the presence of FeCl₃ and in absence of sunlight to form
a) Chlorobenzene
b) Benzyl chloride
c) Benzal chloride
d) Benzene hexachloride
Answer:
a) Chlorobenzene

Question 11.
The name of C₂F₄C₁₂ is
a) Freon – 112
b) Freon – 113
c) Freon – 114
d) Freon – 115
Answer:
c) Freon – 114

Question 12.
Which of the following reagent is helpful to differentiate ethylene dichloride and ehtylidene chloride?
a) Zn / methanol
b) KQH / ethanol
c) aqueous KOH
d) ZnCl₂ / Con HCl
Answer:
c) aqueous KOH

Question 13.
Match the compounds given in Column I with suitable items given in Column II:

Column I (Compound)	Column II (Uses)
A. Iodoform	1. Fire extinguisher
B. Carbon tetra chloride	2. Insecticide
C. CFC	3. Antiseptic
D. DDT	4. Refrigerants

Code
a) A → 2 B → 4 C → 1 D → 3
b) A → 3 B → 2 C → 4 D → 1
c) A → 1 B → 2 C → 3 D → 4
d) A → 3 B → 1 C → 4 D → 2
Answer:
d) A → 3 B → 1 C → 4 D → 2

Question 14.
Assertion:
Inmonohaloarenes, electrophilic substitution occurs at ortho and para positions.
Reason:
Halogen atom is a ring deactivator.
Assertion and Reason type questions.
Directions:
In the following questions, a statement of assertion (A) is followed by a statement of reason (R) mark the correct choice as
(i) If both assertion and reason are true and reason is the correct explanation of assertion.
(ii) If both assertion and reason are true but reason is not the correct explanation of assertion.
(iii) If assertion is true but reason is false.
(iv) If both assertion and reason are false.
a) (i)
b) (ii)
c) (iii)
d) (iv)
Answer:
b) (ii)

Question 15.
Consider the reaction,
CH₃CH₂CH₂Br + NaCN → CH₃CH₂CH₂CN + NaBr This reaction will be the fastest in
a) ethanol
b) methanol
c) DMF (N, N’ – dimethyl formamide)
d) water
Answer:
c) DMF (N, N’ – dimethyl formamide)

Question 16.
Freon – 12 manufactured from tetrachloro methane by
a) Wurtz reaction
b) Swarts reaction
c) Haloform reaction
d) Gattermann reaction
Answer:
b) Swarts reaction

Question 17.
The most easily hydrolysed molecules under S_N¹ condition is
a) allyl chloride
b) ethyl chloride
c) isopropyl chloride
d) benzyl chloride
Answer:
a) allyl chloride

Question 18.
The carbon cation formed in S_N¹ reaction of alkyl halide in the slow step is
a) sp³ hybridized
b) sp² hybridized
c) sp hybridized
d) none of these
Answer:
b) sp² hybridized

Question 19.
The major products obtained when chlorobenzene is nitrated with HNO₃ and con H₂SO₄
a) 1 – chloro – 4 – nitrobenzene
b) 1 – chloro – 2 – nitrobenzene
c) 1 – chloro – 3 – nitrobenzene
d) 1 – chloro – 1 – nitrobenzene
Answer:
a) 1 – chloro – 4 – nitrobenzene

Question 20.
Which one of the following is most reactive towards nucleophilic substitution reaction?
a)

b)

c)

d)
Answer:
d)

Question 21.
Ethylidene chloride on treatment with aqueous KOH gives
a) acetaldehyde
b) ethylene glycol
c) formaldehyde
d) glycoxal
Answer:
a) acetaldehyde

Question 22.
The raw material for Raschig process
a) chloro benzene
b) phenol
c) benzene
d) anisole
Answer:
c) benzene

Question 23.
Chloroform reacts with nitric acid to produce
a) nitro toluene
b) nitro glycerine
c) chloropicrin
d) chloropicric acid
Answer:
c) chloropicrin

Question 24.
Acetone X, X is
a) 2 – propanol
b) 2 – methyl – 2 – propanol
c) 1 – propanol
d) acetonol
Answer:
b) 2 – methyl – 2 – propanol

Question 25.
Silverpropionate when refluxed with Bromine in carbon tetrachloride gives
a) propionic acid
b) chloroethane
c) Bromo ethane
d) chloro propane
Answer:
c) bromo ethane

II. Write brief answer to the following questions:

Question 26.
Classify the following compounds in the form of alkyl, allylic, vinyl, benzylic halides.
i) CH₃ – CH = CH – Cl

ii) C₆H₅CH₂I

iii)

iv) CH₂ = CH – Cl
Answer:
i) CH₃ – CH = CH – Cl = Allylic halide

ii) C₆H₅CH₂I = Benzylic halide

iii) = Alkyl halide

iv) CH₂ = CH – Cl = Vinyl halide

Question 27.
Why chlorination of methane is not possible in dark?
Answer:

• Chlorination of methane is a free radical substitution reaction.
• Before chlorine reacts with methane, the Cl-Cl single bond must break to form free radicals and this can only be done in the presence of ultraviolet light.
• In dark, chlorine-free radicals formation is not possible and so chlorination of methane is not possible in dark.
• The ultraviolet light is a source of energy and is being used to break of Cl-Cl and produce Cl free radical Free radicals which can attack methane. in dark, this is not possible.

Question 28.
How will you prepare n propyl iodide from n – propyl bromide?
Answer:
Finkelstein reaction,
nCH₃ – CH₂ – CH₂ – Br + NaI n – CH₃ – CH₂ – CH₂ – I + NaBr
n – propyl iodide                                              n- propyl bromide

Question 29.
Which alkyl halide from the following pair is
i) chiral
ii) undergoes faster S_N² reaction?

Answer:

It contains one chiral carbon atom.
2 – Bromo butane undergoes S_N² mechanism faster than 1- Chloro butane.

Question 30.
How does chlorobenzene react with sodium in the presence of ether? What is the name of the reaction?
Answer:
Haloarenes react with sodium metal in dry ether, two aryl groups combine to give biaryl products.
This reaction is called fittig reaction.
C₆H₅Cl + 2Na + Cl – C₆H₅ C₆H₅ – C₆H₅ + 2NaCl
Chlorobenzene                                      Biphenyl

Question 31.
Give reasons for the polarity of C – X bond in haloalkane.
Answer:
Carbon halogen bond is a polar bond as halogens are more electronegative than carbon. The carbon atom exhibits a partial positive charge (δ⁺) and halogen atom a partial negative charge (δ^–)

The C -X bond is formed by overlap of sp³ orbital of a carbon atom with half-filled p- orbital of the halogen atom. The atomic size of halogen increases from fluorine to iodine, which increases the C – X bond length. Larger the size, greater is the bond length, and the weaker is the bond formed. The bond strength of C – X decreases from C – F to C – I in CH₃X.

Question 32.
Why is it necessary to avoid even traces of moisture during the use of Grignard reagent?
Answer:
Grignard reagents are mostly reactive and react with the source of the product to give hydrocarbons. Even alcohols, amines, H₂O are sufficiently acidic to convert them to corresponding hydrocarbons.
R Mg X + H₂O → RH +

Due to its high reactivity, it is necessary to avoid even traces of moisture from the Grignard reagent.

Question 33.
What happens when acetyl chloride is treated with an excess of CH₃MgI?
Answer:

Question 34.
Arrange the following alkyl halide in increasing order of bond enthalpy of RX.
CH₃Br, CH₃F, CH₃Cl, CH₃I
Answer:
The order is:
CH₃I < CH₃Br < CH₃Cl < CH₃F.

Question 35.
What happens when chloroform reacts with oxygen in the presence of sunlight?
Answer:
2 CHCl₃ + O₂ → 2 COCl₂ + 2 HCl

Question 36.
Write down the possible isomers of C₅H₁₁Br and give their IUPAC and common names.
Answer:
C₅H₁₁Br – Possible isomers
1. CH₃ – CH₂ – CH₂ – CH₂ – CH₂ – Br → 1 – bromo pentane

2. → 2 – bromo pentane

3. → 3 – bromo pentane

4.  → 1 – bromo 2, 2 – dimethyl propane

5. → 1 – bromo 3 – methyl butane

6. → 2 – bromo 3 – methyl butane

7. → 2 – bromo 2 – methyl butane

8. → 1 – bromo 2- methyl butane

9. → (2S) – 1 – bromo 2 – methyl butane

10. → (2R) – 1 – bromo 2 – methyl butane

Question 37.
Mention any three methods of preparation of haloalkanes from alcohols.
Answer:
Haloalkanes are prepared by the following methods.
From alcohols: Alcohol can be converted into halo alkenes by reacting it with any one of the following reagents.

• Hydrogen halide
• Phosphorous halides
• Thionyl chloride.

a) Reaction with hydrogen halide:

Mixture of con. HCl and anhydrous ZnCl₂ is called Lucas Reagent.

The order of reactivity of halo acids with alcohol is in the order HI > HBr > HCl.
The order of reactivity of alcohols with halo acid is tertiary > secondary > primary.

b) Reaction with phosphorous halides:
Alcohols react with PX₅ or PX₃ to form haloalkanes.
Example:
CH₃CH₂OH + PCl₅ → CH₃CH₂Cl + POCl₃ + HCl
Ethane                      Chloro ethane

3CH₃CH₂OH + PCl₃ → 3 CH₃CH₂Cl + H₃PO₃
Ethanol                       Chloro ethane

c) Reaction with Thionyl chloride(Sulphonyl Chloride)
CH₃CH₃OH + SOCl₂ CH₃CH₂Cl + SO₂↑ + HCl↑
Ethanol                                      Chloro ethane

Question 38.
Compare S_N¹ and S_N² reaction mechanisms.
Answer:

	SN1	SN2
Rate law	Unimolecular (Substrate only)	Biomolecular (substrate and nucleophile)
“Big Barrier”	Carbocation stability	Steric hindrance
Alkyl halide (electrophile)	3° > 2° > 1°	1° > 2° > 3°
Nucleophile	Weak (generally neutral)	Strong (generally bearing a negative charge)
Solvent	Polar protic (e.g., alcohols)	Polar aprotic (e.g., DMSO, acetone)
Stereo Chemistry	Mix of retention and inversion	inversion

Question 39.
Reagents and the conditions used in the reactions are given below. Complete the table by writing down the product and the name of the reaction.

Answer:

Question 40.
Discuss the aromatic nucleophilic substitutions reaction of chlorobenzene.
Answer:
The halogen of haloarenes can be substituted by OH^–, NH₂^– or CN^– with appropriate nucleophilic reagents at high temperature and pressure.
Example:
(i) Chlorobenzene reacts with ammonium at 250 and at 50 atm to give aniline.
C₆H₅Cl + 2NH₃ C₆H₅NH₂ + NH₄Cl
Chlorobenzene                  Aniline

(ii) Chlorobenzcne reacts with CuCN in presence of pyridine at 250 to give phenyl cyanide.
C₆H₅Cl + CuCN C₆H₅CN + CuCl
Chlorobenzene                Phenyl cyanide

(iii) Dows process:
C₆H₅Cl + NaOH C₆H₅OH + NaCl
Chlorobenzene                   Phenol
This reaction is known as Dow’s process.

Question 41.
Account for the following:
(i) t – butyl chloride reacts with aqueous KOH by S_N¹ mechanism while n – butyl chloride reacts with S_N² mechanism.
(ii) p – dichlorobenzene has a higher melting point than those of o – and m – dichlorobenzene.
Answer:
(i) t – butyl chloride reacts with aqueous KOH by S_N¹ mechanism while n – butyl chloride reacts with S_N² mechanism.
In general, the S_N¹ reaction proceeds through the formation, of carbocation, The tert-butyl chloride readily loses Cl ion to form stable 3° carbocation. Therefore, it reacts with aqueous KOH by S_N¹ mechanism as:

On the other hand, n-Butyl chloride does not undergo ionization to form n-Butyl carbocation (1°) because it is not stable. Therefore, it prefers to undergo reaction by an S_N² mechanism, which occurs is one step through a transition state involving the nucleophilic attack of OH^– ion from the backside with simultaneous expulsion of Cl^– ion from the front side.

S_N¹ mechanism follows the reactivity order as 3° > 2°> 1° while S_N² mechanism follows the reactivity order as 1° > 2° > 3°. Therefore, tert-butyl chloride (3°) reacts by S_N¹ mechanism while n-butyl chloride (1°) reacts by an S_N² mechanism. (ii) p – dichlorobenzene has a higher melting point than those of o – and m – dichloro benzene. The higher melting point of p – isomer is due to its symmetry which leads to more close packing of its molecules in the crystal lattice and consequently strong intermolecular attractive force which requires more energy for melting. p – Dihalo benzene > o – Dichloro benzene> m – Dichioro benzene
Melting point: 323 K 256 K 249 K

Question 42.
In an experiment methyl iodide in ether is allowed to stand over magnesium pieces. Magnesium dissolves and product is formed.
a) Name of the product and write the equation for the reaction.
b) Why all the reagents used in the reaction should be dry? Explain.
c) How is acetone prepared from the product obtained in the experiment?
Answer:
a) Name of the product and write the equation for the reaction.
CH₃I + Mg CH₃MgI

b) Why all the reagents used in the reaction should be dry? Explain.
All the reagents used in the reaction should be dry because reagent reacts with H20 to produce alkane. This is the reason that everything has to be very dry during the preparation of Grignard reagents.
CH₃ – MgI + H₂O → CH₄ +
Methane

c) How is acetone prepared from the product obtained in the experiment?

Question 43.
Write a chemical reaction useful to prepare the following.
i) Freon – 12 from Carbon tetrachloride
ii) Carbon tetrachloride from carbon disulphide.
Answer:
i) Freon – 12 from Carbon tetrachloride:
Freon – 12 is prepared by the action of hydrogen fluoride on carbon tetrachloride in the presence of catalytic amount of antimony pentachloride.
CCl₄ + 2HF 2HCl + CCl₂F₂
Carbon tetrachloride            Freon – 12

ii) Carbon tetrachloride from carbon disulphide.
Carbon disulphide reacts with chlorine gas in the presence of anhydrous AlCl3 as catalyst giving carbon tetrachloride.
CS₂ + 3 Cl₂ CCl₄ + S₂Cl₂
Carbon disulfide                Carbon tetrachloride

Question 44.
What are Freons? Discuss their uses and environmental effects.
Answer:
The chloro fluoro derivatives of methane and ethane are called freons.
Nomenclature:
Freon is represented as Freon – cba
Where a = number of carbon atoms – 1;
b = number of hydrogen atoms + 1
a = total number of fluorine atoms

Uses:

• Freons are used as a refrigerant in refrigerators and air conditioners.
• It is used as a propellant for aerosols and foams
• It is used as a propellant for foams to spray out deodorants, shaving creams, and insecticides.

Question 45.
Predict the products when Bromo ethane is treated with the following.
i) KNO₂
ii) AgNO₂
Answer:
i) KNO₂:
Bromo ethane reacts with an alcoholic solution of NaNO2 or KNO2 to form ethyl nitrite.
CH₃CH₂Br + KNO₂ → CH₃CH₂ – O – N = O + KBr
Bromoethane Ethyl nitrite

ii)AgNO₂:
Bromo ethane reacts with an alcoholic solution of AgNO2 to form nitroethane.
CH₃CH₂Br + AgNO₂ → CH₃CH₂ NO₂ + AgBr
Bromoethane                Nitroethane

Question 46.
Explain the mechanism of S_N¹ reaction by highlighting the stereochemistry behind it.
Answer:

In S_N¹ reactions, if the alkyl halide is optically active, the product obtained in a racemic mixture. The intermolecular carbocation formed in slowest step being sp2 hybridized is planar species. Therefore the attack of the nucleophile OH on it, can occur from both the faces with equal case forming a mixture of two enantiomers. Thus S_N¹ reaction of optically active alkyl halides is accompanied by racemization.

Question 47.
Write short notes on the following.
i) Raschig process
ii) Dows process
iii) Darzen’s process
Answer:
i) Raschig process:
Chloro benzene is commercially prepared by passing a mixture of benzene vapour, air and HCl overheated cupric chloride, this reaction is called the Raschig process,

ii) Dows Process:
C₆H₅Cl + NaOH C₆H₅OH + NaCl
This reaction is known as Dows process.

iii) Darzen’s process:
CH₃CH₂OH + SOCl CH₃CH₂Cl + SO₂↑ + HCl↑
Ethanol Chloro ethane
This reaction is known as Darzen’s process.

Question 48.
Starting from CH₃MgI, How will you prepare the following?
i) Acetic acid
ii) Acetone
iii) Ethyl acetate
iv) Isopropyl alcohol
v) Methyl cyanide
Answer:
i) Acetic acid:
Solid carbon dioxide reacts with methyl magnesium iodide to form additional product which on hydrolysis yields acetic acid.

ii) Acetone:
Acetyl chloride reacts with methyl magnesium iodide and followed by acid hydrolysis to give acetone.

iii) Ethyl Acetate:
Ethyl chloroformate reacts with methyl magnesium iodide to form ethyl acetate.

iv) Isopropyl alcohol:
Aldehydes (Acetaldehyde) other than formaldehyde, react with methyl magnesium iodide to give additional product which on hydrolysis yields isopropyl alcohol.

v) Methyl cyanide:
Methyl magnesium iodide reacts with cyanogen chloride to give methyl cyanide.

Question 49.
Complete the following reactions.
i) CH₃ – CH = CH₂ + HBr
ii) CH₃ – CH₂ – Br + NaSH
iii) C₆H₅Cl + Mg
iv) CHCl₃ + HNO₃
v) CCl₄ + H₂O
Answer :
i) CH₃ – CH = CH₂ + HBr CH₃ – CH₂ – CH₂ – Br
Propene                                                   n – propyl bromide

ii) CH₃ – CH₂ – Br + NaSH CH₃ – CH₂ – SH + NaBr
Propyl bromide                                     Ethanethiol

iii) C₆H₅Cl (Chloro benzene) + Mg C₆H₅MgCl (Phenyl magnesium chloride)

iv) CHCl₃ + HNO₃ CCl₃NO₂ + H
Chloroform                               Chloropicrin

v) CCl₄ (Carbon tetrachloride) + H₂O COCl₂ (Carbonyl chloride) + 2HCl

Question 50.
Explain the preparation of the following compounds.
i) DDT
ii) Chloroform
iii) Biphrnyl
iv) Chloropicrin
v) Freon – 12
Answer:
i) DDT:
DDT can be prepared by heating a mixture of chlorobenzene with chloral (Trichloro acetaldehyde) in the presence of con.H₂SO₄.

ii) Chloroform:
Preparation:
Chloroform is prepared in the laboratory by the reaction between ethyl alcohol with bleaching powder followed by the distillation of the product chloroform. Bleaching powder act as a source of chlorine and calcium hydroxide. This reaction is called the haloform reaction. The reaction proceeds in three steps as shown below.

Step – 1: Oxidation
CH₃CH₂OH + Cl₂ → CH₃CHO + 2HCl
Ethyl alcohol           Ethanal (Acetaldehyde)

Step – 2: Chlorination
CH₃CHO + 3Cl₂ → CCl₃CHO + 3HCl
Acetaldehyde        Trichloro acetaldehyde

Step – 3: Hydrolysis
2CCl₃CHO + Ca(OH)₂ → 2CHCl₃ + (HCOO)₂ Ca
Chloral                         chloroform

iii) Biphenyl:
Chloro benzene reacts with sodium metal in dry ether, to give biphenyl. This reaction is called a fitting reaction.
C₆H₅Cl + 2 Na + Cl – C₆H₅ C₆H₅ – C₆H₅ + 2NaCl
Chloro benzene                                          Biphenyl

iv) Chloropicrin:
Chloroform reacts with nitric acid to form chloropicrin. (Trichloro nitromethane)
CHCl₃ + HNO₃ CCl₃NO₂ + H₂O
Chloroform                   Chloropicrin

v) Freon – 12
Freon – 12 is prepared by the action of hydrogen fluoride on carbon tetrachloride in the presence of the catalytic amount of antimony pentachloride
CCl₄ + 2 HF 2 HCl + CCl₂F₂
Carbon tetrachloride                     Freon – 12

Question 51.
An organic compound (A) with molecular formula C₂H₅Cl reacts with KOH gives compounds (B) and with alcoholic KOH gives compound (C). Identify (A), (B), (C).
Answer:

Question 52.
The simplest alkene (A) reacts with HCl to form compound (B). Compound (B) reacts with ammonia to form compound (C) of molecular formula C₂H₇N. Compound (C) undergoes carbylamine test. Identify (A),
(B) and (C).
Answer:
CH₂ = CH₂ + HCl → C₂H₅Cl
(A) Ethylene              (B) Ethyl chloride

C₂H₅Cl + NH₃    →    C₂H₅NH₂ + HCl
(C) Ethyl chloride      (B) Ethyl amine

Question 53.
A hydrocarbon C₃H₆(A) reacts with HBr to form compound (B). Compound (B) reacts with aqueous potassium hydroxide to give (C) of molecular formula C₃H₆O. What are the (A), (B) and (C). Explain the reactions.
Answer:

Question 54.
Two isomers (A) and (B) have the same molecular formula C₂H₄Cl₂. Compound (A) reacts with aqueous KOH gives compound (C) of molecular formula C₂H₄O. Compound (B) reacts with aqueous KOH gives compound (D) of molecular formula C₂H₆O₂. Identify (A), (B), (C) and (D).
Answer:

11th Chemistry Guide Haloalkanes and Haloarenes Additional Questions and Answers

I. Choose the best answer:

Question 1.
Which of the following is an example of polyhalo compounds?
(a) Vnyl iodide
(b) Chiorobenzene
(c) Allyl chloride
(d) Chloroform
Answer:
(d) Chloroform

Question 2.
2° halide among the following
a) isopropyl chloride
b) isobutyl chloride
c) n – propyl chloride
d) n – butyl chloride
Answer:
a) isopropyl chloride

Question 3.
How many isomers are possible for the formula C₄H₉Cl?
(a) 3
(b) 2
(c) 4
(d) 5
Answer:
(c) 4

Question 4.
Which of the following is gemdihalide?
a) CH₃CHBrCH₂Br
b) CH₃CHBr₂
c) CH₃CHBrCH₂CH₂Br
d) BrCH₂CH₂Br
Answer:
b) CH₃CHBr₂

Question 5.
Which of the following is called Lucas reagent?
(a) Conc. H₂SO₄ + Anhydrous CuSO₄
(b) Conc.HCl + Anhydrous ZnCl₂
(c) Dil.HCl + AlCl₃
(d) Conc.HCl + ConcHNO₂
Answer:
(b) Conc.HCl + Anhydrous ZnCl₂

Question 6.
The reagent used to get alkyl halide from alcohol
a) PCl₅
b) SOCl₂
c) Both a and b
d) Cl₂
Answer:
c) Both a and b

Question 7.
The reactivity of alcohols with haloacid is –
(a) 3° > 2° > 1°
(b) 1° > 2° > 3°
(c) 2° > 3° > 1°
(d) 3° > l° > 2°
Answer:
(a) 3° > 2°> 1°

Question 8.
In the preparation of alkyl halide from alkane and halogen which of the following reaction involved
a) Electrophilic addition
b) Nucleophilic addition
c) Electrophilic substitution
d) Nucleophilic substitution
Answer:
a) Electrophilic addition

Question 9.
What is the name of the reaction in which bromoethane is converted to iodoethane by reacting with NaI in acetone?
(a) Hunsdicker reaction
(b) Dow’s process
(c) Finkelstein reaction
(d) Swarts reaction
Answer:
(c) Finkelstein reaction

Question 10.
The Grignard reagent is formed when alkyl halide reacts with which one of the following
a) Mg in alcohol
b) Mg in acid
c) Mg in dry ether
d) MgO
Answer:
c) Mg in dry ether

Question 11.
Which of the following pair functional groups represents ambident nucleophiles?
(a)-SH-&-OH
(b)-CN-&-NO₂
(c)-Br-&-Cl
(d)-O-&-CHO
Answer:
(b)-CN & NO₂

Question 12.
Alkyl halide on reduction with Zn + HCl gives
a) alcohol
b) alkene
c) alkane
d) ether
Answer:
c) alkane

Question 13.
Which one of the following is used for producing pesticìdes?
(a) CHI₃
(b) CHCl₃
(c) CCl₃ NO₂
(d) CCl₄
Answer:
(b) CHCl₃

Question 14.
Which of the reactions are most common in alkyl halides
a) Nucleophilic addition
b) Electrophilic addition
c) Nucleophilic substitution
d) Electrophilic substitution
Answer:
c) Nucleophilic substitution

Question 15.
Which one of the following reacts with CH₃MgI followed by hydrolysis and gives isopropyl alcohol?
(a) CH₃ COCH₃
(b) CH₃ CHO
(c) HCHO
(d) CNCl
Answer:
(b) CH₃ CHO

Question 16.
In chloroethane the carbon-bearing halogen is bonded to ________.
a) three, primary
b) two, secondary
c) one, tertiary
d) two, primary
Answer:
d) two, primary

Question 17.
Which one of the following reacts with CH₃ Mg I followed by acid hydrolysis to yield acetic acid?
(a) CNCl
(b) CH₃COOC₂H₅
(c) HCOOC₂H₅
(d) CO₂
Answer:
(d) CO₂

Question 18.
SN¹reaction occurs through the intermediate formation of
a) carbocation
b) carbanion
c) free radicals
d) transition
Answer:
a) carbocation

Question 19.
Which one of the following is used as a fiber-swelling agent in textile processing?
(a) Chiorohenzene
(b) Chloroform
(c) Chlorai
(d) Chloroethane
Answer:
(a) Chiorobenzene

Question 20.
The most reactive nucleophile among the following is
a) CH₃O^–
b) C₆H₅O^–
c) (CH₃)₂CHO^–
d) (CH₃)₃CO^–
Answer:
a) CH₃O^–

Question 21.
Which of the following reagent is used to distinguish gem-dihalides and vicinal dihalides?
(a) Alcoholic KOH
(b) Aqueous KOH
(c) FeCl₃ / Cl₂
(d) Ethanol
Answer:
(b) Aqueous KOH

Question 22.
In SN₂ reactions the order of reactivity of the halides.
CH₃X, C₂H₅X , n – C₃H₇X, n- C₄H₉X is
a) CH₃X > C₂H₅X > n – C₃H₇X > n – C₄H₉X
b) C₂H₅X > n – C₃H₇X > n – C₄H₉X > CH₃X
c) C₂H₅X > n – C₃H₇X > n – C₄H₉X < CH₃X
d) n – C₄H₉X > n – C₃H₇X > C₂H₅X > CH₃X
Answer:
a) CH₃X > C₂H₅X > n – C₃H₇X > n – C₄H₉X

Question 23.
Which one of the following is used as a metal cleaning solvent?
(a) Isopropylidene chloride
(b) Methylene chloride
(c) Chloroform
(d) lodoform
Answer:
(b) Methylene chloride

Question 24.
In Dow’s process the starting raw material is
a) Phenol
b) Chlorobenzene
c) Aniline
d) Diazobenzene
Answer:
b) Chlorobenzene

Question 25.
Which one of the following is used to test primary amines?
(a) Schiff’s test
(b) Carbylarnine test
(c) Dye test
(d) Silver mirror test
Answer:
(b) Carbyianiine test

Question 26.
Chloro benzene is ________ reactive than benzene towards electrophilic substitution and directs incoming electrophile to the ______ position.
a) more, ortho & para
b) less, ortho & para
c) more, meta
d) less, meta
Answer:
b) less, ortho & para

Question 27.
The raw material for raschig; process is
a) chloro benzene
b) phenol
c) benzene
d) anisol
Answer:
c) benzene

Question 28.
Chloro benzene on treatment with sodium in dry ether gives diphenyl. The name of the reaction is
a) Fitting reaction
b) Wurtz fittig reaction
c) Wurtz reaction
d) Sandmeyer reaction
Answer:
a) Fitting reaction

Question 29.
Which one of the following compounds does not undergo nucleophilic substitution reactions at all?
(a) Ethyl bromide
(b) Vinyl chloride
(c) Benzyl chloride
(d) isopropyl chloride
Answer:
(b) Vinyl chloride

Question 30.
The raw materials for the commercial manufacture of DDT are
a) chloro benzene and chloroform
b) chlorobenzene and chloromethane
c) chloro benzene and chloral
d) chloro benzene and iodoform
Answer:
c) chloro benzene and chloral

Question 31.
Iodoform is used as
a) anesthetic
b) antiseptic
c) analgesic
d) anti fibrin
Answer:
b) antiseptic

Question 32.
The following is used in paint removing
a) CHCl₃
b) CH₂Cl₂
c) CCl₄
d) CH₃CI
Answer:
b) CH₂Cl₂

Question 33.
In fire extinguishers, following is used
a) CHCl₃
b) CS₂
c) CCl₄
d) CH₂Cl₂
Answer:
c) CCl₄

Question 34.
The following is used for metal cleaning and finishing
a) CHCl₃
b) CHI₃
c) CH₂Cl₂
d) C₆H₆
Answer:
c) CH₂Cl₂

Question 35.
First chlorinated insecticide
a) DDT
b) Gammaxene
c) Iodoform
d) Freon
Answer:
a) DDT

Question 36.
The following is used as anaesthetic
a) C₂H₄
b) CHCl₃
c) CH₂Cl₂
d) DDT
Answer:
b) CHCl₃

Question 37.
Freon – 12 is
a) CF₃Cl
b) CHCl₂F
c) CF₂Cl₂
d) DDT
Answer:
c) CF₂Cl₂

Question 38.
The name of DDT
a) p, p’ – dichloro diphenyl trichloro ethane
b) p, p’ – dichloro diphenyl trichloro ethene
C) p, p’ – dichloro diphenyl tnchloro benzene
d) p, p’ – tetra chloro ethane
Answer:
a) p, p’ – dichloro diphenyl trichloro ethane

Question 39.
Freon R – 22 is
a) CHClF₂
b) CCl₂F₂
c) CH₃Cl
d) CH₂Cl₂
Answer:
a) CHClF₂

Question 40.
Molecular formula of DDT has
a) 5 Cl atoms
b) 4 Cl atoms
c) 3 Cl atoms
d) 2 Cl atoms
Answer:
a) 5 Cl atoms

Question 41.
What is DDT among the following
a) Green house gas
b) A fertilizer
c) Bio degradable pollutant
d) Non – Bio degradable pollutant
Answer:
d) Non – Bio degradable pollutant

Question 42.
The IUPAC name of (CH₃)₃CHCH₂Br is
a) 1 – bromo – 2 – methyl propane
b) 2 – bromo – 2 -methyl propane
c) 1 – bromo – 1 – methyl propane
d) 2 – bromo – 1 -methylpropane
Answer:
a) 1 – bromo – 2 – methyl propane

Question 43.
IUPAC name of allyl chloride is
a) 1 – chloro ethane
b) 3 – chloro- 1 – propyne
c) 3 – chloro – 1 – propene
d) 1 – chloro propane
Answer:
c) 3 – chloro – 1 – propene

Question 44.
The number of structural isomers possible with the formula C₄H₉Cl are
a) 5
b) 4
c) 3
d) 2
Answer:
b) 4

Question 45.
Density is highest for
a) CH₃Cl
b) CH₂Cl₂
c) CHCl₃
d) CCl₄
Answer:
d) CCl₄

Question 46.
C₂H₅OH X. In this reaction ‘X’ is
a) Ethanol
b) Ethylene chloride
c) ethylidene chloride
d) ethyl chloride
Answer:
d) ethyl chloride

Question 47.
Thionyl chloride is preferred in the preparation of chloro compound from alcohol since
a) Both the byproducts are gases and they escape out leaving product in pure state
b) It is a chlorinating agent
c) It is an oxidising agent
d) All other reagents are unstable
Answer:
a) Both the byproducts are gases and they escape out leaving product in pure state

Question 48.
The only alkene which gives primary alkyl halides on hydro halogenation
a) C₂H₄
b) C₃H₆
c) C₄H₈
d) C₅H₁₀
Answer:
a) C₂H₄

Question 49.
– OH cannot be replaced by – Cl if we use
a) PCl₅
b) PCl₃
c) S₂Cl₂
d) SOCl₂
Answer:
c) S₂Cl₂

Question 50.
In the hydrohalogenation of ethylene for adding HCl, the catalyst used is
a) Anhydrous AlCl₃
b) Conc. Sulphuric acid
c) Dilute Sulphuric acid
d) Anhydrous ZnCl₂
Answer:
a) Anhydrous AlCl₃

Question 51.
Which one of the following has the lowest boiling point?
a) CH₃Cl
b) C₂H₅Cl
c) C₂H₅Br
d) C₂H₅I
Answer:
a) CH₃Cl

Question 52.
Chloroethane is reacted with alcoholic potassium hydroxide. The product formed is
a) C₂H₆O
b) C₂H₆
c) C₂H₄
d) C₂H₄O
Answer:
c) C₂H₄

Question 53.
What is X in the following reaction? C₂H₅Cl + X → C₂H₅OH + KCl
a) KHCO3
b) alc. KOH
c) aq. KOH
d) K₂CO₃
Answer:
c) aq. KOH

Question 54.
Which of the following acids will give maximum yield of alkyl chloride in Hunsdiecker reaction
a) CH₃CH₂CH₂COOH
b) (CH₃)₂CHCOOH
c) (CH₃)₃CCOOH
d) C₆H₅CH (CH₃)COOH
Answer:
a) CH₃CH₂CH₂COOH

Question 55.
In the reaction sequence
C₂H₅Cl + KCN X. What is the molecular formula of X is
a) C₂H₅CN
b) C₂H₅NC
c) C₂H₅OH
d) C₂H₄O
Answer:
a) C₂H₅CN

Question 56.
Ethyl chloride on heating with silver cyanide forms a compound X. The functional isomer of X is
a) C₂H₅NC
b) C₂H₅NCN
c) H₃C – NH – CH₃
d) C₂H₅NH₂
Answer:
b) C₂H₅NCN

Question 57.
With Zn – Cu couple and C₂H₅OH, ethyl Iodide reacts to give
a) ethers
b) diethyl ether
c) Iodoform
d) Ethane
Answer:
d) Ethane

Question 58.
Ethyl bromide on boiling with alcoholic solutions of sodium hydroxide forms
a) Ethane
b) ethylene
c) ethyl alcohol
d) all of these
Answer:
b) ethylene

Question 59.
Following major compound is formed when ethyl chloride reacts with silver nitrite
a) Nitroethane
b) Ethyl nitrite
c) Ethylene
d) Acetaldehyde
Answer:
b) Ethyl nitrite

Question 60.
Which of the following represents Williamson’s synthesis?
a) CH₃COOH + PCl₃ →
b) CH₃ – CH₂ – Cl + CH₃COOH →
c) CH₃ – CH₂ – ONa + CH₃ – CH₂ – Cl →
d) CH₃ – CH₂ – OH + Na →
Answer:
c) CH₃ – CH₂ – ONa + CH₃ – CH₂ – Cl →

Question 61.
The reaction of an alkyl halide with benzene in presence of anhydrous A1Cl₃ gives alkyl benzene the reaction is known as
a) Friedel – craft’s reaction
b) Carbylamine reaction
c) Gattermann reaction
d) Wurtz reaction
Answer:
a) Friedel – craft’s reaction

Question 62.
A Grignard’s reagent reacts with water to give
a) ether
b) alkanes
c) amine
d) Alcohol
Answer:
b) alkanes

Question 63.
C₂H₅Cl + Mg → C₂H₅ MgCl in this reaction the solvent is
a) C₂H₅OH
b) Water
c) Dry ether
d) Acetone
Answer:
c) Dry ether

Question 64.
For a nucleophilic substitution reaction the rate was found in the order RI > RBr > RCl > RF then the reaction could be
a) S_N¹ only
b) S_N² only
c) Either S_N¹ or S_N²
d) Neither S_N¹ or S_N²
Answer:
c) Either S_N¹ or S_N²

Question 65.
S_N² reaction leads to
a) inversion of configuration
b) retention of configuration
c) partial racemisation
d) no racemisation
Answer:
a) inversion of configuration

Question 66.
Which of the following alkyl halide is hydrolysed by S_N¹ mechanism
a) CH₃Cl
b) CH₃ – CH₂ – Cl
c) CH₃ – CH₂ – CH₂ – Cl
d) (CH₃)₃CCl
Answer:
d) (CH₃)₃CCl

Question 67.
S_N¹ reaction is favoured by
a) non – polar solvents
b) Bulky group on the carbon atom attached to the halogen atom
c) Small groups on the carbon atom attached to halogen atom
d) All of the above
Answer:
b) Bulky group on the carbon atom attached to the halogen atom

Question 68.
Which of the following is not stereospecific
a) S_N¹
b) S_N²
c) E₂
d) Addition of Br₂ to ethylene in CCl₄
Answer:
a) S_N¹

Question 69.
Which of the following factors does not favour S_N¹ mechanism
a) Strong base
b) Polar solvent
c) Low. conc. of nucleophile
d) 3° halide
Answer:
c) Low. conc. of nucleophile

Question 70.
The order of reactivity of various alkyl halides toward SN1 reaction is
a) 3° > 2° > 1°
b) 1° > 2° > 3°
c) 3° = 2° = 1°
d) 1° > 3° > 2°
Answer:
a) 3° > 2° > 1°

Question 71.
In aryl halides carbon atom holding halogen is
a) sp² hybridised
b) sp hybridised
c) sp³ hybndised
d) sp³d hybridised
Answer:
a) sp² hybridised

Question 72.
Chloro benzene can be prepared by reacting benzene diazonlum chloride with
a) HCl
b) Cu₂Cl₂ / HCl
c) Cl₂ / AlCl₃
d) HNO₂
Answer:
b) Cu₂Cl₂ / HCl

Question 73.
+ Cl₂ X, X is
a) Chlorobenzene
b) m – dichloro benzene
c) benzene hexachioride
d) p – dichlorobenzene
Answer:
a) Chlorobenzene

Question 74.
The following is an example of Sandmeyer reaction
a) C₆H₅N₂⁺ Cl^– C₆H₅Cl
b) C₆H₅N₂⁺ Cl^– C₆H₅OH
c) C₆H₅N₂⁺ Cl^– C₆H₅F
d) C₆H₅N₂⁺ Cl^– C₆H₅Cl
Answer:
a) C₆H₅N₂⁺ Cl^– C₆H₅Cl

Question 75.
Chlorobenzene on reaction with CH₃Cl in presence of AlCl₃ gives
a) toulene
b) m – chloro toulene
c) only o – chloro toulene
d) mixture of o – and p – chlorotoulene
Answer:
d) mixture of o – and p – chlorotoulene

Question 76.
2C₆H₅Cl + 2Na → X, X is
a) toulene
b) biphenyl
C) phenyl ethane
d) 1 – chloro – 2 – phenyl ethane
Answer:
b) biphenyl

Question 77.
Chlorobenzene on fusing with solid NaOH gives
a) benzene
b) benzoic acid
c) phenol
d) benzene chloride
Answer:
c) phenol

Question 78.
Chlorobenzene on nitration gives major product of
a) 1 – chloro – 4 – nitro benzene
b) 1 – chloro – 3 – nitro benzene
c) 1, 4 – dinitro benzene
d) 2, 4, 6 – tri nitro benzene
Answer:
a) 1 – chloro – 4 – nitro benzene

Question 79.
The reaction C₆H₅I + 2 Na + CH₃I → C₆H₅CH₃ + 2 NaI is
a) Wurtz reaction
b) Fittig reaction
c) Wurtz-Fittig reaction
d) Sandmeyer reaction
Answer:
c) Wurtz-Fittig reaction

Question 80.
R – Cl + Nal R – I + NaCl. This reaction is
a) Wurtz reaction
b) Fittig reaction
c) Finkelstein reaction
d) Frankland reaction
Answer:
c) Finkelstein reaction

Question 81.
C₂H₅OH + SOCl₂ x + y + z. In this reaction x, y and z respectively are
a) C₂H₄Cl₂, SO₂, HCl
b) C₂H₅Cl, SO₂, HCl
c) C₂H₅Cl, SOCl, HCl
d) C₂H₄, SO₂, Cl₂
Answer:
b) C₂H₅Cl, SO₂, HCl’

Question 82.
C₂H₅Cl + AgOH → A + AgCl.
A + CH₃COCl → C + HCl. “C” is
a) Ethyl acetate
b) Methyl acetate
c) butanone – 2
d) propanone
Answer:
a) Ethyl acetate

Question 83.
The compound (B) in the below reaction is:
C₂H₅Cl A B
a) ethylene chloride
b) acetic acid
c) propionic acid
d) ethyl cyanide
Answer:
c) propionic acid

Question 84.
Chloro ethane reacts with X to form diethyl ether. What is X?
a) NaOH
b) H₂SO₄
c) C₂H₅ONa
d) Na₂S₂O₃
Answer:
c) C₂H₅ONa

Question 85.
1 – chlorobutane on reaction with alcoholic potash gives
a) 1 – butene
b) 1 – butanol
c) 1 – butyne
d) 2 – butanol
Answer:
a) 1 – butene

Question 86.
Propane nitrile may be prepared by heating
a) Propyl alcohol with KCN
b) ethyl chloride with KCN
c) Propyl chloride with KCN
d) ethyl chloride with KCN
Answer:
d) ethyl chloride with KCN

Question 87.
CH₃CH = CH₂ A B, B is
a) propanol – 2
b) propanal – 1
c) propanol – 1
d) propanal – 2
Answer:
c) propanol – 1

Question 88.
Y is

Answer:
c)

Question 89.
The correct order of increasing boiling points is
a) 1 – chloropropane < isopropylchloride < 1 – chlorobutane
b) isopropylchloride < 1 – chloropropane < 1 – chlorobutane
c) 1 – chlorobutane < isopropylchloride < 1 – chloropropane
d) 1 – chlorobutane < 1 – chloropropane < isopropylchloride
Answer:
b) isopropylchloride < 1 – chloropropane < 1 – chlorobutane

Question 90.
The correct order of decreasing S_N² reactivity
a) RCH₂X > R₂CHX > R₃CX
b) RCH₂X > R₃CX > R₂CHX
c) R₂CHX > R₃CX > RCH₂X
d) R₃CX > R₂CHX > RCH₂X
Answer:
a) RCH₂X > R₂CHX > R₃CX

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

Question 1.
What are haloalkanes? Give example.
Answer:
Mono halogen derivatives of alkanes are called haloalkanes. Haloalkanes are represented by general formula R – X, Where, R is an alkyl group (C_nH_{2n + 1}) – and X is a halogen atom (X = F, Cl, Br or I). Haloalkanes are further classified into primary, secondary, tertiary haloalkane on the basis of type of carbon atom to which the halogen is attached.
Example:

Question 2.
How will you convert methane into tetra chloro methane?
Answer:
Chlorination of methane gives different products which have differences in the boiling points. Hence, these can be separated by fractional distillation.

Question 3.
What is Finkelstein’s reaction?
Answer:
Chloro or bromoalkane on heating with a concentrated solution of sodium iodide in dry acetone gives iodo alkanes. This reaction is called as Finkelstein reaction.
CH₃CH₂Br + NaI CH₃CH₂I + NaI
Bromo ethane Iodoethane

Question 4.
What is Swartz reaction?
Answer:
Chloro or bromo alkanes on heating with metallic fluorides like AgF or SbF₃ gives fluoro alkanes. This reaction is called Swarts reaction.
CH₃CH₂Br + AgF CH₃CH₂F + AgBr
Bromo ethane Fluoro ethane

Question 5.
What is Hunsdiecker’s reaction?
Answer:
Silver salts of fatty acids when refluxed with bromine in CCl₄ gives bromo alkane.
CH₃CH₂COOAg + Br₂ CH₃CH₂Br + CO₂ + AgBr
Silver propionate Bromo ethane

Question 6.
How is ehtyl magnesium bromide prepared from ethyl bromide?
Answer:
When a solution of ethyl bromide in ether is treated with magnesium, we get ethyl magnesium bromide.
CH₃ – CH₂ – Br + Mg CH₃CH₂MgBr
Ethyl bromide Ethyl magnesium bromide

Question 7.
How will you convert ethyl bromide into ethyl lithium?
Answer:
Ethyl bromide reacts with active metals like sodium, lead etc in the presence of dry ether to form ethyl lithium.
CH₃ CH₂ Br + 2Li CH₃ CH₂Li + LiBr
Ethyl bromide Ethyl Lithium

Question 8.
How is TEL prepared from ethyl bromide?
Answer:
When ethyl bromide reacts with Na / Pb alloy to give TEL.
4CH₃ CH₂ Br + 4Na/Pb → (CH₃CH₂)₄Pb + 4NaBr + 3Pb
Ethyl bromide Sodium-lead alloy Tetraethyl lead (TEL)

Question 9.
Haloalkanes have higher boiling point and melting point than the parent alkane. Justify this statement.
Answer:
Haloalkanes have higher boiling point than the parent alkane having the same number of carbon atoms because the intermolecular forces of attraction and dipole-dipole interactions are comparatively stronger in haloalkanes.

Question 10.
How is methane prepared from Grignard reagent?
Answer:
Compounds like water, alcohols and amines which contain active hydrogen atom react with Grignard reagents to form alkanes.
CH₃MgI + HO – H → CH₄ + Mg I (OH)

CH₃MgI + C₂H₅OH          CH₄ + MgI (OC₂H₅)
Ethyl alcohol                Methane

Question 11.
What are Haloarenes? Give a suitable example.
Answer:
Haloarenes are the compounds in which the halogen is directly attached to the benzene ring.
Example:

Question 12.
How is chloro benzene prepared from benzene by direct halogenation?
Answer:
Chloro benzene is prepared by the direct chlorination of benzene in the presence of lewis acid catalyst like FeCl₃.

Question 13.
Write a note on Sand Meyer reaction.
Answer:
When aqueous solution of benzene diazonium chloride is warmed with Cu₂Cl in HCl gives chioro benzene.

Question 14.
How is Iodo benzene prepared from benzene diazonium chloride?
Answer:
Iodo benzene is prepared by warming benzene diazonium chloride with an aqueous KI solution.
C₆H₅N₂Cl + Kl          C₆H₅I + N₂ + KCl
Benzene dizonium chloride        Iodo benzene

Question 15.
What happens when ethylidene dichloride is treated with Zinc dust in methanol?
Answer:
Gem dihalides and vic – dihalides on treatment with zinc dust in methanol give alkenes.
CH₃ – CHCl₂ + Zn CH₂ = CH₂ + ZnCl₂
Ethylidene dichloride                     Ethylene

Question 16.
How will you convert chloroform into methylene chloride?
Answer:
a) Reduction of chloroform in the presence of Zn + HCl gives methylene chloride.
CHCl₃ (chloroform) CH₂Cl (Methylene chloride) + HCl
b) Reduction of chloroform using H2/Ni
CHCl₃ (chloroform) CH₂Cl₂ (Methylene chloride) + HCl

Question 17.
Write the chlorination reaction of methane.
Answer:
Chlorination of methane gives methylene chloride
CH₄ CH₃Cl CH₂Cl₂
Methane                                               Methylene chloride

Question 18.
How is chloroform prepared from carbon tetrachloride?
Answer:
Carbon tetrachloride is reduced by iron powder in dilute HCl medium to form chloroform
CCl₄ (carbon tetrachloride) + 2[H] CHCl₃ (chloroform) + HCl

III. Short question and answers (3 Marks):

Question 1.
write the IUPAC name of the following.
i)

ii)

iii)
Answer:
i)

ii)

iii)

Question 2.
Write the structure of the following compounds.
i) 1 – Bromo – 4 – ethyl cyclohexane
ii) 1, 4 – Dichlorobut – 2 – ene
iii) 2 – chloro – 3 – methyl pentane
Answer:
i) 1 -Bromo – 4 – ethyl cyclohexane:

ii) 1, 4 – Dichlorobut – 2 – ene
– Cl – CH₂ – CH = CH – CH₂ – Cl

iii) 2 – chloro – 3 – methyl pentane

Question 3.
Write any three methods of preparation of chloro ethane from ethanol.
Answer:
a) Reaction with hydrogen halide:
Ethanol is heated with HCl in presence of anhydrous ZnCl₂ to give ethyl chloride.
CH₃CH₂OH + HCl CH₃CH₂Cl + H₂O
Ethanol Chloroethane

b) Reaction with phosphorous halides:
Ethanol reacts with PCl₅ or PCl₃ it gives ethyl chloride.
CH₃CH₂OH + PCl₅ → CH₃CH₂Cl + POCl₃ + HCl
Ethanol                      Chloro ethane

3 CH₃CH₂OH + PCl₃ → 3CH₃CH₂Cl + H ₃PO₃
Ethanol                        Chloro ethane

C) Reaction with thionyl chloride (Sulphonyl chloride):
When ethanol reacts with SOCl₂ in presence of pyridine, it gives chloro ethane.
CH₃CH₂OH + SOCl₂ CH₃CH₂Cl + SO₂ + HCl
Ethanol                                            Chloro ethane

Question 4.
What happens when haloalkane reacts with aqueous alkali or most silver oxide?
Answer:
Ethyl bromide reacts with aqueous solution of KOH or moist silver oxide (Ag₂O/H₂O) to form ethanol.
CH₃ – CH₂ – Br + KOH (aq) CH₃CH₂ – OH + KBr
Bromoethane                                   Ethyl alcohol

CH₃ – CH₂ – Br   +   AgOH(aq)    CH₃CH₂ – OH + AgBr
Bromoethane         Moist silver oxide            Ethanol

Question 5.
Explain the ammonolysis reaction of bromo ethane.
Answer:
Ethyl bromide reacts with alcoholic ammonia solution to form ethylamine.
CH₃ – CH₂ – Br + H – NH₂ → CH₃ CH₂ – NH₂ + HBr
Bromo ethane                        Ethyl amine

However, with excess of ethyl bromide, secondary and tertiary amines along with quaternary ammonium salts are obtained.

Question 6.
Explain the hydrolysis of 2-bromobutane with aqueous KOB.
Answer:

• 2-bromobutane is optically active and it undergoes S_N¹ reaction with aqueous KOH.
• The product obtained will be an optically inactive racemic mixture.
• As nucleophilic reagent OH^– ion can attack the carbocation from both sides to form equal proportions of dextro and levorotatory optically active isomers, it results in the formation of an optically inactive racemic mixture.

Question 7.
How are the following compounds prepared from bromo ethane?
i) Ethane thiol
(ii) Diethyl ether
Answer:
i) Ethane thiol:
Ethyl bromide reacts with sodium or potassium hydrogen sulphide to form thio alcohols.
CH₃CH₂Br CH₃CH₂SH + NaBr
Bromo ethane                           Ethane thiol

ii) Diethyl ether:
Ethyl bromide, when boiled with sodium alkoxide gives diethyl ether. This method can be used to prepare mixed (unsymmetrical) ethers also.
CH₃ – CH₂Br +      NaOCH₂CH₃          CH₃CH₂ – O – CH₂CH₃ + NaBr
Bromo ethane        Sodium ethoxide                 diethyl ether

Question 8.
How is ethane prepared from the following compounds?
i) Bromo ethane ii) Iodo ethane
Answer:
i) From bromo ethane:
Ethyl bromide is reduced to ethane by treating with H₂ in the presence of metal catalyst like nickel, palladium etc or with hydroiodic acid in the presence of red phosphrous.
Ni(or)Pd
CH₃CH₂Br + H₂ CH₃ – CH₃ + HBr
Bromo ethane                              Ethane

ii) From iodo ethane:
Iodo ethane is reduced with H₂ in presence of Red P it gives ethane.
CH₃CH₂I + HI CH₃ – CH₃ + I₂
Iodo ethane                         Ethane

Question 9.
What are the uses of carbon tetrachioride?
Answer:

• Carbon tetrachioride in used as a dry cleaning agent.
• It is used as a solvent for oils, fats and waxes.
• As the vapours of CCl₄ is non-combustible, it is used under the name pyrene for extinguishing the fire caused by oil (or) petrol.

Question 10.
Write the uses of Grignard reagents.
Answer:

1. Grignard reagents are synthetically very useful compounds. These reagents are converted to various organic compounds like alcohols, carboxylic acids, aldehydes and ketones.
2. The alkyl group being electron rich acts as a carbanion or a nucleophile.
3. They would attack polarized molecules at a point of low electron density. The following reactions illustrate the synthetic uses of the Grignard reagent.

Question 11.
What is Balz – Schiemann reaction?
Answer:
Fluoro benzene is prepared by treating benzene diazonium chloride with fluoro boric acid. This reaction produces diazonium fluoroborate which on heating produces fluorobenzene. This reaction is called Balz – Schiemann reaction.
C₆H₅N₂Cl + HBF₄ C₆H₅N₂ + BF₄^– C₆H₅ F + BF₃ + N₂
Benzene diazonium chloride                                          Fluorobenzene

Question 12.
What are the uses of chlorobenzene?
Answer:

• Chlorobenzene is used in the manufacture of pesticides like DDT.
• It is used as high boiling solvent in organic synthesis.
• It is used as a fibre – swelling agent in textile processing.

Question 13.
How is ethylidene dichloride prepared from
(i) Acetaldehyde
(ii) Acetylene
Answer:
i) Treating acetaldehyde with PCl₅:
CH₃CHO (Acetaldehyde) + PCl₅ → CH₃CHCl₂ (Ethylidene dichloride) + POCl

ii) Adding hydrogen chloride to acetylene

Question 14.
How is ethylene dichloride prepared from
(i) Chloride?
(ii) PCl₅?
Answer:
i) Addition of chlorine to ethylene

ii) Action of PCl₅ (or HCl) on ethylene glycol

Question 15.
What happens when the following compounds are treated with alcoholic KOH?
i) Ethylidene dichloride
ii) Ethylene dichloride
Answer:
i) Ethylidene dichloride:

ii) Ethylene dichloride:

Question 16.
What are the uses of methylene chloride?
Answer:
Methylene chloride is used as:

• Aerosol spray propellant.
• Solvent in paint remover.
• Process solvent in the manufacture of drugs.
• A metal cleaning agent.

Question 17.
How are the following compounds prepared from chlorobenzene?
i) Benzene
ii) Phenyl magnesium chloride
Answer:
i) C₆H₆ (Benzene):
Chlorbenzene undergoes reduction with Ni – Al alloy in the presence of NaOH gives benzene.
C₆H₅Cl + 2(H) C₆H₆ + HCl
Chloro benzene              Benzeneide

ii) Phenyl Magnesium chloride:
Chloro benzene reacts with magnesium to form phenyl magnesium chloride in tetra hydrofuran (THF).
C₆H₅Cl (Chloro benzene) + Mg C₆H₅MgCl (Phenyl magnesium chloride)

IV. Long question and answers (5 Marks):

Question 1.
Explain the S_N² mechanism of haloalkanes with a suitable example.
Answer:
S2 stands for bimolecular nucleophilic substitution
“S” stands for substitution
“N” stands for nucleophilic
“2” stands for bimolecular (two molecules are involved in the rate-determining step)
The rate of S_N² reaction depends upon the concentration of both alkyl halide and the nucleophile.
Rate of reaction = k₂ [alkyihalide] [nucleophile]
This S_N² reaction follows second-order kinetics and occurs in one step.

This reaction involves the formation of a transition state in which both the reactant molecules are partially bonded to each other. The attack of nucleophile occurs from the backside and the halide ion leaves from the front side. The carbon at which substitution ocurs ha inverted configuration during the course of reaction just as an umbrella has tendency to invert in a wind storm. This inversion of configuration is called Walden inversion; after paul Walden who first discovered the inversion of configuration of a compound in S_N² reaction.

We understand S_N² reaction mechanism by taking an example of reaction between chloromethane and aqueous KOH. S_N² reaction of an optically active haloalkane is always accompanied by inversion of configuration at the asymmetric centre.

 

Question 2.
Explain the S_N¹ mechanism of haloalkanes with suitable examples.
Answer:
S_N¹ stands for unimolecular nucleophilic substitution
‘S’ stands for substitution
‘N’ stands for nucleophilic
‘1’ stands for unimolecular (one molecule is involved in the rate-determining step)
The rate of the following S_N¹ reaction depends upon the concentration of aikyl halide (RX) and is independent of the concentration of the nucleophile (OH-).
Rate of the reaction = k[alkyl halide]
R – Cl + OH^– → R – OH + Cl^–
This S_N¹ reaction follows first order kinetics and occurs in two steps.
We understand S_N¹ reaction mechanism by taking a reaction between tertiary hutyl bromide with aqueous KOH.

This reaction takes place in two steps aas shown below.

Step – 1:
Formation of carbocation:

t – butyl bromide

The polar C – Br bond breaks forming a carbocation and bromide ion. This step is slow and hence it is the rate-determining step.

Step 2:
Nucleophilic attack on carbocation
The carbocation immediately reacts with the nucleophile. This step is fast and hence does not affect the rate of the reactions.

As shown above, the nucleophilic reagent OH~ can attack carbocation from both the sides, they will be a mirror image of each other.

In the above example, the substrate tert-butyl bromide is not optically active, hence the obtained product is optically inactive. If halo alkane substrate is optically active then, the product obtained will be an optically inactive racemic mixture. As nucleophilic reagent OH- can attack carbocation from both the sides, to form an equal proportion of dextro and levorotatory optically active isomers which form optically inactive racemic mixture.

Example: Hydrolysis of optically active 2- Bromo butane gives racemic mixture of ±butan-2-ol.

Question 3.
Explain the E₂ reaction mechanism with a suitable example.
Answer:
E₂ stands for bimolecular elimination reaction
‘E’ stands for elimination
‘2’ stands for bimolecular
The rate of E₂ reaction depends on the concentration of alkyl halide and base

Rate = k₂ [alkyl halide] [base]
It is therefore, a second-order reaction generally primary alkyl halide undergoes this reaction in the presence of alcoholic KOH. E₂ is a one-step process in which the abstraction of the proton from the p carbon and expulsion of halide from the carbon occurs simultaneously. The mechanism is shown below.

Question 4.
Explain the E₁ reaction mechanism with a suitable example.
Answer:
E₁ stands for unimolecular elimination reaction
‘E’ stands for elimination ‘1’ stands for unimolecular
The rate of E₁ reaction depends on the concentration of alkyl halide only and hence
rate = k [alkyl halide]
Generally, tertiary alkyl halide which undergoes elimination reaction by this mechanism in the presence of alcoholic KOH following first order kinetics is a two step process. The mechanism is shown by taking following example.

step – 1:
Heterolytic fission to yield a carbocation:

Step – 2:
Elimination of a proton from the β – carbon to produce an alkene:

Question 5.
How are the following compounds prepared from Methyl magnesium iodide?
(i) Ethanol
(ii) Tert-butyl alcohol
(iii) Acetaldehyde
Answer:
(i) Ethanol:
Formaldehyde reacts with methyl magnesium iodide to give addition products which on hydrolysis yield ethanol.

(ii) Tert-butyl alcohol:
Acetone reacts with methyl magnesium iodide to give an additional product which on hydrolysis yields tert butyl alcohol.

(iii) Acetaldehyde:
Ethyl formate reacts with methyl magnesium iodide and followed by acid hydrolysis, it forms Acetaldehyde.

Question 6.
How is ethylene dichloride converted into
i) Acetaldehyde
ii) Ethylene glycol
ii) Ethylene
Answer:
i) Acetaldehyde:
(Hydrolysis with aqueous NaOH or KOH)
Ethylidene chloride reacts with aqueous KOH to give acetaldehyde.

ii) Ethylene glycol :
Ethylene chloride reacts with aqueous KOH, it gives glycol.

iii) Ethylene:
Ethylene dichloride is heated with Zinc in presence of methanol gives ehtylene.

Question 7.
How are the following compounds prepared from chloroform?
(i) Phosgene
(ii) Methylene chloride
(iii) Methyl isocyanide
Answer:
(i) Phosgene:
Chloroform undergoes oxidation in the presence of light and air to form phosgene (carbonyl chloride)

Since phosgene is very poisonous, its presence makes chloroform unfit for use as an anaesthetic.

(ii) Methylene chloride:
Chloroform undergoes reduction with zinc and HCl in the presence of ethyl alcohol to form methylene chloride.
CHCl₃ (Chloroform) + 2(H) CH₂Cl₂ (Methylene chloride)+ Cl₂

(iii) Methyl isocyanide:
Chloroform reacts with an aliphatic or aromatic primary amine and alcoholic caustic potash, to give foul-smelling alkyl isocyanide (carbylamines).
CH₃NH₂    +     CHCl₂ + 3KOH → CH₃NC + 3KCl + 3H₂O
Methylamine Chloroform          Methyl isocyanide

Question 8.
Discuss the aromatic electrophilic substitutions reaction of chlorobenzene.
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Ex 8.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Ex 8.3

Choose the correct Answer:

Question 1.
A ……………… may be finite or infinite according as the number of observation or items in it is finite or infinite
(a) Population
(b) census
(c) parameter
(d) none of these
Solution:
(a) population

Question 2.
A …………….. of statistical individuals in a population is called a sample.
(a) Infinite set
(b) finite subset
(c) finite set
(d) entire set
Solution:
(b) finite subset

Question 3.
A finite subset of statistical individuals in a population is called ………………
a) a sample
(b) a population
(c) universe
(d) census
Solution:
(a) a sample

Question 4.
Any statistical measure computed from sample data is known as ……………..
(a) Parameter
(b) random sample
(c) Infinite measure
(d) uncountable
Solution:
(b) random sample

Question 5.
A ………………. is one where each item in the universe has an equal chance of known opportu¬nity of being selected
(a) Parameter
(b) random sample
(c) statistic
(d) entire data
Solution:
(b) random sample

Question 6.
A random sample is a sample selected in such a way that every item in the population has an equal chance of being included
(a) Harper
(b) fisher
(c) karl pearson
(d) Dr. yates
Solution:
(a) Harper

Question 7.
Which one of the following is probability sampling
(a) Purposive sampling
(b) judgement sampling
(c) sample random sampling
(d) Convenience sampling
Solution:
(c) sample random sampling

Question 8.
In simple random sampling of drawing any unit, the probability of drawing any unit at the draw is ?
(a) \(\frac { n }{N}\)
(b) \(\frac { 1 }{N}\)
(c) \(\frac { N }{n}\)
(d) n
Solution:
(b) \(\frac { 1 }{N}\)

Question 9.
In ……………. the heterogeneous groups divided into homogeneous groups
(a) Non-probability sample
(b) a sample random sample
(c) a stratified random sample
(d) Systematic sample
Solution:
(c) a stratified random sample

Question 10.
Errors in sampling are of
(a) Two types
(b) three types
(c) four types
(d) five types
Solution:
(a) Two types

Question 11.
The method of obtaining the most likely value of the population parameter using statistic is called
(a) estimate
(b) estimate
(c) biased estimate
(d) standard error
Solution:
(d) standard error

Question 12.
An estimator is a sample statistic used to estimate a
(a) population parameter
(b) biased estimate
(c) sample size
(d) census
Solution:
(a) population parameter

Question 13.
…………… is a relative property, which states that one estimate is efficient relative to another.
(a) efficiency
(b) sufficiency
(c) unbiased
(d) consistency.
Solution:
(a) efficiency

Question 14.
If probability p[|\(\bar { θ }\) – θ|< ∈|< ∈|] 1 → µ as n → α for any positive then \(\bar { θ }\) is said to estimator of θ
(a) efficient
(b) sufficient
(c) unbiased
(d) consistent
Solution:
(d) Consistent

Question 15.
An estimator is said to be ………….. if it contains all the information in the data about the parameter it estimates.
(a) efficient
(b) sufficient
(c) unbiased
(d) consistent
Solution:
(b) sufficient

Question 16.
An estimate of a population parameter given by two numbers between which the parameter would be expected to lie called an ………….. interval estimate of the parameter
(a) point estimate
(b) interval estimate
(c) standard error
(d) confidence
Solution:
(b) interval estimate

Question 17.
A ……………… is a statement or an assertion about the population parameter
(a) hypothesis
(b) statistic
(c) sample
(d) census
Solution:
(a) hypothesis

Question 18.
Type I error is
(a) Accept H₀ when it is true
(b) Accept H₀ when it is false
(c) Reject H₀ when it is true
(d) Reject H₀ when it is false
Solution:
(c) Reject H₀ when it is true

Question 19.
Type II error is?
(a) Accept H₀ when it is wrong
(b) Accept H₀ when it is when it is true
(c) Reject H₀ when it is true
(d) Reject H₀ when it is false
Solution:
(a) Accept H₀ when it is wrong

Question 20.
The standard error of sample mean is?
(a) \(\frac { σ }{\sqrt{2n}}\)
(b) \(\frac { σ }{n}\)
(c) \(\frac { σ }{√n}\)
(d) \(\frac { σ^2 }{√n}\)
Solution:
(c) \(\frac { σ }{√n}\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Miscellaneous Problems

Question 1.
Explain the types of sampling.
Solution:
The different types of sampling are

• simple random sampling
• Stratified random sampling and
• Systematic sampling

(i) In simple random sampling, every item of the population has an equal chance for being selected. The sampling can be done with replacement (or) without replacement. A random sampling from a finite population with replacement is equivalent to sampling from an infinite population without replacement. This technique will give useful results only if the population is homogeneous. The following are some of the methods of selecting a random sample.

(a) Use of an unbiased die or coin: If we have to choose between two alternatives, a coin is tossed, and depending on the head or tail course of action is taken. A die can be employed if there are six different alternatives.

(b) Lottery sampling: Here a random sample is selected by identifying each element of the population by means of a card of a pack of uniform cards or (by writing the number on pieces of paper) and to select a required number of cards after thorough mixing of the cards.

(c) Random numbers: Random numbers are formed of ‘random digits’ and arranged in the form of a table having a number of rows and columns. Tippett’s numbers form one such table wherein 40,000 digits were selected at random from census reports and combined by groups of four into 10,000 numbers.

(ii) In stratified random sampling, a population of units is divided into L sub-populations of N₁, N₂, …… N_L. The sub-populations being non-overlapping and mutually exhaustive so that N = N₁ + N₂ + …….. + N_L. Each subpopulation is known as a stratum. If we select n₁, n₂, ……. n_l items, respectively, from these strata, we get a stratified sample. If a simple random sample is taken from each stratum, the whole procedure is referred to as stratified random sampling.

(iii) Systematic sampling is a form of restricted random selection which is highly useful in surveys concerning enumerable population. In this method, every member of the population is numbered in serial order and every ith element, starting from any of the first items is chosen. For example, suppose we require a 5% sample of students from a college where there are 2000 students, we select a random number from 1 to 20. If it is 12, then our sample consists of students with numbers 12, 32, 52, 72, …… 1992.

Question 2.
Write a short note on sampling distribution and standard error.
Solution:
sampling distribution:
The sampling distribution of a statistic is the frequency distribution which is formed with various values of a statistic computed from different samples of the same size drawn from the same population.

For instance, if we draw a sample a size n from a given finite population of N, then the total number of possible samples is Nc_n = \(\frac { N! }{n!(N-n)!}\) = k(say)

Standard Error:
The standard deviation of the sampling distribution of a statistic is known as its standard Error abbreviated as S.E. The standard error (S.E) of some of the well-known statistics, for large samples, are given below, where n is the sample size, σ² is the population variance.

Question 3.
Explain the procedures of testing of hypothesis
Solution:
The following are the steps involved in hypothesis testing problems:
1. Null hypothesis: Set up the null hypothesis H₀

2. Alternative hypothesis: Set up the alternative hypothesis. This will enable us to decide whether we have to use two-tailed test or single-tailed test (right or left tailed)

3. Level of significance: Choose the appropriate level of significance (a) depending on the reliability of the estimates and permissible risk. This is to be fixed before the sample is drawn, i.e., a is fixed in advance.

4. Test statistic: Compute the test statistic
Z = \(\frac { t-E(t) }{\sqrt{var(t)}}\) = \(\frac { t-E(t) }{S.E(t)}\) N(0, 1) as n → ∞

5. Conclusion: We compare the computed value of Z in step 4 with the significant value or critical value or table value Zα at the given level of significance.
(i) If |Z | < Zα i.e., if the calculated value is less than the critical value we say it is not significant. This may due to fluctuations of sampling and sample data do not provide us sufficient evidence against the null hypothesis which may therefore be accepted.

(ii) If |Z |> Zα i.e., if the calculated value of Z is greater than critical value Zα then we say it is significant and the null hypothesis is rejected at the level of significance α.

Question 4.
Explain in detail the test of significance for a single mean.
Solution:
Let xi, (i = 1, 2, 3, …, n) is a random sample of size from a normal population with mean µ and variance σ² then the sample mean is distributed normally with mean and variance
\(\frac { σ^2 }{n}\), i.e \(\bar { x }\) N(µ, \(\frac { σ^2 }{n}\))

Thus for large samples, the standard normal variate corresponding to \(\bar { x }\) is
Z = \(\frac {\bar { x } -µ }{\frac{σ}{√n}}\) N (0, 1)

Under the null hypothesis that the sample has been drawn from a population with mean and variance σ², i.e., there is no significant difference between the sample mean (\(\bar { x }\)) and the population mean (α), the test statistic (for large samples) is:
Z = \(\frac {\bar { x } -µ }{\frac{σ}{√n}}\)

Question 5.
Determine the standard error of proportion for a random sample of 500 pineapples was taken from a large consignment and 65 were found to be bad?
Solution:
sample size n = 500
No. of bad pine apples = 65
sample proportion = P = \(\frac { 65 }{500}\) = 0.13
Q = 1 – p ⇒ Q = 1 – 0.13
∴ Q = 0.87
The S.E for sample proportion is given by
S.E = \(\sqrt{\frac { PQ }{N}}\) = \(\sqrt{\frac { (0.13)(0.87) }{500}}\)
= \(\sqrt{\frac { 0.1131 }{500}}\) = \(\sqrt{0.0002262}\)
= 0.01504
∴ S.E = 0.015
Hence the standard error for sample proportion is S.E = 0.015

Question 6.
A sample of 100 students are drawn from a school The mean weight and variance of the sample are 67.45 kg and 9 kg respectively find (a) 95% and (b) 66% confidence intervals for estimating the mean weight of the students.
Solution:
sample size n = 100
The sample mean = \(\bar { x }\) = 67.45
The sample variance S² = 9
The sample standard deviation S = 3
S.E = \(\frac { S }{√n}\) = \(\frac { 3 }{\sqrt{100}}\) = \(\frac { 3 }{10}\) = 0.3

(a) The 95% confidence limits for µ are given by
\(\bar { x }\) – z_{\(\frac { α }{2}\)} S.E < µ < \(\bar { x }\) + z_{\(\frac { α }{2}\)} S.E
67.45 – (1.96 × 0.3) ≤ µ ≤ 67.45 + (1.96 × 0.3) 67.45 – 0.588 ≤ µ ≤ 67.45 + 0.588
66.862 ≤ µ ≤ 68.038
The confidential limits is (66.86, 68.04)

(b) The 99% confidence limits for estimating µ are given by
\(\bar { x }\) – z_{\(\frac { α }{2}\)} S.E ≤ µ ≤ \(\bar { x }\) + z_{\(\frac { α }{2}\)} S.E
67.45 – (2.58 × 0.3) ≤ µ ≤ 67.45 + (2.58 × 0.3)
67.45 – 0.774 ≤ µ ≤ 67.45 + 0.774
66.676 ≤ µ ≤ 68.224
∴ The 99% confidence limits is (66.68, 68.22)

Question 7.
The mean I.Q of a sample of 1600 children was 99. it is likely that this was a random sample from a population with a mean I.Q 100 and standard deviation of 15? (Test at 5% level of significance)
Solution:
sample size n = 1600
\(\bar { x }\) = 99
sample mean
Population mean µ = 100
population S.D σ = 15
under the Null hypothesis H₀ : µ = 100
Alternative hypothesis H₁ : µ = 100 (two tails)
Level of significance µ = 0.05

z = -2.666
z = -2.67
Calculated value |z| = 2.67
critical value at 5% level of significance is
z_{\(\frac { α }{2}\)} = 1.96
Inference:
Since the calculated value is greater than table value i.e z ⇒ z_{\(\frac { α }{2}\)} at 5% level of significance, the null hypothesis is rejected. Therefore we conclude that the sample mean differs, significantly from the population mean.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Ex 9.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Ex 9.2

Question 1.
Define Index Number.
Solution:
“An Index Number is a device which shows by its variations the Changes in a magnitude which is not capable of accurate measurements in itself or of direct valuation in practice”. – Wheldon

“An Index number is a statistical measure of fluctuations in a variable arranged in the form of a series and using a base period for making comparisons” – Lawrence J Kaplan

Question 2.
State the uses of Index Number.
Solution:

1. The index number is an important tool for formulating decision and management p policies.
2. It helps in studying the trends and tendencies.
3. It determines the inflation and deflation in an Economy.

Question 3.
Mention the classification of Index Number.
Solution:
Classification of Index Numbers:
Index number can be classified as follows:

1. Price Index Number: It measures the general changes in the retail or wholesale price level of a particular or group of commodities.
2. Quantity Index Number: These are indices to measure the changes in the number of goods manufactured in a factory.
3. Cost of living Index Number: These are intended to study the effect of change in the price level on the cost of living of different classes of people.

Question 4.
Define Laspeyre’s Price index number.
Solution:
Laspeyres Price index number
P\(_{ 01 }^{L}\) ⥪ \(\frac { Σp_1q_0 }{Σp_0q_0}\) × 100
where P₁ = Current year price
p₀ = base year price
q₀ = base year quantity

Question 5.
Explain Paasche’s price index number.
Solution:
Paasches price index number
P\(_{ 01 }^{L}\) = \(\frac { Σp_1q_1 }{Σp_0q_1}\) × 100
where P₁ = current year price
q₁ = current year quantity
p₀ = base year price
q₀ = base year quantity

Question 6.
Write note on Fisher’s price index number.
Solution:
Fishers price index number
P^F = \(\sqrt { P^L×P^P}\)
P\(_{ 01 }^{F}\) = \(\sqrt { \frac{Σp_1p_0×Σp_1q_1}{Σp_0q_0×Σp_0q_1}}\) × 100
where P₁ = current year price
q₁ = current year quantity
p₀ = base year price
q₀ = base year quantity

Question 7.
State the test of adequacy of index number.
Solution:
There are two tests which are used to test the adequacy for an index number. The two tests are as follows
(i) Time Reversal Test
(ii) Factory Reversal Test

Question 8.
Define Time Reversal Test.
Solution:
It is an important test for testing the consistency of a good index number. This test maintains time consistency by working both forward and backward with respect to time (here time refers to base year and current year). Symbolically the following relationship should be satisfied. P₀₁ × p₁₀ = 1

Fisher’s index number formula satisfies the above relationship
P\(_{ 01 }^{F}\) = \(\sqrt { \frac{Σp_1p_0×Σp_1q_1}{Σp_0q_0×Σp_0q_1}}\)

when the base year and current year are interchanged, we get
P\(_{ 10 }^{F}\) = \(\sqrt { \frac{Σp_0q_1×Σp_0q_0}{Σp_1q_1×Σp_1q_0}}\)
P\(_{ 01 }^{F}\) × P\(_{ 10 }^{F}\) = 1

Question 9.
Explain factor reversal test.
Solution:
This is another test for testing the consistency of a good index number. The product of price index number and quantity index number from the base year to the current year should be equal to the true value ratio. That is, the ratio between the total value of current period and total
value of the base period is known as true value ratio. Factor Reversal Test is given by,

where P₀₁ is the relative change in price
Q₀₁ is the relative change in quantity.

Question 10.
Define true value ratio.
Solution:
The ratio between the total value of the current period and the total value of the base period is known as the true value ratio.
(i.e) true value ratio = \(\frac{\sum p_{1} q_{1}}{\sum p_{0} q_{0}}\)

Question 11.
Discuss the Cost of the Living Index Number.
Solution:
Cost of living index numbers are generally designed to represent the average change over time in the prices paid by the ultimate consumer for a specified quantity of goods and services cost of living index number is also known as the consumer price index number.

It is well known that a given change in the level of prices (retail) affects the cost of living of different classes of people in different manners. The general index number fails to reveal this. Therefore it is essential to construct a cost of living index number which helps us in determining the effect of rise and fall in prices on different classes of consumers living in different areas.

Question 12.
Define family budget method.
Solution:
Family Budget Method:
In this method, the weights are calculated by multiplying prices and quantity of the base year.
(i.e.) V = Σp₀q₀. The formula is given by,
Cost of Living Index Number = \(\frac{\sum \mathrm{PV}}{\sum \mathrm{V}}\)
where P = \(\frac{p_{1}}{p_{0}} \times 100\) is the price relative
V = Σp₀q₀ is the value relative

Question 13.
State the uses of cost of Living Index Number.
Solution:
(i) It indicates whether the real wages of workers are rising or falling for a given time.
(ii) It is used by the administrators for regulating dearness allowance or grant of bonus to the workers.

Question 14.
Calculate by a suitable method, the index number of price from the following data:

Solution:

Question 15.
Calculate price index number for 2005 by (a) Laspeyre’s (b) Paasche’s method

Solution:

Question 16.
Calculate price index number for 2005 by (a) Laspeyre’s (b) Paasche’s method

Solution:

Question 17.
Using the following data, construct Fisher’s Ideal index and show how it satisfies Factor Reversal Test and Time Reversal Test?

Solution:

Question 18.
Using the following data, construct Fisher’s Ideal index and show how it satisfies Factor Reversal Test and Time Reversal Test?

Solution:

Question 19.
Calculate Fisher’s index number to the following data. Also show that it satisfies Time Reversal Test.

Solution:

Question 20.
Th following are the group index numbers and the group weights of an average working class family’s budget. Construct the cost of living index number:

Solution:

Question 21.
Construct the cost of living Index number for 2015 on the basis of 2012 from the following data using family budget method.

Solution:

Question 22.
Calculate the cost of living index by aggregate expenditure method:

Solution: