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Students can download Maths Chapter 7 Mensuration Additional Questions and Answers, Notes, Samacheer Kalvi 10th 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.

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Additional Questions

I. Multiple Choice Questions:

Question 1.
The curved surface area of a right circular cylinder of radius 1 cm and height 1 cm is equal to ______
(1) π cm²
(2) 2π cm²
(3) 3π cm²
(4) 2 cm²
Answer:
(2) 2π cm²
Hint:
C.S.A of a cylinder = 2πrh sq. units = 2 × π × 1 × 1 cm² = 2π cm²

Question 2.
The total surface area of a solid right circular cylinder whose radius is half of its height h is equal to ______ sq. units.
(1) \(\frac{3}{2} \pi h\)
(2) \(\frac{2}{3} \pi h^{2}\)
(3) \(\frac{3}{2} \pi h^{2}\)
(4) \(\frac{2}{3} \pi h\)
Answer:
(3) \(\frac{3}{2} \pi h^{2}\)
Hint:
T.S.A = 2πr(h + r)
[radius is half of the height]
= \(2 \pi \times \frac{h}{2}\left(\frac{h}{2}+h\right)\)
= \(=\pi h\left(\frac{3 h}{2}\right)=\frac{3}{2} \pi h^{2}\) sq. units

Question 3.
Base area of a right circular cylinder is 80 cm². If its height is 5 cm, then the volume is equal to _______
(1) 400 cm³
(2) 16 cm³
(3) 200 cm³
(4) \(\frac{400}{3}\) cm³
Answer:
(1) 400 cm³
Hint:
Volume of a cylinder = πr²h cu. units
[Base area (πr²) = 80 cm² = 80 × 5 cm³ = 400 cm³

Question 4.
If the total surface area of a solid right circular cylinder is 200π cm² and its radius is 5 cm, then the sum of its height and radius is ______
(1) 20 cm
(2) 25 cm
(3) 30 cm
(4) 15 cm
Answer:
(1) 20 cm
Hint:
T.S.A of a cylinder = 200π cm²
2πr (h + r) = 200π
2 × 5 (h + r) = 200
(h + r) = 20 cm

Question 5.
The curved surface area of a right circular cylinder whose radius is a units and height is b units, is equal to ______
(1) πa²b sq.cm
(2) 2πab sq.cm
(3) 2π sq.cm
(4) 2 sq.cm
Answer:
(2) 2πab sq.cm .
Hint:
C.S.A. of a cylinder = 2πrh sq. units = 2 × π × a × b sq. cm = 2πab sq. cm

Question 6.
Radius and height of a right circular cone and that of a right circular cylinder are respectively, equal. If the volume of the cylinder is 120 cm³, then the volume of the cone is equal to _______
(1) 1200 cm³
(2) 360 cm³
(3) 40 cm³
(4) 90 cm³
Answer:
(3) 40 cm³
Hint:
Volume of the cone = \(\frac{1}{3}\) × volume of the cylinder
= \(\frac{1}{3}\) × 120 cm³
= 40 cm³

Question 7.
If the diameter and height of a right circular cone are 12 cm and 8 cm respectively, then the slant height is
(1) 10 cm
(2) 20 cm
(3) 30 cm
(4) 96 cm
Answer:
(1) 10 cm
Hint:
Slant height of a cone

Question 8.
If the circumference at the base of a right circular cone and the slant height are 120π cm and 10 cm respectively, then the curved surface area of the cone is equal to ______
(1) 1200π cm²
(2) 600π cm²
(3) 300π cm²
(4) 600 cm²
Answer:
(2) 600π cm²
Hint:
Circumference (2πr) = 120π cm
Slant height (l) = 10 cm;
Curved surface area of a cone = πrl sq. units
= \(\frac{120 \pi}{2}\) × 10 cm² = 600π cm²

Question 9.
If the volume and the base area of a right circular cone are 48π cm and 12π cm respectively, then the height of the cone is equal to ______
(1) 6 cm
(2) 8 cm
(3) 10 cm
(4) 12 cm
Answer:
(4) 12 cm
Hint:
Volume of a cone = 48π cm³
[Base area (πr²) = 12π]
\(\frac{1}{3}\) πr²h = 48π
\(\frac{1}{3}\) × 12π × h = 48π
[Substitute πr² = 12π]
h = \(\frac{48}{4}\) = 12 cm

Question 10.
If the height and the base area of a right circular cone are 5 cm and 48 sq.cm respectively, then the volume of the cone is equal to _______
(1) 240 cm³
(2) 120 cm³
(3) 80 cm³
(4) 480 cm³
Answer:
(3) 80 cm³
Hint:
Volume of a cone (V) = \(\frac{1}{3}\) πr²h sq. units
Base area (πr²) = 48 sq. cm
V = \(\frac{1}{3}\) × 48 × 5 = 80 cm³

Question 11.
The ratios of the respective heights and the respective radii of two cylinders are 1 : 2 and 2 : 1 respectively. Then their respective volumes are in the ratio _______
(1) 4 : 1
(2) 1 : 4
(3) 2 : 1
(4) 1 : 2
Answer:
(3) 2 : 1
Hint:
h₁ : h₂ = 1 : 2
r₁ : r₂ = 2 : 1
Ratio of their volumes
= \(\frac{1}{3} \pi r_{1}^{2} h_{1}: \frac{1}{3} \pi r_{2}^{2} h_{2}\)
= 22 × 1 : 12 × 2 = 4 : 2 = 2 : 1

Question 12.
If the radius of a sphere is 2 cm, then the curved surface area of the sphere is equal to ________
(1) 8π cm²
(2) 16 cm²
(3) 12π cm²
(4) 16π cm²
Answer:
(4) 16π cm²
Hint:
C.S.A of a sphere = 4πr² sq. units
[radius = 2 cm]
= 4 × π × 22 cm²
= 16π cm²

Question 13.
The total surface area of a solid hemisphere of diameter 2 cm is equal to _______
(1) 12 cm²
(2) 12π cm²
(3) 4π cm²
(4) 3π cm²
Answer:
(4) 3π cm²
Hint:
Radius of a hemisphere = \(\frac{2}{2}\) = 1 cm
Total surface area of a hemisphere = 3πr² sq. units = 3 × π × 12 cm² = 3π cm²

Question 14.
If the volume of a sphere is \(\frac{9}{16} \pi\) cu.cm, then its radius is ________
(1) \(\frac{4}{3}\) cm
(2) \(\frac{3}{4}\) cm
(3) \(\frac{3}{2}\) cm
(4) \(\frac{2}{3}\) cm
Answer:
(2) \(\frac{3}{4}\) cm
Hint:
Volume of the sphere = \(\frac{9}{16} \pi\)

Question 15.
The surface areas of two spheres are in the ratio of 9 : 25. Then their volumes are in the ratio _______
(1) 81 : 625
(2) 729 : 15625
(3) 27 : 75
(4) 27 : 125
Answer:
(4) 27 : 125
Hint:
Ratio of their surface area = 9 : 25

Question 16.
The total surface area of a solid hemisphere whose radius is a units, is equal to ________
(1) 2πa² sq. units
(2) 3πa² sq. units
(3) 3πa sq. units
(4) 3a² sq. units
Answer:
(2) 3πa² sq. units
Hint:
T.S.A. of a solid hemisphere = 3πr² sq. units
= 3 × π × a × a sq.units
= 3πa² sq. units

Question 17.
If the surface area of a sphere is 100π cm², then its radius is equal to ______
(1) 25 cm
(2) 100 cm
(3) 5 cm
(4) 10 cm
Answer:
(3) 5 cm
Hint:
Surface area of a sphere = 100π cm²
4πr² = 100π
r² = 25
r = √25 = 5 cm

Question 18.
If the surface area of a sphere is 36π cm², then the volume of the sphere is equal to _______
(1) 12π cm³
(2) 36π cm³
(3) 72π cm³
(4) 108π cm³
Answer:
(2) 36π cm³
Hint:
Surface area of a sphere = 36π cm²
4πr² = 36π
r² = 9
r = 3 cm
Volume of a sphere = \(\frac{4}{3} \pi r^{3}\) cu. units
= \(\frac{4}{3} \pi\) × 3 × 3 × 3 cm³ = 36π cm³

Question 19.
If the total surface area of a solid hemisphere is 12π cm² then its curved surface area is equal to ______
(1) 6π cm²
(2) 24π cm²
(3) 36π cm²
(4) 8π cm²
Answer:
(4) 8π cm²
Hint:
T.S.A of a hemisphere = 12π cm²
3πr² = 12π
r² = 4
r = 2
Curved surface area of a hemisphere = 2πr² = 2 × π × 4 = 8π cm²

Question 20.
If the radius of a sphere is half of the radius of another sphere, then their respective volumes are in the ratio _____
(1) 1 : 8
(2) 2 : 1
(3) 1 : 2
(4) 8 : 1
Answer:
(1) 1 : 8
Hint:
\(r_{1}=\frac{r_{2}}{2} \Rightarrow \frac{r_{1}}{r_{2}}=\frac{1}{2} \Rightarrow r_{1}: r_{2}=1: 2\)

II. Answer the following questions:

Question 1.
Curved surface area and circumference at the base of a solid right circular cylinder are 4400 sq.cm and 110 cm respectively. Find its height and diameter.
Answer:
Given, Circumference of the base of a cylinder = 110 cm
2πr = 110 ……. (1)
Curved surface area = 4400 cm²
2πrh = 4400 cm² ……. (2)
From (1) & (2), \(\frac{(2)}{(1)} \Rightarrow \frac{2 \pi r h}{2 \pi r}=\frac{4400}{110}=40 \mathrm{cm}\)
Height of the cylinder (h) = 40 cm
From (1), 2πr = 110
2 × \(\frac{22}{7}\) × r = 110
r = \(\frac{35}{2}\)
We know that, diameter (d) = 2 × radius
d = 2 × \(\frac{35}{2}\) = 35 cm
Diameter of the Circular cylinder = 35 cm

Question 2.
A mansion has 12 right cylindrical pillars each having radius 50 cm and height 3.5 m. Find the cost of painting the lateral surface of the pillars at ₹ 20 per square metre.
Answer:
Given, Radius of a cylinder (r) = 50 cm = 0.5 m
Height of a cylinder (h) = 3.5 m
Curved surface area of a pillar = 2πrh sq. units
Curved surface area of 12 pillars = 12 × 2πrh
= 12 × 2 × \(\frac{22}{7}\) × 0.5 × 3.5 m²
= 132 sq. m.
Cost for painting the lateral surface of pillars per metre = ₹ 20
Cost of painting = 132 × ₹ 20 = ₹ 2640

Question 3.
The total surface area of a solid right circular cylinder is 231 cm². Its curved surface area is two thirds of the total surface area. Find the radius and height of the cylinder.
Answer:
Given, Total surface area of a cylinder (T.S. A) = 231 sq.cm
Curved surface area = \(\frac{2}{3}\) × T.S.A = \(\frac{2}{3}\) × 231 = 154 cm²
2πrh = 154 cm² …… (1)
Total surface area = 231 cm²
2πr (h + r) = 231
2πrh + 2πr²= 231
154 + 2πr² = 231 [from (1)]
2πr² = 231 – 154 = 77

Radius of the cylinder = 3.5 cm
Height of the cylinder = 7 cm

Question 4.
The total surface area of a solid right circular cylinder is 1540 cm². If the height is four times the radius of the base, then find the height of the cylinder.
Answer:
Given, Let the radius of the cylinder be ‘r’
Height of a cylinder (h) = 4r (by given condition)
Total surface area = 1540 cm²
2πr(h + r) = 1540 cm²

Height of the cylinder = 4r = 4 × 7 = 28 cm

Question 5.
If the vertical angle and the radius of a right circular cone are 60° and 15 cm respectively, then find its height and slant height.
Answer:
Given, In the figure, OAB is a cone and OC ⊥ AB
∠AOC = \(\frac{60^{\circ}}{2}\) = 30°
In the right ∆OAC, tan 30° = \(\frac{\mathrm{AC}}{\mathrm{OC}}\)

Slant Height of the cone (l) = 15 × 2 = 30 cm

Question 6.
The central angle and radius of a sector of a circular disc are 180° and 21 cm respectively. If the edges of the sector are joined together to make a hollow cone, then find the radius of the cone.
Answer:

Given, Radius of a sector (r) = 21 cm
The angle of the sector (θ) = 180°
Let “R” be the radius of the cone.
Circumference of the base of a cone = Arc length of the sector

Radius of the cone (R) = 10.5 cm

Question 7.
If the curved surface area of a solid hemisphere is 2772 sq.cm, then find its total surface area.
Answer:
Given, the Curved surface area of a solid hemisphere = 2772 cm²
2πr² = 2772

Total surface area = 3πr² sq. units
= 3 × \(\frac{22}{7}\) × 21 × 21
= 4158 sq.cm
Aliter:
C.S.A of a hemisphere = 2772 cm²
2πr² = 2772 cm²
πr² = \(\frac{2772}{2}\) = 1386 cm
T.S.A of a hemisphere = 3πr² sq.units = 3 × 1386 cm² = 4158 cm²

Question 8.
An inner curved surface area of a hemispherical dome of a building needs to be painted. If the circumference of the base is 17.6 m, find the cost of painting it at the rate of ₹ 5 per sq. m.
Answer:
Given, Circumference of the dome = 17.6 m
2πr = 17.6
\(r=\frac{17.6 \times 7}{2 \times 22}=\frac{8.8 \times 7}{22}=2.8 \mathrm{m}\)
Curved surface area of the dome = 2πr² sq. units
= 2 × \(\frac{22}{7}\) × 2.8 × 2.8 m²
= 49.28 m²
Cost of painting for one sq.metre = ₹ 5
Cost of painting the curved surface = 49.28 × ₹ 5 = ₹ 246.40

Question 9.
Volume of a solid cylinder is 62.37 cu.cm. Find the radius if its height is 4.5 cm.
Answer:
Given, Height of a cylinder (h) = 4.5 cm
Volume of a solid cylinder = 62.37 cu. cm

Radius of a cylinder (r) = 2.1 cm

Question 10.
A rectangular sheet of metal foil with dimension 66 cm × 12 cm is rolled to form a cylinder of height 12 cm. Find the volume of the cylinder.
Answer:
After rolling the rectangular sheet into a cylinder

Volume of the cylinder = 4158 cm³

Question 11.
The circumference of the base of a 12 m high wooden solid cone is 44 m. Find the volume.
Answer:
Given, Height of the wooden solid cone (h) = 12 m
Circumference of the base = 44 m
2πr = 44
r = \(\frac{44 \times 7}{2 \times 22}\) = 7 m
Volume of the wooden solid = \(\frac{1}{3} \pi r^{2} h\) cu. units
= \(\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 12 \mathrm{m}^{3}\)
= 88 × 7
= 616 m³
Volume of the solid = 616 m³

Question 12.
Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 14 cm.
Answer:
Given, Edge of the cube = 14 cm
The largest circular cone is cut out from the cube.
Radius of the cone (r) = \(\frac{14}{2}\) = 7 cm
Height of the cone (h) = 14 cm
Volume of a cone

Volume of a cone = 718.67 cm³

Question 13.
The thickness of a hemispherical bowl is 0.25 cm. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. (Take π = \(\frac{22}{7}\))
Answer:
Let r, R and w be the inner and outer radii and thickness of the hemispherical bowl respectively.

Given that r = 5 cm, w = 0.25 cm
R = r + w = 5 + 0.25 = 5.25 cm
Now, outer surface area of the bowl = 2πR²
= 2 × \(\frac{22}{7}\) × 5.25 × 5.25
= 173.25 sq. cm
Thus, the outer surface area of the bowl = 173.25 sq. cm

Question 14.
Volume of a hollow sphere is \(\frac{11352}{7}\) cm3. If the outer radius is 8 cm, find the inner radius of the sphere. (Take π = \(\frac{22}{7}\))
Answer:
Let R and r be the outer and inner radii of the hollow sphere respectively.
Let V be the volume of the hollow sphere.

Hence, the inner radius r = 5 cm

Question 15.
How many litres of water will a hemispherical tank whose diameter is 4.2 m?
Answer:
Radius of the tank = \(\frac{4.2}{2}\) = 2.1 m
Volume of the hemisphere
= \(\frac{2}{3} \pi r^{3}\) cu.units
= \(\frac{2}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 \mathrm{m}^{3}\)
= 19.404 m³
= 19.404 x 1000 lit
= 19,404 litres

III. Answer the following questions.

Question 1.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Answer:

For cylindrical part:
Radius (r) = 7 cm
Height (h) = 6 cm
Curved surface area = 2πrh = 2 × \(\frac{22}{7}\) × 7 × 6 cm² = 264 cm²
For hemispherical part:
Radius (r) = 7 cm
Surface area (h) = 2πr²
= 2 × \(\frac{22}{7}\) × 7 × 7 cm²
= 308 cm²
Total surface area = (264 + 308) = 572 cm²

Question 2.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Answer:

Question 3.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Answer:

For cylinderical part:
Height (h) = 2.4 cm
Diameter (d) = 1.4 cm
Radius (r) = 0.7 cm
Total surface area of the cylindrical part

For conical part:
Base area (r) = 0.7 cm
Height (h) = 2.4 cm

Question 4.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Answer:
Diameter of the cylindrical well = 7 m
Radius of the cylinder (r) = \(\frac{7}{2}\) m
Depth of the well (h) = 20 m
Volume = πr²h
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20 \mathrm{m}^{3}\)
= 22 × 7 × 5 m³
Volume of the earth taken out = 22 × 7 × 5 m³
Now this earth is spread out to form a cuboidal platform having
Length (l) = 22 m
Breadth (b) = 14 m
Let ‘h’ be the height of the platform.
Volume of the platform = 22 × 14 × h m³
Volume of the platform = Volume of the earth taken out
22 × 14 × h = 22 × 7 × 5
\(h=\frac{22 \times 7 \times 5}{22 \times 14}=\frac{5}{2} \mathrm{m}=2.5 \mathrm{m}\)
Thus, the required height of the platform is 2.5 m.

Question 5.
The perimeters of the ends of the frustum of a cone are 207.24 cm and 169.56 cm. If the height of the frustum is 8 cm, find the whole surface area of the frustum. [Use π = 3.14]
Answer:

Let the radii of circular ends are R and r [R > r]
Perimeter of circular ends are 207.24 cm and 169.56 cm
2πR = 207.24 cm

The whole surface area of the frustum = π [(R² + r²) + (R + r) l]
Required whole surface area of the frustum
= 3.14 [33² + 27² + (33 + 27) × 10] cm²
= 3.14 [1089 + 729 + 600] cm²
= 3.14 [2418] cm²
= 7592.52 cm²

Question 6.
A cuboid-shaped slab of iron whose dimensions are 55 cm × 40 cm × 15 cm is melted and recast into a pipe. The outer diameter and thickness of the pipe are 8 cm and 1 cm respectively. Find the length of the pipe. (Take π = \(\frac {22}{7}\))
Answer:

Let h₁ be the length of the pipe
Let R and r be the outer and inner radii of the pipe respectively.
Iron slab:
Volume = lbh = 55 × 40 × 15 cm³
Iron pipe:
Outer diameter, 2R = 8 cm
Outer radius, R = 4 cm
Thickness, w = 1 cm
Inner radius, r = R – w = 4 – 1 = 3 cm
Now, the volume of the iron pipe = Volume of the iron slab

Time is taken by the pipe to empty half of the tank = 3 hours 12 minutes.

Question 7.
The perimeter of the ends of a frustum of a cone are 44 cm and 8.4π cm. If the depth is 14 cm., then find its volume.
Answer:
Given let the radius of the top of the frustum be “R” and the radius of the bottom of the frustum be “r”

Question 8.
A tent is in the shape of a right circular cylinder surmounted by a cone. The total height and the diameter of the base are 13.5 m and 28 m. If the height of the cylindrical portion is 3 m, find the total surface area of the tent.
Answer:

Given, Total height of solid = 13.5 cm
Diameter of the cylinder (d) = 28 m
Height of a cylinder (h) = 3 m
Height of a conical portion = 13.5 – 3 = 10.5 m
From the diagram, Radius of a cone = Radius of a cylinder
Radius (r) = 14 m

Tamilnadu State Board New Syllabus Samacheer Kalvi 10th English Guide Pdf Poem 6 No Men Are Foreign Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 10th English Solutions Poem 6 No Men Are Foreign

10th English Guide No Men Are Foreign Textbook Questions and Answers

A. Based on the understanding of the poem, read the following lines and answer the questions given below.

1. “Beneath all uniforms, a single body breathes
Like ours: the land our brothers walk upon
Is earth like this, in which we all shall lie”
(a) What is found beneath all uniforms?
(b) What is same for every one of us?
(c) Where are we all going to lie finally?
(d) What is the alliterated words in the 2nd line?
(e) What is the figure of speech in the 2nd and 3rd line?
Answers:
(a) Human body is found beneath all uniforms.
(b) The earth we walk up on is the same for every one of us.
(c) We are all going to lie beneath the earth.
(d) beneath; body; breathes
(e) Simile

2. “They, too, aware of sun and air and water,
Are fed by peaceful harvests, by war’s long winter starv ’d.
(a) What is common for all of us? (or) What are they aware of?
(b) How are we fed?
(c) Mention the season referred here.
Answer:
(a) The sun, air and water are common for all of us.
(b) We are fed by peaceful harvest.
(c) The winter season

3. Their hands are ours, and in their lines we read A labour not different from our own.
(a) Who does “their” refer to?
(b) What does the poet mean by lines we read?
(c) What does not differ?
Answers:
(a) ‘Their’ refers to the other people of the world whom we consider as strange and foreign.’
(b) The poet by the words, Tines we read’ means that their destiny is similar to ours.The lines of their hands also show their capacity of doing hard work or labour.
(c) Labour does not differ.

No Men Are Foreign MCQ Questions Class 9 English with Answers.

4. “Let us remember, whenever we are told
To hate our brothers, it is ourselves
That we shall dispossess, betray, condemn ”
(a) Who tells us to hate our brothers?
(b) What happens when we hate our brothers?
(c) What do we do to ourselves?
Answers:
(a) The evil rulers tell us to hate our brothers.
(b) When we hate our brothers, we hate ourselves.
(c) We dispose, betray and blame ourselves.

5. “Our hells of fire and dust outrage the innocence
Of air that is everywhere our own,
Remember, no men are foreign, and no countries Strange ”
(a) What outrages the innocence?
(b) Who are not foreign?
(c) What is not strange?
(d) Who defiles the earth?
Answers:
(a) War, which is futile spoiling the very earth with hells of fire and dust outrages the innocence.
(b) Any human being who breathes the same air is not foreign.
(c) The world which becomes a more difficult place to live in and any country in this world is not strange.
(d) The men who fight with each other defile our earth.

Additional Questions and Answers

1. Remember they have eyes like ours that wake Or sleep, and strength that can be won By love
(a) What do they have like ours?
(b) What do the eyes do?
(c) How can strength be won?
Answers:
(a) They have eyes like ours.
(b) The eyes wake up or sleep.
(c) Strength can be won by love.

2. In every land is common life
That all can recognize and understand
(a) What is common in every land?
(b) What can all recognise and understand?
Answers:
(a) Life is common in every land.
(b) All can recognise and understand that life is common in everyland.

3. Remember we who take arms against each other
It is the human earth that we defile
(a) Who defiles the earth?
(b) Whose earth is this?
Answers:
(a) We who take arms against each other defile the earth.
(b) This is our earth. It is the human earth.

4. Or sleep, and strength that can be won.
Pick out the words that are in alliteration in this line.
Answer:
The alliterated words are: Sleep, Strength

5. Remember, we who take arms against each other.
Write down the words that are in assonance here.
Answer:
The words in assonance are : arm, against.

6. Beneath all uniforms, a single body breathes like ours;
(a) Who does all refer to?
(b) What does the poet denote?
Answer:
(a) All refers to the people from the countries.
(b) The poet denotes universal brotherhood and equality.

7. Are fed by peaceful harvests, by war’s long winter starv’d
What is the poetic device employed here?
Answer:
The poetic device employed here is transferred epithet’. It is used in the phrase – “winter starv’d”.

8. Our hells of fire and dust outrage the innocence.
(a) What is the figure of speech used here?
Answer:
Metaphor is used here. ‘Hells of fire’ is a metaphor.

9. Remember, no men are strange, no countries foreign
Remember, they have eyes like ours that wake
Remember, we who take arms against each other
Remember, no men are foreign, and no countries strange
What is the figure of speech used in these lines?
Answer:
The figure of speech used here is ‘repetition’.

B. Based on your understanding of the poem complete the following by choosing the appropriate words/phrases given in brackets:

This poem is about the ………………….. (1) …………………… of all men. The subject of the poem is the …………………. (2) …………………. race, despite of the difference in colour, caste, creed, religion, country, etc. All human beings are the same. We walk on the ……………… (3) ……………….. and w/-e will be buried under it. Each and every one of us is related to the other. We all are born the same and die in the same way. We may wear different uniforms like ………………….. (4) ………………… during wars the opposing side will also have the same …………………… (5) ……………….. like ours. We as human do the same labour with ……………….. (6) ………………… and look at the world with the …………………. (7) ………………… Waging war against others as they belong to a different country is like attacking our own selves. It is the ………………… (8) ………………….. we impair. We all share the same ……………………. (9) …………………….. We are similar to each other. So the poet concludes that we shouldn’t have wars as it is …………………… (10) ……………………. to fight against us.
(unity of human, dreams and aspirations, same land, our hands, unnatural, breathing body, same eyes, brotherhood, language, human-earth)

Answers:

1. brotherhood
2. unity of human
3. same land
4. language
5. breathing body
6. our hands
7. same eyes
8. dreams and aspirations
9. human earth
10. unnatural

C. Based on your understanding of the poem answer the following questions in a paragraph of about 100-150 words.

Question 1.
What is the central theme of the poem “No Men are Foreign”.
Answer:
James Kirkup gives a positive message of hope to mankind. In spite of obvious divisions and variances, all are united together by the common bond of civilization and mankind. For their entrusted interests, some selfish people divide lands and people. They collaborate to create hatred and divisions among people. The poet validates the statement that people living in different countries are essentially the same by proclaiming that ‘no men are strange and no men are foreign’.

That is the part of the title of the poem and it is the central theme too. Every single body breathes and functions in the same way as ours. Each one of us equally needs the sun, air and water. Human hands too are used for the similar purpose of labouring for livelihood. Even eyes perform similar purpose of sleeping and waking up. Love wins us all and we all identify its power.

In peace times, we all flourish and wars starve us. Hatred leads us astray and when we take up arms against each other, the entire earth is defiled and destroyed. Therefore, we all like peace which showers abundance and prosperity on us. Therefore, fundamentally we all are the same.

We should understand and try to recognise that the same soul runs through all the people. Let us work for the unity and affluence of all lands and all people. Let us not pollute and taint the earth which is ours. Hatred and narrow ideas pollute the minds of the people.

Conflicts and wars bring destruction and violence. We should remember that raising our arms against anyone means fighting against ourselves. The poet reminds us to remember, recognise and strengthen the common bond that unites mankind and humanity which is the main theme of this poem.

‘Sometimes one feels better speaking to a stranger than someone known.’

(OR)

Poem: No men are foreign
Poet: James Falconer Kirkup
Theme: Universal brotherhood

Human beings are same. We walk on the same land. We will be buried under it. We are related to each other. We all utilise the sunlight, air and water. We live by the food crops of the earth. All have their dreams and aspiration like us. We do the same labour with our hands. We look at the world with the same eyes. We hate ourselves, when we hate our brothers. We are similar to each other. We shouldn’t have war and fight against them.

(OR)

1. Human beings are same.
2. In the world, all are brothers.
3. We live and die in the same earth.
4. The sun, air, water are common to all.
5. We live by the earth food crops.
6. Land is common to all.
7. Love strengthens us.
8. During war, we hate our brothers.
9. So, learn to live in peace with all.

Question 2.
The poem “No Men are Foreign” has a greater relevance ¡n today’s world – Elucidate.
Answer:
Introduction:
The poem “No men are foreign” has a greater relevance ¡n today’s world. Let us see what are the relevance in today’s world.

Theme of the poem:
The poem tells us about unity of human race, despite the differences ¡n colour, caste, creed, etc. The poet tells that there are a lot of conflicts and disasters in and out of every country.

Today’s world:
This has totally affected the world peace and harmony. We defile our earth by means of war. Enmity and hatred must be given to peace and harmony. The earth is full of fire and dust created by means of war. We should not hate our brothers.

Conclusion:
In such situation of considering the earth as the single living place for all, we shall live together, strengthened by love. Thus the earth will be a better place and there will be no more of fire and dust.

(OR)

The poem ‘No Men are Foreign’ has a greater relevance in today’s world. There are a lot of fights and disasters in and out of every countries. The wars are always there between countries for one matter or the other. This has totally affected the world peace and harmony. No one wants to be defeated by the other.

No one bothers about the damage caused to the countries and the loss of lives of lot of people. People pay more attention to the differences and shoot troubles always. As there is no love and spirit of brotherhood, they wage war every now and then. Enmity and hatred must be given to maintain peace and harmony.

Let us consider the earth as the single living place for all who live on it. In such a situation we shall live together, strengthened by love, admiring one another with better understanding. Thus the earth will be a paradise and there will be no more hell of fire and dust.

(OR)

1. This poem has greater relevance to the present world.
2. There is endless war everywhere.
3. There is no spirit of brotherhood.
4. We look at others as different creations.
5. But we are the same and live on the same earth.
6. We have fights and confusions because of our nationalism and racism.
7. We can overcome this if we think about the oneness of mankind.
8. Our inventions destroy us.
9. Only understanding each other can save us.
10. Love alone can bind us and bring out of our blind thoughts.
11. The earth will be our paradise if we give up the differences.
12. Let “No Men are Foreign” be our motto to promote us to peaceful life.

No Men Are Foreign Summary of the poem

The poem ‘No men are foreign’ tells about that no people unity of the human race. The people of one country shouldn’t think of the people of other countries as strangers. The poet tells that they are all humans the same as we should not hate it, brothers. At last, the poet asks us to remember that no men are strange and no countries foreign.

Glossary:

Condemn – express complete disapproval
Labour – hard work
Betray – disloyal
Defile – damage the purity or appearance
Outrage – the extremely strong reaction of anger shock

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

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

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

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

(iii) How many two – digit numbers can be formed using 1 , 2,3,4,5 without repetition of digits?
Answer:

The given digits are 1, 2, 3, 4, 5
A two-digit number has a unit place and 10’s place. We are given 5 digits (1, 2, 3, 4, 5). The unit place can be filled (using the 5 digits) in 5 ways. After filling the unit place since repetition is not allowed one number (filled in the unit place) should be excluded. So the 10’s place can be filled (using the remaining 4 digits) in 4 ways.
∴ Unit place and 10’s place together can be filled in 5 × 4 = 20 ways. So the number of two-digit numbers = 20

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

(v) In how ways 5 persons can be seated In a row?
Answer:
Number of persons = 5

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

Question 2.
(i) A mobile phone has a passcode of 6 distinct digits. What is the maximum number of attempts one makes to retrieve the passcode?
Answer:
Number of distinct digit in a passcode of a mobile phone = 6

First digit can be tried in 10 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Second digit can be tried in 9 ways
Third digit can be tried in 8 ways
Fourth digit can be tried in 7 ways
Fifth digit can be tried in 6 ways
Sixth digit can be tried in 5 ways
Therefore, the maximum number of attempts made to retrieve the passcode = 10 × 9 × 8 × 7 × 6 × 5 = 151200

(ii) Given four flags of different colours, how many different signals can be generated if each signal requires the use of three flags, one below the other?
Answer:

Number of flags = 4
Number of flags required for a signal = 3
The total number of signals is equal to the number of ways of filling 3 places in succession by 4 flags of different colours. Number of ways of filling the top place using 4 different colour flags is 4 ways. Number of ways of filling the middle place using the remaining 3 different colour flags is 3 ways. Number of ways of filling the bottom place using the remaining 2 different colour flags is 2 ways.

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

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

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

Question 4.
Count the number of three – digit numbers which can be formed from the digits 2, 4, 6, 8 if
(i) Repetitions of digits is allowed.
(ii) Repetitions of digits is not allowed?
Answer:
(i) Repetitions of digits is allowed
The given digits are 2, 4, 6, 8

A number of ways of filling the unit place using the 4 digits 2, 4, 6, 8 is 4 ways. Number of ways of filling the tens place using the 4 digits 2, 4, 6, 8 in 4 ways Number of ways of filling the hundred’s place using the 4 digits 2, 4, 6, 8 is 4 ways

Therefore, by fundamental principle of multiplication, the total number of 3 digit numbers = 4 × 4 × 4
= 64 ways

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

Question 5.
How many three-digit numbers are there with 3 in the unit place?
(i) with repetition
(ii) without repetition.
Answer:
(i) With repetition:

The given digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
The unit place can be filled in only one way using the digit 3. The ten’s place can be filled in 10 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The hundred’s place can be filled in 9 ways using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9.

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

(ii) Without repetition:

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

Question 6.
How many numbers are there between 100 and 500 with the digits 0,1,2,3,4,5 if
(i) Repetition of digit is allowed
(ii) Repetition of digits is not allowed.
Answer:
(i) Repetition of digit is allowed:

The numbers between 100 and 500 will have 3 digits. The unit place can be filled in 6 ways using the digits 0, 1, 2, 3, 4, 5. The ten’s place can be filled in 6 ways using the digits 0, 1, 2, 3, 4, 5. The hundred’s place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 ( Excluding 0 and 5). Therefore, by fundamental principle of multiplication, the number of 3 digit, numbers between 100 and 500 with repetition of digits using the digits 0, 1, 2, 3, 4, 5 is = 6 × 6 × 4 = 144

(ii) Repetition of digits is not allowed:

The 100’s place can be filled (by using 1, 2, 3, 4) 10’s in 4 ways
The 10’s place can be filled in (6 – 1) 5 ways and the unit place can be filled in (5 – 1) 4 ways
So the number of 3 digit number 4 × 5 × 4 = 80

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

(i) Repetition of digits is not allowed:

Since we need 3 – digit odd numbers the unit place can be filled in 3 ways using the digits 1, 3 or 5. Hundred’s place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 excluding 0 and the number placed in unit place. Ten’s place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 excluding the digit placed in the hundred’s place.

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

(ii) Repetition of digits is allowed:

The unit place can be filled in 3 ways. We are given 6 digits.
So 10’s place can be filled in 6 ways and the 100’s place can be filled in (6 – 1) (excluding zero) 5 ways
So the Number of 3 digit numbers = 3 × 6 × 5 = 90

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

(i) No restriction:

Thousand’s place can be filled in 9 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 excluding 0. Since there is no restriction, the hundred’s place, Ten’s place, and the unit place can be filled in 10 ways using the digit 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

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

(ii) No digit is repeated:

Thousand’s place can be filled in 9 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 excluding 0. Since repetition is not allowed. The unit place can be filled in 9 ways using the digits 0,1, 2, 3, 4, 5, 6, 7, 8, 9 excluding the digit placed in the thousand’s place. Since repetition of digits is not allowed, the ten’s place can be filled in 8 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 excluding the digits used in thousand’s a place and unit place. Since repetition of digits is not allowed, the hundred’s place can be filled in 7 ways using the digits 0,1,2,3,4, 5,6,7,8,9 excluding the digits used in thousand’s a place and unit place.

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

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

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

(i)The repetition of digits are not allowed:

Since the 3 – digit number is divisible by 5, the unit place can be filled in 2 ways using the digits 0 or 5

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

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

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

(ii) The repetition of digits are allowed:

The digits are
0 1 2 3 4 5
To get a number divisible by 5 we should have the unit place as 5 or 0 So the unit place (using 0 or 5) can be filled in 2 ways.
The 10’s place can be filled (Using 0, 1, 2, 3, 4, 5) in 6 ways and the 100’s place (Using 1, 2, 3, 4, 5) can be filled in 5 ways.
So the number of 3 digit numbers ÷ by 5 (with repetition) = 2 × 6 × 5 = 60

Question 10.
To travel from place A to place B, there are two different bus routes B₁, B₂, two different train routes T₁, T_2, and one air route A₁. From place B to place C, there is one bus route say B’₁, two different train routes say T’₁, T’_2, and one air route A’₁. Find the number of routes of commuting from place A to place C via place B without using a similar mode of transportation.
Answer:
Route map diagram for the given data.

The possible choices for a number of routes commuting from A to place C via place B without using similar mode transportation are
(B₁, T’₁), (B₁, T’₂), ( B₁, A₁), ( B₂, T’₁), (B₂, T’₂)
(B₂, A’₁), (T₁, B’₁), (T₁, A’₁), ( T₂, B’₁), ( T₂, A’₁) (A₁, B’₁) , (A₁, T’₁) and (A₁, T’₂)
Therefore, the Required number of routes is 13.

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

Three-digit numbers:

The unit place can be filed in 4 ways using the digits 1, 3, 7, 9. Hundred’s place can be filled in 9 ways excluding 0. Ten’s place can be filled in 10 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Therefore, the required number of 3 digit numbers neither divisible by 2 nor by 5 is = 9 × 10 × 4 = 360.
There is only one 4 – digit number, but it is divisible by 2 and 5.
Therefore, required numbers using fundamental principle of addition = 4 + 36 + 360 = 400

Question 12.
How many strings can be formed using the letters of the word LOTUS if the word
(i) either start with L or end with S?
(ii) neither starts with L nor ends with S?
Answer:
(i) Either starts with L or ends with S

The first box is filled with the letter L. The second box can be filled with the remaining letters O, T, U, S in 4 ways. The third box can be filled with the remaining letters excluding L and the letter placed in box 2 in 3 ways. The fourth box can be filled with the remaining letters excluding L and the letters placed in a box – 2 and box – 3 in 2 ways. The fifth box can be filled with the remaining one letter excluding L and the letters placed in a box – 2 and box – 3, box – 4 in 1 way.
Therefore, by fundamental principle of multiplication, the number of words start with L is = 1 × 4 × 3 × 2 × 1 = 24

Since the word ends with S, the fifth box can be filled in one way with the letter S. The remaining four boxes can be filled 4 × 3 × 2 × 1 way.
Therefore, the number of words ending with S = 4 × 3 × 2 × 1 × 1 = 24

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

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

MCQ Questions For Class 11 hindi with answers· 

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

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

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

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

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

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

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

(v) \(\frac{12 !}{9 ! \times 3 !}\)

= 2 × 11 × 10 = 220

(vi) \(\frac{(n+3) !}{(n+1) !}\)

= (n + 3) (n + 2)
= n² + 3n + 2n + 6
= n² + 5n + 6

Question 15.
Evaluate \(\frac{\mathbf{n} !}{\mathbf{r} !(\mathbf{n}-\mathbf{r}) !}\) when
(i) n = 6 , r = 2
(ii) n = 10, r = 3
(iii) For any n with r = 2.
Answer:
(i) n = 6 , r = 2

(ii) n = 10, r = 3

(iii) For any n with r = 2.

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

(ii) \(\frac{1}{8 !}+\frac{1}{9 !}=\frac{n}{10 !}\)

Students can Download Samacheer Kalvi 10th Social Science Model Question Paper 3 English Medium Pdf, Samacheer Kalvi 10th Social Science Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

CBSE Class 10th SST MCQ – Download PDF & Exam updates.

Tamil Nadu Samacheer Kalvi 10th Social Science Model Question Paper 3 English Medium

General Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the questions in each part. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III, and IV are to be attempted separately.
4. Question numbers 1 to 14 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by writing the correct answer along with the corresponding option code and the corresponding answer
5. Question numbers 15 to 28 in Part II are of two marks each. Any one question should be answered compulsorily.
6. Question numbers 29 to 42 in Part III are of five marks each. Any one question should be answered compulsorily.
7. Question numbers 43 to 44 in Part IV are of Eight marks each. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 100

Part – I

Answer all the questions. Choose the correct answer [14 × 1 = 14]

Question 1.
What were the three major empires shattered by the end of First World War?
(a) Germany, Austria-Hungary, and the Ottomans
(b) Germany, Austria-Hungary, and Russia
(c) Spain, Portugal and Italy
(d) Germany, Austria-Hungary, Italy
Answer:
(a) Germany, Austria-Hungary, and the Ottomans

Question 2.
Which quickened the-process of liberation in South America?
(a) Support of US
(b) Napoleonic Invasion
(c) Simon Bolivar’s involvement
(d) French Revolution
Answer:
(d) French Revolution

Question 3.
When was people’s Political consultative conference held in China?
(a) September 1959
(b) September 1948
(c) September 1954
(d) September 1949
Answer:
(d) September 1949

Question 4.
Who were driven out of their homeland during the process of creation of zamins under permanent Settlement?
(a) Santhals
(b) Titu Mir
(c) Munda
(d) Kol.
Answer:
(a) Santhals

Question 5.
Which among the following was declared as ‘Independence Day’?
(a) 26th January 1930
(b) 26th December 1929
(c) 16th June 1946
(d) 15th January 1947
Answer:
(d) 15th January 1947

Question 6.
The extent of Himalayas in the east-west is about …………………
(a) 2,500 km
(b) 2,400 km
(c) 800 km
(d) 2,200 km
Answer:
(a) 2,500 km

Question 7.
Western disturbances cause rainfall in …………………
(a) Tamil Nadu
(b) Kerala
(c) Punjab
(d) Madhya Predesh
Answer:
(c) Punjab

Question 8.
The soil formed by the river are …………………
(a) Red soils
(b) Black soils
(c) Desert soils
(d) Alluvial soils
Answer:
(d) Alluvial soils

Question 9.
Retreating monsoon wind pick up moisture from …………………
(a) Arabian sea
(b) Bay of Bengal
(c) Indian ocean
(d) Timor sea
Answer:
(b) Bay of Bengal

Question 10.
Second staple food of the people of Tamil Nadu is …………………
(a) Pulses
(b) Millets
(c) Oilseeds
(d) Rice
Answer:
(b) Millets

Question 11.
Which one of the following rights was described by Dr. B. R. Ambedkar as the heart and soul of the constitution?
(a) Right to freedom of religion
(b) Right to equality
(c) Right to constitutional remedies
(d) Right to property
Answer:
(c) Right to constitutional remedies

Question 12.
Which of the following country is not the founder member of NAM?
(a) Yugoslavia
(b) Indonesia
(c) Egypt
(d) Pakistan
Answer:
(d) Pakistan

Question 13.
………………… is the process of providing or obtaining the food necessary for health and growth.
(a) Health
(b) Nutrition
(c) Sanitation
(d) Security
Answer:
(b) Nutrition

Question 14.
Gross value added at current prices for services sector is estimated at lakh crore
in 2018-19.
(a) 91.06
(b) 92.26
(c) 80.07
(d) 98.29
Answer:
(b) 92.26

Part – II

Answer any 10 questions. Question No. 28 is compulsory. [10 x 2 = 20]

Question 15.
Access the role of Ayyankali in fighting for the cause of “Untouchables”.
Answer:
(i) Ayyankali brought tremendous social changes especially in caste structure. The discrimination he faced as a child turned him into a leader of an anti-caste movement and who later fought for basic rights including access to public spaces and entry to schools.

(ii) Ayyankali challenged many caste conventions such as clothing style, he wore clothes associated with upper castes that were prohibited for lower castes.

Question 16.
How did Hitler get the support from the people of germany?
Answer:
(i) Hitler was well aware of the discontent among the Germans. He used his oratorical skills to sway the common people and promised them to return the glorious military past of Germany. He founded the National Socialist Party, generally known as “the Nazis”.

(ii) The fundamental platform on which Hitler built his support was the notion of the racial superiority of the Germans as a pure, ‘Aryan’ race and a deep-seated hatred of the Jews. Hitler came to power in 1933 and ruled Germany for twelve long years.

Question 17.
Why was Heron dismissed from service?
Answer:

• Colonel Heron was urged to deal with Puli Thevar as he continued to defy the authority of the company. Puli Thevar wielded much influence over the western Palayakkarars.
• Heron had to abandon the plan for want of cannon and of supplies and pay to soldiers. He retired to Madurai. He was then recalled and dismissed from service.

Question 18.
Why did Gandhi withdraw the Non- cooperation Movement?
Answer:

• The Non-cooperation Movement started in 1920. It soon became a nation-wide movement because it got support of the people across the country. But in February 1922, a violent incident occurred at Chauri Chaura, a village near Gorakhpur in Uttar Pradesh.
• In this incident a procession of nationalists provoked by the police turned violent. The police finding themselves outnumbered shut themselves inside the police station.
• The mob burnt the police station in which 22 policemen lost their lives. The incident hurt Gandhiji too much and he immediately withdrew the movement.

Question 19.
What are “Jet Stream”?
Answer:
In the upper layers of the atmosphere, there are strong westerly winds concentrated in a relatively narrow and shallow streams known as “Jet streams” They cause heavy rainfall in North-west India.

Question 20.
State any two characteristics of block cotton soil.
Answer:

• This soil is rich in calcium carbonate, magnesium, potash, lime and iron but deficient in phosphorous. It is clayey and impermeable which has great capacity to retain moisture for a long time.
• It becomes sticky when wet but develops cracks during dry summer season. The soil is suited for dry farming due to its high moisture retentivity.

Question 21.
Name the tributaries of river Thamirabarani.
Answer:
Karaiyar, Servalar, Manimuthar, Gadananathi, Pachaiyar, Chittar and Ramanathi are its main tributaries.

Question 22.
List out the three leads of the relations between the centre and the states.
Answer:
There are –

• Legislative relations
• Administrative relations
• Financial relations.

Question 23.
What do you know about kaladan Multi-Model Transit Transport?
Answer:
(i) India is building the Kaladan Multi-Model Transit Transport, a road-river-port cargo transport project to link Kolkata to Sittwe in Myanmar.

(ii) A project aiming to connect Kolkata with Ho Chi Minh City on the South Sea for the formation of an economic zone will have a road pass through Myanmar, Cambodia and Vietnam and work on the first phase connecting Guwahati with Mandalay is currently underway.

Question 24.
What is national emergency?
Answer:
National emergency is a situation beyond the ordinary. The President declares this emergency if he is satisfied that India’s security is threatened due to war, external aggression or armed rebellion or if there is an imminent danger or threat.

Question 25.
How the state of Jammu and Kashmir differ from the other states of India.
Answer:

• The Constitution of India grants special status to Jammu and Kashmir among Indian States, and it is the only state in India to have a separate Constitution.
• The Directive Principles of the State Policy and Fundamental Duties of the Constitution are not applicable to the State of Jammu and Kashmir.
• Rights to property, which is denied as a Fundamental Right to rest of India is still guaranteed in Jammu and Kashmir.

Question 26.
what are the factors supporting to develop the Indian economy?
Answer:
Factors supporting to develop the Indian economy :

• A fast growing population of working age.
• India has a strong legal system and many English language speakers
• Wage costs are low here.
• India’s economy has successfully developed highly advanced and attractive clusters of business in the technology space.

Question 27.
What is the main objective of WTO?
Answer:
The main objective of WTO is to set and enforce rules for international trade and to provide a forum for negotiating and monitoring further trade liberalisations.

Question 28.
Name the important multipurpose projects of Tamil Nadu.
Answer:
Mettur Dam, Amaravathi Dam, Papanasam Dam, Bhavani Sagar Dam.

Part – III

Answer any 10 questions. Question No. 42 is compulsory. [10 x 5 = 50]

Question 29.
Fill in the blanks:
(i) ………………… treaty signed on February 7, 1922 created the EU.
(ii) The riverine island of Srirangam is located between …………………and ………………… branches of Cauvery.
(iii) …………………was India’s policy in the face of the bipolar order of the cold war.
(iv) In the year ………………… National Food Security Act was passed by the Indian Parliament.
(v) Tamil Nadu ranks …………………in India with a share of over 20% in total road projects under
operation in the Public Private Partnership (PPP).
Answers
(i) The Maastricht
(ii) Northern, Southern
(iii) Non-Alignment
(iv) 2013
(v) Second

Question 30.
Match the following:

Answers:

Question 31.
Match the following:

Answers:

Question 32.
(a) Distinguish between
(i) North East Monsoon and South West Monsoon
(ii) Rabi and Kharif Crop Season
Answer:
(a) (i) North East Monsoon and South East Monsoon:
North East Monsoon :

1. North east monsoon occurs between October and December.
2. It is Winter Monsoon.
3. The coromandal coast of Tamil Nadu gets, heavy rainfall from north east monsoon.

South West Monsoon :

1. Southwest Monsoon occurs between June to September.
2. It is Summer Monsoon.
3. The districts of Nilgiris, Kanyakumari, Salem, Coimbatore and Erode get rainfall.

(ii) Rabi and Kharif crop season:
Rabi Crop Season :

1. Rabi crops are sown in the beginning of the winter (i.e.) in November.
2. The crops are harvested in the beginning of summer (i.e) in March. days of November.
3. Major crops are wheat, tobacco, mustard, pulses, linseed and grains.

Kharif Crop Season :

1. Kharif crops are sown in the beginning of monsoon (i.e.) in June.
2. The crops are harvested in the early
3. Major crops are paddy, maize, cotton, millets, jute and sugarcane.

(b) Give reason: Alluvial soil is fertile
Answer:
Alluvial soil are formed by the deposition of silt by the rivers. Alluvial soils are generally fertile as they are rich in minerals such as lime, potassium, magnesium, nitrogen and phosphoric acid. It is deficient in nitrogen and humus. It is porous and loamy.

Question 33.
Discuss the main causes of the First World War.
Answer:
The causes of the First World War are given below:

• Formation of European alliances and counter alliances
• Emergence of violent forms of nationalism in countries like England, France and Germany
• Aggressive attitude of the German Emperor Kaiser Wilhelm II
• Hostility of France towards Germany
• Opportunity for imperial power politics in the Balkans
• The Balkans wars
• Immediate cause which included the assassination of Archduke Franz Ferdinand,nephew and heir to Franz Joseph, Emperor of Austria-Hungary, by Princip, a Bosnian Serb, on 28 June 1914.

Question 34.
Discuss the causes and consequences of the Revolt of 1857?
Answer:
The Great Rebellion of 1857 is a unique example of resistance to the British authorities, in India. There were several reasons that triggered the Revolt:

(i) The annexation policy of British India created dissatisfaction among the native rulers. The British claimed themselves as paramount, exercising supreme authority. New territories were annexed on the grounds that the native rulers were corrupt, and inept.

(ii) The British annexed several territories such as Satara, Sambalpur, parts of Punjab, Jhansi and Nagpur through the Doctrine of Lapse. This also angered many Indian rulers.

(iii) Indian sepoys were upset with discrimination in salary and promotion. They were paid much less than their European counterparts. They felt humiliated and racially abused by their seniors.

Consequences:

• India was pronounced as one of the many crown colonies to be directly governed by the Parliament. This resulted in the transfer of power from the East India company to the British crown.
• Queen Victoria proclaimed to the Indian people that the British government would not interfere in traditional institutions and religious matters. It was promised that Indians would be absorbed in government services.
• There came significant changes in the Indian army. The number of Indians was reduced. Indians were restrained from holding important ranks and position.
• It was also decided that instead of recruiting soldiers from Rajputs, Brahmins and North Indian Muslims, more soldiers would be recruited from the Gorkhas, Sikhs and Pathans.

Question 35.
Bring out the characteristics of Intensive and Plantation farming.
Answer:
Intensive farming:

• It involves various types of agriculture with higher levels of input, such as capital and labour, per unit of agricultural land area.
• It aims to maximize yields from available land through various means, such as heavy use of pesticides and chemical fertilizers. ,
• Intensive farming is practiced in Punjab, parts of Rajasthan, Uttar Pradesh, and Madhya Pradesh in India.

Plantation farming:

• It is a single crop farming, practised on a large area.
• Crops are mainly grown for the market.
• It is both labour intensive and capital intensive.
• It has an interface of agriculture and industry.
• Developed network of transport and communication connecting the plantation processing industries and markets play an important role in the development of plants. Example- tea, coffee, rubber, sugarcane, etc.

Question 36.
Write an account on river Cauvery.
Answer:

• The main river of Tamil Nadu is Cauvery which originates at Talacauvery in the Brahmragiri hills of Kodagu (Coorg) district of Karnataka in the Western Ghats.
• About 416 km of its course falls in Tamil Nadu. It serves as the boundary between Karnataka
  and Tamil Nadu for a distance of 64 km. A tributary called Bhavani joins Cauvery on the right bank about 45 km from the Mettur Reservoir.
• Thereafter, it takes easterly course to enter into the plains of Tamil Nadu. Cauvery and its distributaries in its lower course drain the districts of Nagapattinam, Thanjavur, Thivarur and Thiruchirapalli.
• The Cauvery, Kollidam and the vellar jointly drain central part of the Tamil Nadu. The head of the Cauvery delta is near the islands of Srirangam. Kollidam branches off from cauvery at Grand Anaicut, also called as Kallanai was built across the river Cauvery.
• After Kallanai, the river breaks into a large number of distributaries and forms a network all over the delta. The network of tributaries within the delta of cauvery in the coast is called as the ‘Garden of Southern India’.
• It merges into Bay of Bengal to the south of Cuddalore. Cauvery along with its tributaries Bhavani, Noyyal, Mayar and Amaravathi is the most important source of canal irrigation.

Question 37.
Write briefly on the Right to Constitutional Remedies.
Answer:
(i) Our Constitution guarantees six Fundamental Rights to its citizens. It safeguards all these rights by granting us the Right to Constitutional Remedies. It is possible that the Government or private bodies may violate one of our Fundamental Rights.

(ii) Right to Constitutional Remedies protects us from such violations. It allows us to file a case against the Government or private bodies in the High Courts of the States and the Supreme Court of the India.

(iii) Both the Supreme Court and High Courts are empowered to issue five kinds of writs of habeas corpus, mandamus, prohibition two warrants and certiorari to protect the Fundamental Rights of the citizens. That is why the Supreme Court is called the “Guardian of the Constitutions”.

Question 38.
Briefly discuss the Functions of the State Legislature.
Answer:
The powers and functions of the State Legislature are almost the same as that of the Parliament.

• The State Legislature can pass laws on all subjects mentioned in the State List as per the constitutions. It can also pass laws on concurrent subjects.
• The Legislature controls the finances of the State. The Lower House enjoys greater power than the Upper House in money matters. Money Bills can be introduced only in the Lower House of the Assembly.
• The Legislature controls the Executive. The council of Ministers is responsible to the Assembly. The ministers have to answer questions asked by the members of the Legislature.
• The Council cannot vote for grants.
• No new tax can be levied without the sanction and permission of the Assembly.

Question 39.
Briefly explain the evolution of M[NC and its advantages and disadvantages.
Answer:
Multinational companies first started their activities in the extractive industries and controlled raw materials in the host countries during 1920s and then entered the manufacturing and service sectors after 1950s. Most of the MNCs at present belong to the four major exporting countries i.e., USA, UK, France and Germany. However, the largest is America. In 1971, the American Corporations held 52% of the total world stock of foreign direct inverstment.
Great Britain held 14.5% followed by France 5% and Federal Republic of Germany 4.4% and Japan 2.7%.

Advantages of MNCs:

• MNCs produce the same quality of goods at lower cost and without transaction cost.
• They reduce prices and increase the purchasing power of consumers world wide.
• They are able to take advantage of tax variation.
• They spur job growth in the local economy.

Disdvantages of MNCs:

(i) They have led to the downfall of smaller, local business.

(ii) With more companies transferring offices and centering operations in other countries, jobs for the people living in developed countries are threatened.

(iii) MNCs often invest in developing countries where they can take advantage of cheaper labour. Some MNCs prefer to put up branches in these parts of the world where there are no stringent policies in labour and where people need jobs because these MNCs can demand for cheaper labour and lesser healthcare benefits.

Question 40.
Write a note on history of industrialisation in Tamil Nadu.
Answer:

Industrialisation in the Colonial Period:

• The introduction of cotton cultivation in western and southern Tamil Nadu by the colonial government led to the emergence of a large-scale textile sector in these parts, which involved ginning, pressing, spinning and weaving operations.
• Introduction of railways also expanded the market for cotton yam and helped develop the sector.
• There was increase in trade during this period which led to industrial development. The two active ports in the region were Chennai and Tuticorin.
• In Western Tamil Nadu, the emergence of textiles industries also led to demand and starting of textile machinery industry in the region.

Post-Independence to early 1990s:

• After independence, several large enterprises were set up by both the central and state governments.
• The Integral Coach Factory in Chennai made railway coaches and the Bharat Heavy Electricals Limited (BHEL) in Tiruchirapalli manufactured boilers and turbines.
• Ashok Motors and Standard Motors together helped form an automobile cluster in the Chennai region.
• The 1970s and 1980s saw the setting up of emergence of powerloom weaving clusters in the Coimbatore region as well as expansion of cotton knitwear cluster in Tiruppur and home furnishings cluster in Karur.
• The Hosur industrial cluster is a successful case of how such policy efforts to promote industrial estates helped develop industries in a backward region.

Industrialisation in Tamil Nadu – Liberalization Phase:

• The final phase of industrialisation is the post-reforms period since the early 1990s.
• Because of trade liberalisation measures, exports of textiles, home furnishings and leather products began to grow rapidly.
• Efforts to attract investments led to entry of leading multinational firms (MNCs) into the state, especially in the automobile sector.
• Chennai region also emerged as a hub for electronics industry with MNCs such as Nokia, Foxconn, Samsung and Flextronics opening plants on the city’s outskirts.
• A significant share of these investments has come up in special economic zones in the districts bordering Chennai.
• The major industries are automobiles, autocomponents, light and heavy engineering, machinery, cotton, etc.
• This diffused process of industrialisation and corresponding urbanisation has paved the way for better rural-urban linkages in Tamil Nadu than in most other states.

Question 41.
Draw a time line for the following:
Write any five important events between 1905-1920
Answer:

Question 42.
Mark the following places on the world map.
(i) Portugal
(ii) Spain
(iii) Morocco
(iv) France
(v) Great Britian
Answer:

Part – IV

Answer both questions. [2 x 8 = 16]

Question 43.
(a) Battle of Stalingrad
(i) When did Germany attack Stalingrad?
(ii) What were the main manufactures of Stalingrad?
(iii) What was the name of the plan formulated by Hitler to attack Stalingrad?
(iv) What is the significance of the Battle of Stalingrad?
Answer:
(a) Battle of Stalingrad:
(i) In August 1942, Germany attacked Stalingrad.
(ii) The main manufactures of Stalingrad were armaments and tractors.
(iii) Fall Blau or Operation Blue
(iv) The people of Russia were grateful for Stalin’s conduct of the war. They regarded him as ‘a prodigy of patience, tenacity and vigilance, almost omnipresent, almost omniscient.’

(b) Political developments in South America.
(i) By which year did the whole of South America become free from European domination?
(ii) How many republics came into being from the Central America?
(iii) In which year was Cuba occupied by the USA?
(iv) What made oligarchic regimes unpopular in South America?
Answer:
(b) Political developments in South America:
(i) By 1830 the whole of South America was free from European domination.
(ii) Five republics came into being from the Central America.
(iii) The USA occupied Cuba in the year 1898.
(iv) Economic growth, urbanisation and industrial growth in countries like Argentina, Chile, Brazil, and Mexico helped consolidate the hold of middle class and the emergence of militant working class oganisations. At the same time American power and wealth came to dominate Central and South America. These factors made olgarchic regimes unpopular in South America.

[OR]

(c) Gandhi and Mass nationalism.
(i) Which incident is considered a turning point in the life of Gandhi?
(ii) Name the works that influenced Gandhi?
(iii) How did Gandhi use satyagraha as a strategy in South Africa?
(iv) What do you know about the Champaran Satyagraha?
Answer:
(c) Gandhi and Mass nationalism:
(i) On his journey from Durban to Pretoria, at the Pietermaritzburg railway station, he was physically thrown out of the first class compartment in which he was travelling despite having a first class ticket. This incident is considered a turning point in the life of Gandhi.
(ii) Tolstoy’s The Kingdom of God is Within You, Ruskin’s Unto This Last and Thoreau’s Civil Disobedience.
(iii) Gandhi developed satyagraha (truth-force) as a strategy, in which campaigners went on peaceful marches and presented themselves for arrest in protest against unjust laws.
(iv) The Champaran Satyagraha of 1916 was the first satyagraha movement inspired by Gandhi. It was a farmer’s uprising that took place in Champaran district of Bihar, India during the British colonial period.

(d) Periyar E. V. R.
(i) When did Periyar found Dravidar Kazhagam?
(ii) What were the Newspapers and Journals run by Periyar?
(iii) Why was Periyar known as Vaikom hero?
(iv) Which was the most important work of Periyar?
Answer:
(d) Periyar E. V. R.
(i) Periyar found Dravidar Kazhagam in 1944.
(ii) The newspapers and journals started by Periyar were – Kudi Arasu, Revolt, Puratchi, Paguththarivu and Viduthalai.
(iii) In Vaikom, people protested against the practice of no access to the temples by the lower caste people. After the local leaders were arrested Periyar led the Temple Entry Movement and was imprisoned. So, people hailed him as Vaikom Virar or hero of Vaikom.
(iv) Right from 1929, when the Self-respect Conferences began to voice its concern over the plight of women, Periyar had been emphasising women’s right to divorce and property. Periyar’s most important work on this subject is Why the Woman is Enslaved.

Question 44.
Mark the following places on the given outline map of India.
(i) Himalayas
(ii) Nilgiris
(iii) Narmada
(iv) Lakshadweep
(v) Deccan Plateau
(vi) Southwest Monsoon
(vii) Paddy growing area
(viii) Chennai to Mumbai air route
Answer:

[OR]

Mark the following places are given outline map of Tamil Nadu:
(i) Coromandel coast
(ii) Paddy growing area
(iii) Nilgiri Hills
(iv) Nagapattinam
(v) Meenambakkam
(vi) Thamirabarani
(vii) Magnesite region (any one place)
(viii) Vaigai Dam
Answer:

Map for Q. 42
(i) Portugal
(ii) Spain
(iii) Morocco
(iv) France
(v) Great Britain
Answer:

Map for Q. 44
(i) Himalayas
(ii) Nilgiris
(if) Narmada
(iv) Lakshadwccp
(v) Deccan Plateau
(vi) Southwest Monsoon
(vii) Paddy growing area
(viii) Chennai to Mumbai air route
Answer:

Map for Q. 44
(i) Coromandel coast
(ii) Paddy growing area
(iii) Nilgiri Hills
(iv) Nagapattinam
(v) Meenambakkam
(vi) Thamirabarani
(vii) Magnesite region (any one place)
(viii) Vaigai Dam
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 4 Life Mathematics Ex 4.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 4 Life Mathematics Ex 4.3

This calculator computes any of the values in the half-life formula given the rest values.

Question 1.
Fill in the blanks:
(i) The compound interest on ₹ 5000 at 12% p.a for 2 years, compounded annually is ________ .
Answer:
₹ 1272
Hint:
Compound Interest (CI) formula is
CI = Amount – Principal

∴ 6272 – 5000 = ₹ 1272

(ii) The compound interest on ₹8000 at 10% p.a for 1 year, compounded half yearly is ________ .
Answer:
₹ 820
Hint:
Compound interest (CI) = Amount – Principal
Amount = p \(\left(1+\frac{r}{100}\right)^{2 n}\) [2n as it is compounded half yearly]
r = 10% p.a, for half yearly r = \(\frac{10}{2}\) = 5
∴ A = 8000 \(\left(1+\frac{5}{100}\right)^{2 \times 1}\) = 8000 × \(\left(\frac{105}{100}\right)^{2}\) = 8820
CI = Amount – principal = 8820 – 8000 = ₹ 820

(iii) The annual rate of growth in population of a town is 10%. If its present population is 26620, then the population 3 years ago was ________ .
Answer:
₹ 20,000
Hint:
Rate of growth of population r = 10%
Present population = 26620
Let population 3 years ago be x
∴ Applying the formula for population growth which is similar to compound interest,

The population 3 years ago was ₹ 20,000

(iv) If the compound interest is calculated quarterly, the amount is found using the formula ________ .
Answer:
A = \(P\left(1+\frac{r}{400}\right)^{4 n}\)
Hint:
Quarterly means 4 times in a year.
∴ The formula for compound interest is
A = \(P\left(1+\frac{r}{400}\right)^{4 n}\)

(v) The difference between the C.I and S.I for 2 years for a principal of ₹ 5000 at the rate of interest 8% p.a is ________ .
Answer:
₹ 32
Hint:
Difference between S.I & C.I is given by the formula
CI – SI = \(\left(\frac{r}{100}\right)^{2}\)
Principal (P) = 5000. r = 8% p.a
∴ CI – SI = 5000\(\left(\frac{8}{100}\right)^{2}\) = 5000 × \(\left(\frac{8}{100}\right)^{2}\) × \(\left(\frac{8}{100}\right)^{2}\) = ₹ 32

CBSE Class 10 Maths formulas and equations are available chapter wise.

Question 2.
Say True or False.
(i) Depreciation value is calculated by the formula, \(P\left(1-\frac{r}{100}\right)^{n}\).
Answer:
True
Hint:
Depreciation formula is \(P\left(1-\frac{r}{100}\right)^{n}\)

(ii) If the present population ola city is P and it increases at the rate of r% p.a, then the population n years ago would be \(P\left(1-\frac{r}{100}\right)^{n}\)
Answer:
False
Hint:
Let the population ‘n’ yrs ago be ‘x’
∴ Present popuLation (P) = x × \(\left(1+\frac{r}{100}\right)^{n}\)
∴ x = \(\frac{P}{\left(1+\frac{r}{100}\right)^{n}}\)

(iii) The present value of a machine is ₹ 16800. It depreciates at 25% p.a. Its worth after 2 years is ₹ 9450.
Answer:
True
Hint:
Present value of machine = ₹ 16800
Depreciation rate (n) = 25%

(iv) The time taken for ₹ 1000 to become ₹ 1331 at 20% p.a, compounded annually is 3 years.
Answer:
False
Hint:
Pnncipal money = 1000
rate of interest = 20%
Amount = 1331, applying in formula we get

(v) The compound interest on ₹ 16000 for 9 months at 20% p.a, compounded quarterly is ₹ 2522.
Answer:
True
Hint:
Principal (P) = 16000
n = 9 months = \(\frac{9}{12}\) years
r = 20% p.a
For compounding quarterly, we have to use below formula,
Amount(A) = P × \(\left(1+\frac{r}{100}\right)^{4 n}\)
Since quarterly we have to divide ‘r’ by 4

∴ Interest A – P = 18522 – 16000 = 2522 (True)

Question 3.
Find the compound interest on ₹ 3200 at 2.5 % p.a for 2 years, compounded annually.
Answer:
Principal (P) = ₹ 3200
r = 2.5% p.a
n = 2 years comp. annually
∴ Amount (A) = \(\left(1+\frac{r}{100}\right)^{n}\)
= 3200 \(\left(1+\frac{25}{100}\right)^{2}\)
= 3200 × (1.025)² = 3362
Compound interest (CI) = Amount – Principal
= 3362 – 3200 = 162

Question 4.
Find the compound interest for 2\(\frac { 1 }{ 2 }\) years on ₹ 4000 at 10% p.a, if the interest is compounded yearly.
Answer:
Principal (P) = ₹ 4000
r = 10 %p.a
Compounded yearly
n = 2\(\frac { 1 }{ 2 }\) years. Since it is of the form a \(\frac{b}{c}\) years

∴ CI = Amount – principal = 5082 – 4000 = 1082

Question 5.
A principal becomes ₹ 2028 in 2 years at 4% p.a compound interest. Find the principal.
Answer:
n = 2 years
r = rate of interest = 4% p.a
Amount A = ₹ 2028

Question 6.
In how many years will ₹ 3375 become ₹ 4096 at 13 % p.a if the interest is compounded half-yearly?
Answer:
Principal = ₹ 3375
Amount = ₹ 4096
r = 13\(\frac{1}{3}\)%p.a = \(\frac{40}{3}\)%p.a

Let no. of years be n
for compounding half yearly, formula is

Taking cubic root on both sides,

Question 7.
Find the CI on ₹ 15000 for 3 years if the rates of interest are 15%, 20% and 25% for the I, II and III years respectively.
Answer:
Principal (P) = ₹ 15000
rate of interest 1 (a) = 15% for year I
rate of interest 2 (b) = 20% for year II
rate of interest 3 (c) = 25% for year III
Formula for amount when rate of interest is different for different years is

Substituting in the above formula, we get

∴ Compound Interest(CI) = A – P = 25,875 – 15,000 = ₹ 10,875
CI = ₹ 10.875

Question 8.
Find the difference between C.I and S.I on ₹ 5000 for 1 year at 2% p.a, if the interest is compounded half yearly.
Answer:
Principal (P) = ₹ 5000
time period (n) = 1 yr.
Rate of interest (r) = 2%p.a
for half yearly r = 1%
Difference between CI & SI is given by the formula

Question 9.
Find the rate of interest if the difference between C.I and S.I on ₹ 8000 compounded annually for 2 years is ₹ 20.
Answer:
Principal (P) = ₹ 8000
time period (n) = 2 yrs.
rate of interest (r) = ?
Difference between CI & SI is given by the formula
CI – SI = \(p\left(1+\frac{r}{100}\right)^{n}\)
Difference between CI & SI is given as 20
∴ 20 = 8000 × \(\left(\frac{r}{100}\right)^{2}\)
∴ \(\left(\frac{r}{100}\right)^{2}=\frac{20}{8000}=\frac{1}{400}\)
Taking square root on both sides
\(\frac{r}{100}=\sqrt{\frac{1}{400}}=\frac{1}{20}\)
∴ r = \(\) = 5 %

Question 10.
Find the principal if the difference between C.I and S.l on it at 15% p.a for 3 years is ₹ 1134.
Answer:
Rate of interest (r) = 15% p.a
time period (n) = 3 years
Difference between CI & SI is given as 1134
Principal = ? → required to find
Using formula for difference
Simple Interest SI = \(\frac{P n r}{100}\)
Compound Interest CI = p(1 + i)ⁿ – p

1134 = P[(1.15)³ – 1 – 0.45] = P(1.52 – 1.45) = P (0.07)
∴ p = \(\frac{1134 \times 100}{0.07 \times 100}=\frac{113400}{7}\)
P = ₹ 16200

Objective Type Questions

Question 11.
The number of conversion periods in a year, if the interest on a principal is compounded every two months is _________ .
(A) 2
(B) 4
(C) 6
(D) 12
Answer:
(C) 6
Hint:
Conversion period is the time period after which the interest is added to the principal. If principal is compounded every two months then in a year, there will be 6\(\left(\frac{12}{2}\right)\) conversion periods.

Question 12.
The time taken for ₹ 4400 to become ₹ 4851 at 10%, compounded half yearly is _________ .
(A) 6 months
(B) 1 year
(C) 1\(\frac{1}{2}\) years
(D) 2 years
Answer:
(B) 1 year
Hint:
Principal = ₹ 4400
Amount = ₹ 4851
Rate of interest = 10% p.a
for half yearly, divide by 2,
r = \(\frac{10}{2}\) = 5 %
Compounded half yearly, so the formula is
A = P\(P\left(1+\frac{r}{100}\right)^{2 n}\)
Substituting in the above formula, we get

Taking square root on both sides, we get
\(\left(\frac{21}{20}\right)^{2 n}=\left(\frac{21}{20}\right)^{2}\)
Equating power on both sides
∴ 2n = 2, n = 1

Question 13.
The cost of a machine is ₹ 18000 and it depreciates at 16\(\frac{2}{3}\)% annually. Its value after 2 years will be _________ .
(A) ₹ 2000
(B) ₹ 12500
(C) ₹ 15000
(D) ₹ 16500
Answer:
(B) ₹ 12500
Hint:
Cost of machine = 18000
Depreciation rate = 16\(\frac{2}{3}\)% = \(\frac{50}{3}\)%p.a
time period = 2 years
∴ As per depreciation formula,
Depriciated value = Original value \(\left(1-\frac{r}{100}\right)^{n}\)
Substituting in above formula, we get
Depreciated value after 2 years

Question 14.
The sum which amounts to ₹ 2662 at 10% p.a in 3 years, compounded yearly is _________ .
(A) ₹ 2000
(B) ₹ 1800
(C) ₹ 1500
(D) ₹ 2500
Answer:
(A) ₹ 2000
Hint:
Amount = ₹ 2662
rate of interest = 10 % p.a
Time period = 3 yrs. Compounded yearly
Principal (P) → required to find?

Question 15.
The difference between compound and simple interest on a certain sum of money for 2 years at2%p.ais U. The sum of money is _________ .
(A) ₹ 2000
(B) ₹ 1500
(C) ₹ 3000
(D) ₹ 2500
Answer:
(D) ₹ 2500
Hint:
Difference between CI and SI is given as Re I
Time period (n) = 2 yrs.
Rate of interest (r) = 2 % p.a
Formula for difference is
CI – SI = \(P \times\left(1+\frac{r}{100}\right)^{n}\)
Substituting the values in above formula, we get
1 = p × \(\left(\frac{2}{100}\right)^{2}\)
∴ p = 1 × \(\left(\frac{100}{2}\right)^{2}\) = 1 × (50)² = ₹ 2500

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

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

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

Question 1.
Resolve the following rational expressions into partial fractions : \(\frac{1}{x^{2}-a^{2}}\)
Answer:

1 = A (x – a) + B (x + a) ——– (1)
Put x = a in equation (1)
1 = A (0) + B (a + a)
1 = B(2a)
⇒ B = \(\frac{1}{2 a}\)
Put x = – a in equation (1)
1 = A(- a – a) + B(- a + a)
1 = – 2a A + 0
⇒ A = \(-\frac{1}{2 a}\)
∴ The required partial fraction is

Question 2.
\(\frac{3 x+1}{(x-2)(x+1)}\)
Answer:

3x + 1 = A(x + 1) + B (x – 2) ——— (1)
Put x = 2 in equation (1)
3(2) + 1 = A (2 + 1) + B (2 – 2)
6 + 1 = 3A + 0
⇒ A = \(\frac{7}{3}\)
Put x = – 1 in equation (1)
3(-1) + 1 = A (-1 + 1 ) + B (- 1 – 2)
– 3 + 1 = A × 0 – 3B
– 2 = 0 – 3B
⇒ B = \(\frac{2}{3}\)
∴ The required partial fractions are

Question 3.
\(\frac{x}{\left(x^{2}+1\right)(x-1)(x+2)}\)
Answer:

x = Ax (x + 1) (x + 2) + B(x – 1)(x + 2) + C(x² + 1)(x + 2) + D(x² + 1)(x – 1) ——— (1)
Put x = 1 in equation (1)
1 = A(1)(1 – 1) (1 + 2) + B ( ( 1 – 1 ) (1 + 2) + C(1² + 1 ) ( 1 + 2) + D(1² + 1) (1 – 1)
1 = A × 0 + B × 0 + C(2)(3) + D × 0
1 = 6C
⇒ C = \(\frac{1}{6}\)

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

Put x = 0 in equation (1)
0 = A(0) (0 – 1) (0 + 2) + B(0 – 1)(0 + 2)+ C(0² + 1) (0 + 2) + D(0² + 1) (0 – 1)
0 = 0 + B (- 2 ) + C (2) + D (- 1)
0 = – 2B + 2C – D

In equation (1), equate the coefficient of x³ on both sides
0 = A + C + D

Question 4.
\(\frac{x}{(x-1)^{3}}\)
Answer:

X = A(x – 1)² + B(x – 1) + C ——— (1)
Put x = 1 in equation (1)
⇒ 1 = A(1 – 1)² + B(1 – 1) + C
1 = 0 + 0 + C
⇒ C = 1

In equation (1), equating the coefficient of x² on both sides
0 = A ⇒ A = 0
Put x = 0 in equation (1) ⇒ 0 = A(0 – 1)² + B(0 – 1) + C
⇒ 0 = A – B + C
0 = 0 – B + 1
⇒ B = 1

Question 5.
\(\frac{1}{x^{4}-1}\)
Answer:

1 = Ax (x + 1)(x – 1) + B (x + 1) (x – 1) + C (x<sup2 + 1)(x – 1) + D(x + 1)(x² + 1) —— (1)
Put x = 1 in equation (1)
1 = A(1 ) (1 + 1) (1 – 1) + B(1 + 1) (1 – 1) + C(1<sup2 + 1) (1 – 1) + D(1 + 1) (1<sup2 + 1)
1 = A × 0 + B × 0 + C × 0 + D(2)(2)
⇒ 1 = – 4D
⇒ D = \(\frac{1}{4}\)

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

Put x = 0 in equation (1)
I = A(0) (0 + 1 )(0 – 1) + B(0 + 1 )(0 – 1) + C(0² + 1 )(0 – 1) + D(0 + 1)(0² + 1)
1 = A × O + B(-1) + C(- 1) + D(1)
⇒ 1 = – B – C + D
1 = – B + \(\frac{1}{4}\) + \(\frac{1}{4}\)
⇒ B = \(\frac{1}{2}\) – 1 = – \(\frac{1}{2}\)
⇒ B = – \(\frac{1}{4}\)
In equation (1), equate the coefficient of x3 on both sides
0 = A + C + D
⇒ 0 = A – \(\frac{1}{4}+\frac{1}{4}\)
⇒ A = 0

Question 6.
\(\frac{(x-1)^{2}}{x^{3}+x}\)
Answer:

(x – 1)² = A(x² + 1) + Bx² + Cx ——- (1)
Put x = 0 in equation (I)
(0 – 1)² = A(0² + 1) + B × 0 + C × 0
1 = A + 0 + 0
⇒ A = 1
Equating the coefficient of x² on both sides
1 = A + B
1 = 1 + B
⇒ B = 0
Put x = 1 in equation (1)
(1 – 1)² = A(1² + 1) + B × 1² + C × 1
0 = 2A + B + C
0 = 2 × 1 + 0 + C
⇒ C = – 2
∴ The required partial fraction is

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

6x – 5 = A(x – 3) + B(x – 2) ——- (3)
Put x = 3 in equation (3)
6(3) – 5 = A(3 – 3) + B(3 – 2)
18 – 5 = 0 + B
⇒ B = 13
Put x = 2 in equation (3)
6(2) – 5 = A(2 – 3) + B(2 -2)
12 – 5 = – A + 0
7 = -A
⇒ A = – 7
Substituting the values of A and B in equation (2)

Question 8.
\(\frac{x^{3}+2 x+1}{x^{2}+5 x+6}\)
Answer:
\(\frac{x^{3}+2 x+1}{x^{2}+5 x+6}\)
Since the numerator is of degree greater than that of the denominator divide the numerator by the denominator.

21x + 31 = A(x + 3) + B(x + 2) ——- (3)
Put x = – 3 . in equation (3)
21(- 3) + 31 = A(- 3 + 3) + B(- 3 + 2)
– 63 + 31 = 0 – B
– 32 = – B
⇒ B = 32
Put x = – 2 , in equation (3)
21(- 2) + 31 = A(- 2 + 3) + B(- 2 + 2)
– 42 + 31 = A + 0
⇒ A = – 11
Substituting the values of A and B in equation (2)

Question 9.
\(\frac{x+12}{(x+1)^{2}(x-2)}\)
Answer:

x + 12 = A(x + 1 ) (x – 2) + B(x – 2) + C(x + 1)² ——– (1)
Put x = 2 in equation (l)
2 + 12 = A(2 + 1 ) (2 – 2) + B(2 – 2) + C(2 + 1)²
14 = A(3) (0) + B × 0 + C (3 )²
4 = 0 + 0 + 9C
⇒ C = \(\frac{14}{9}\)

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

Put x = 0 in equation (1)
0 + 12 = A (0 + 1)(0 – 2) + B(0 – 2) + C(0 + 1)²
12 = – 2A – 2B + C

Question 10.
\(\frac{6 x^{2}-x+1}{x^{3}+x^{2}+x+1}\)
Answer:

6x² – x + 1 = Ax (x + 1) + B (x + 1) + C(x² + 1) ——— (1)
Put x = – 1 in equation (1)
6 x (- 1)² – (- 1) + 1 = A (- 1 ) (- 1 + 1 ) + B(- 1 + 1) + C ( (- 1)² + 1 )
6 + 1 + 1 = A × 0 + B × 0 + C (2)
8 = 2C
⇒ C = 4

Put x = 0 in equation (1)
6 × 0² – 0 + 1 = A(0)(0 + 1 ) + B(0 + 1) + C(0² + 1)
1 = 0 + B + C
1 = B + 4
B = 1 – 4 = – 3
Equating the coefficient of x² in equation (1) we have
6 = A + C
6 = A + 4
⇒ A = 6 – 4
⇒ A = 2
∴ The required partial fraction is

Question 11.
\(\frac{2 x^{2}+5 x-11}{x^{2}+2 x-3}\)
Answer:
\(\frac{2 x^{2}+5 x-11}{x^{2}+2 x-3}\)
Since the degree of the numerator is equal to the degree of the denominator divide the numerator by the denominator

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

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

Substituting the values of A and B in equation (2) we have

∴ The required partial fraction is

Question 12.
\(\frac{7+x}{(1+x)\left(1+x^{2}\right)}\)
Answer:

7 + x = A( 1 + x²) + Bx (1 + x) + C(1 + x) ——- (1)
Put x = -1 , in equation (1)
7 – 1 = A(1 + (-1)²) + B (- 1) (1 – 1) + C(1 – 1)
6 = A(1 + 1) + 0 + 0
A = \(\frac{6}{2}\) = 3
⇒ A = 3
Put x = 0 , in equation (1)
7 + 0 = A(1 + 0²) + B × 0 (1 + 0) + C(1 + 0)
7 = A + 0 + C
7 = 3 + C
⇒ C = 4
Equating the coefficient of x² in equation (I) we have
0 = A + B
0 = 3 + B
⇒ B = – 3
∴ The required partial fraction is

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

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

11th Chemistry Guide Solutions Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

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

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

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

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

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

Question 6.
Which one of the following is incorrect for an ideal solution?
a) ∆H_mix = 0
b) ∆U_mix =0
c) ∆P = P_observed – P_{calculated by Raoults law} = 0
d) ∆G_mix = 0
Answer:
d) ∆G_mix = 0

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

b) α² = \(\frac{n(1-i)}{(n-1)}\)

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

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

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

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

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

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

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

II. Write brief answer to the following questions:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 42.
Calculate the molality of a solution containing 7.5 g glycine(NH₂-CH₂-COOH) dissolved in 500g of water.
Solution:

Question 43.
Which solution has the lower freezing point? 10 g of methanol (CH₃OH) in 100g g of water (or) 20 g of ethanol (C₂H₅OH) in 200 g of water.
Solution:
∆T_f = K_f i.e
∆T_f α m
m_CH3-OH = \(\frac{\left(\frac{10}{32}\right)}{0.1}\)
= 3.125 m

m_C2H5-OH = \(\frac{\left(\frac{20}{46}\right)}{0.2}\)
= 2.174 m

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

Question 44.
How many moles of solute particles are present in one liter of 10^-4 M potassium sulphate?
Answer:
In 10^-4 M K₂SO₄ solution, there are 10^-4 moles of potassium sulphate.
K₂SO₄ molecule contains 3 ions (2K+ and 1 SO₄^2-)
1 mole of K₂SO₄ molecule contains 3 × 6.023 × 10²³ ions
10^-4 mole of K₂SO₄ contains 3 × 6.023 × 10²³ × 10^-4 ions
= 18. 069 × 10¹⁹

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

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

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

Question 47.
The vapour pressure of pure benzene (C₆H₆) at a given temperature is 640 mm Hg. 2.2 g of non – volatile solute is added to 40 g of benzene. The vapour pressure of the solution is 600 mm Hg. Calculate the molar mass of the solute?
Solution:
P°C₆H₆ = 640 mm Hg
W₂ = 2.2 g (non volatile solute)
W₁ = 40 g (benzene)
P_solution = 600 mm Hg
M₂ =?

11th Chemistry Guide Solutions Additional Questions and Answers

I. Choose the best answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

III. Short Question and answers(3 Marks):

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

Question 2.

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

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

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

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

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

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

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

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

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

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

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

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

Question 9.
0.24 g of a gas dissolves in 1L of water at 1.5 atm pressure. Calculate the amount of dissolved gas when the pressure is raised to 6.0 atm at constant temperature?
Solution:
P_solute = K_H X_{solute in solution}
At pressure 1.5 atm,
P₁ = K_HX₁ …………..(1)
At pressure 6.0 atm,
p₂ = K_HX₂ …………….(2)
Dividing equation (1) by (2)
From equation = \(\frac{P_{1}}{P_{2}}=\frac{X_{1}}{X_{2}}\)

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

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

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

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

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

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

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

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

Question 13.
At 400 K 1.5 g of an unknown substance is dissolved in solvent and the solution is made to 1.5 L. Its osmotic pressure is found to be 0.3 bar. Calculated the molar mass of the unknown substance.
Solution :

Question 14.
How would you compare Raoult’s law and Henry’s law.
Answer:
1. According to Raou It’s law, for a solution containing a non volatile solute.
P_solution = P⁰_solvent . x_solute

2. According to henry’s law, P_solution = K_H . x_solute in solution

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

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

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

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

IV. Long question and answers(5 Marks):

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

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

Effect of temperature:

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

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

Solute(solid) ⇌ Solute(dissolved)

According to the Le-Chatelier principle, if the dissolution process is endothermic, the increase in temperature will shift the equilibrium towards left i.e solubility increases, for an exothermic reaction, the increase in temperature decreases the solubility. The solubilities, of ammonium nitrate, calcium Chloride, ceric sulphate nano-hydrate, and sodium chloride in water at different temperatures are given in the following graph.

Plot of solubility versus temperature for selective compounds

The following conclusions are drawn from the above graph:

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

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

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

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

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

According to Dalton’s law of partial pressure the total pressure in a closed vessel will be equal to the sum of the partial pressures of the individual components.
Hence,
P_total = P_A + P_B
Substituting the values of P_A and P_B from equations in the above equation,
P_total = X_A P°_A + X_BP°_B
We know that X_A + X_B = 1 or X_A = 1 – X_B
Therefore,
P_total = (1 – X_B)P°_A + X_BP°_B
P_total = P°_A + X_B(P°_B – P°_A)
The above equation is of the straight¬line equation form y = mx + c. The plot of P_totalversus x_B will give a straight line with (P°_B – P_A) as slope and P°_A as the y intercept.

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

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

Solution of benzene in toluene obeying Raoult’s law

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

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

For example, when sodium chloride is added to the water, the vapor pressure of the salt solution is lowered. The vapour pressure of the solution is determined by the number of molecules of the solvent present in the surface at any time and is proportional to the mole fraction of the solvent.

Rate of vapourization reduced by presence of nonvolatile solute.

P_solution ∝ X_A
Where X_A is the mole fraction of the solvent
P_solution = kX_A
When X_A = 1, k = P°_solvent
(P°_solvent is the partial pressure of pure solvent)
P_Solution = P°_Solvent
\(\frac{P_{\text {solution }}}{P_{\text {solvent}}^{a}}\) = X_A

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

\(\frac{p_{\text {Solvent }}^{o}-p_{\text {Solurion }}}{P_{\text {Solvent }}^{0}}\) = X_B
Where X_B is the fraction of the solute
(∴ x_A + x_B = 1, X_B = 1 – X_A)

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

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

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

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

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

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

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

Here, P_A > p°_A X_A and p_B > P°_B X_B
Hence P_total > p°_A X_A + P_B X_B

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

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

Example:
Ethyl alcohol cyclohexane, Benzene & acetone, Carbon tetrachloride & chloroform, Acetone & ethyl alcohol, Ethyl alcohol & water.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Molar mass can be calculated using

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

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

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

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

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

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

Textual Questions:

I. Choose the best answer:

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

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

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

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

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

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

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

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

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

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

Question 9.
When 22.4 litres of H₂ (g) is mixed with 11.2 litres of Cl₂ (g), each at 273 K at 1 atm the moles of HCl (g), formed is equal to
(a) 2 moles of HCl (g)
(b) 0.5 moles of HCl (g)
(c) 1.5 moles of HCl (g)
(d) 1 moles of HCl (g)
Answer:
(d) 1 moles of HCl (g)
Solution:
H_2(g) + Cl_2(g) → 2HCl_(g)

Content	CH4	02	CO2
Stoichiometric coefficient	1	1	2
No. of moles of reactants allowed to react at 273 K and 1 atm pressure	22.4 L (1 mol)	11.2 L (0.5 mol)	_
No. of moles of a reactant reacted and product formed	0.5	0.5	_

Amount of HCl formed = 1 mol

Question 10.
Hot concentrated sulphuric acid is a moderately strong oxidising agent. Which of the following reactions does not show oxidising behaviour?
(a) Cu + 2H₂ → CuSO₄ + SO₂ +2 H₂O
(b) C + 2 H₂SO₄ → CO₂ + 2 SO₂ +2 H₂O
(c) BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl
(d) none of the above
Answer:
(c) BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl
Solution:

Question 11.
Choose the disproportionation reaction among the following redox reactions.
(a) 3Mg_(s) + N_2(g) → Mg₃N_2(s)
(b) P_4(s) + 3NaOH + 3H₂O → PH_3(g) + 3NaH₂PO_2(aq)
(c) Cl_2(g) + 2KI_(aq) → 2KCl_(aq) + I_2(s)
(d) Cr₂O_3(s) + 2Al_(s) → Al₂O_3(s) + 2Cr_(s)
Answer:
b) P_4(s) + 3NaOH + 3H₂O → PH_3(g) + 3NaH₂PO_2(aq)
Solution:

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

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

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

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

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

Question 16.
The correct increasing order of the oxidation state of sulphur in the anions
SO₄^2-, SO₃^2-, S₂O₄^2-, S₂O₆^2- is

(a) SO₃^2- < SO₄^2- < S₂O₄^2- < S₂O₆^2-

(b) SO₄^2- < S₂O₄^2- < S₂O₆^2- < SO₃^2-

(c) S₂O₄^2- < SO₃^2- < S₂O₆^2- < SO₄^2-

(d) S₂O₆^2- < S₂O₄^2- < SO₄^2- < SO₃^2-
Answer:
(c) S₂O₄^2- < SO₃^2- < S₂O₆^2- < SO₄^2-
Solution:

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

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

(c) \(\frac{\text { molar mass of ferrous oxalate }}{3}\)
(d) none of these
Answer:
(c) \(\frac{\text { molar mass of ferrous oxalate }}{3}\)
Solution:

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

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

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

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

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

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

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

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

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

II. Write brief answer to the following questions:

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

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

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

Question 29.
Distinguish between oxidation and reduction.
Answer:

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

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

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

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

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

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

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

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

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

Compound	Given no. of moles	No. of oxygen atoms
Ethanol – C2H5OH	1	1 × 6.022 × 1023
Formic acid -HCOOH	1	2 × 6.022 × 1023
Water – H2O	1	1 × 6.022 × 1023

Answer: Formic acid

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

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

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

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

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

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

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

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

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

there in 0.320 kg
Given:
mass of 1 atom = 6.645 × 10^-23 g
∴ mass of 1 mole of atom = 6.645 × 10^-2323 g × 6.022 × 10²³ = 40 g
∴ Number of moles of element in 0.320 kg = \(\frac{1 \mathrm{~mol}}{40 \mathrm{~g}}\) × 0.320 kg
= \(\frac{1 \mathrm{~mol} \times 320 \mathrm{~g}}{40 \mathrm{~g}}\) = 8 mol

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

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

Molar mass:

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

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

Compound	Molecular formula	Empirical formula
Fructose	C6H12O6	CH2O
Caffeine	C8H10N4O2	C4H5N2O

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

Molar mass of Al₂O₃ formed = 6mol × 102 g mol^-1 = 612 g
[Al₂O₃ = (2 × 27) + (3 × 16) = 54 + 48 = 102]
Excess reagent = Fe₂O₃.
Amount of excess reagent left at the end of the reaction = 1 mol × 160 g mol^-1 = 160 g
[Fe₂O₃ = (2 × 56) + (3 × 16) = 112 + 48 = 160 g] = 160 g

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

Question 41.
Hydrogen peroxide is an oxidising agent. It oxidises ferrous ion to ferric ion and reduced itself to water. Write a balanced equation.
Answer:

Question 42.
Calculate the empirical and molecular formula of a compound containing 76.6% carbon, 6.38 % hydrogen and rest oxygen its vapour density is 47.
Answer:

Empirical formula = C₆H₆O
η = Molar mass / Calculated empirical formula mass = \(\frac{2 \times \text { vapour density }}{94}\)
= \(\frac{2 \times 47}{94}\) = 1
∴ Molecular formula(C₆H₆O) × 1 = C₆H₆O

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

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

Question 44.
Balance the following equations by oxidation number method
(i) K₂Cr₂O₇ + KI + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + I₂ + H₂O
Answer:

K₂Cr₂O₇ + 6KI + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + I₂ + H₂O
K₂Cr₂O₇ + 6KI + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 3I₂ + H₂O
K₂Cr₂O₇ + 6KI + 7H₂SO₄ → 4K₂SO₄ + Cr₂(SO₄) + I₂ + 7H₂O

(ii) KMnO₄ + Na₂SO₃ → MnO₂ + Na₂SO₄ + KOH
Answer:

⇒ 2KMnO₄ + 3Na₂SO₃ → MnO₂ + Na₂SO₄ + KOH
⇒ 2KMnO₄ + 3Na₂SO₃ → 2MnO₂ + 3Na₂SO₄ + KOH
⇒ 2KMnO₄ + 3Na₂SO₃ → MnO₂ + Na₂SO₄ + 2KOH

(iii) Cu + HNO₃ → Cu(NO₃)₂ + NO₂ + H₂O
Answer:

Cu + 2HNO₃ → Cu(NO₃)₂ + NO₂ + H₂O
Cu + 2HNO₃ + 2HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

(iv) KMnO₄ + H₂C₂O₄ + H₂SO₄ → K₂SO + MnSO₄ + CO₂ + H₂O
Answer:

KMnO₄ + 5H₂C₂O₄ + H₂SO₄ → K₂SO + MnSO₄ + CO₂ + H₂O
2KMnO₄ + 5H₂C₂O₄ + H₂SO₄ → 2MnSO₄ + 10CO₂ + H₂O
2KMnO₄ + 5H₂C₂O₄ + 3H₂SO₄ → K₂SO + 2MnSO₄ + 10CO₂ + 8H₂O

Question 45.
Balance the following equations by ion electron method.
(i) KMnO₄ + SnCl₂ + HCl → MnCl₂ SnCl₄ + H₂O + KCl
Answer:

(ii) C₂O₄^2- + Cr₂O₇^2- →
Cr³⁺ + CO₂ (in acid medium)
Answer:

(iii) Na₂S₂O₃ + I₂ → Na₂S₄O₆ + NaI ₂ (in acid medium)
Answer:
S₂O₃^2- → S₄O₆^2- ………….(1)
Half reaction ⇒ I₂ → I_– …………….(2)

(iv) Zn + NO₃^– → Zn⁺² + NO
Half reactions are

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

I. Choose the best answer:

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

Question 2.
_______ are made up of molecules which contain two or more atoms of different elements.
(a) Mixtures
(b) Compounds
(c) Pure substances
(d) Elements
Answer:
(b) Compounds

Question 3.
Match the correct pair:

A. Compound	(i) S8
B. Mixture	(ii) Glucose
C. Element	(iii) Air

(a) A – ii, B – i, C – iii
(b) A – i, B – ii, C – iii
(c) A – ii, B – iii, C – i
(d) A – iii, B – ii, C – i
Answer:
(c) A – ii, B – iii, C – i

Question 4.
_______ atom is considered as standard by the IUPAC for calculating atomic masses.
(a) C – 13
(b) C – 15
(c) C – 14
(d) C – 12
Answer:
(d) C – 12

Question 5.
The value of unified mass is equal to
(a) 1.6605 × 10^-27 kg
(b) 1.6736 × 10^-27 kg
(c) 1.6605 × 10²⁷ kg
(d) 1.6736 × 10^-29 kg
Answer:
(a) 1.6605 × 10^-27 kg

Question 6.
Chlorine consists of two naturally occurring isotopes ₁₇Cl³⁵ and ₁₇Cl³⁷ in the ratio 77 : 23. The average relative atomic mass of chlorine is
(a) 36.45 u
(b) 35.56 u
(c) 35.46 u
(d) 35.65 u
Answer:
(c) 35.46 u

Question 7.
The relative atomic masses of hydrogen, oxygen, and carbon are 1.008 u, 16 u, and 12 u respectively. The relative molecular mass of glucose (C₆ H₁₂ O₆ ) is
(a) 170.096 u
(b) 189.096 u
(c) 180.096 u
(d) 190.086 u
Answer:
(c) 180.096 u

Question 8.
The specific amount of a substance is represented in SI unit is
(a) amu
(b) mole
(c) atomic mass
(d) equivalent mass
Answer:
(b) mole

Question 9.
Commonly used medicines for treating heart bum and acidity are called
(a) antipyretics
(b) analgesics
(c) antiseptics
(d) antacids
Answer:
(d) antacids

Question 10.
The typical concentration of hydrochloric acid in gastric acid is
(a) 0.1 M
(b) 0.01 M
(c) 0.082 M
(d) 0.82 M
Answer:
(c) 0.082 M

Question 11.
Antacids used to treat acidity contain mostly
(a) Al(OH)₃
(b) Mg(OH)₂
(c) (a) or (b)
(d) Ca(OH)₂
Answer:
(c) (a) or (b)

Question 12.
The value of Avogadro number is
(a) 6.022 × 10²³
(b) 6.022 × 10²²
(e) 6.022 × 10^-25
(d) 6.022 × 10^-23
Answer:
(a) 6.022 × 10²³

Question 13.
The volume occupied by one mole of any substance in the gaseous state st 273 K and 1 atm pressure is _______ (in litres)
(a) 24.5
(b) 22.4
(e) 22.71
(d) 21.18
Answer:
(b) 22.4

Question 14.
The equivalent mass of KMnO₄(Molar mass 158 g mol ^-1 based on the equation
MnO₄ + 8H⁺ + 5e → Mn²⁺ + 4H₂O
(a) 158
(b) 52.66
(c) 31.6
(d) 40
Answer:
(c) 31.6

Question 15.
The molecular formula of acetic acid is C₂H₄O₂ Its empirical formula is
(a) C₂H₄O₂
(b) C₂H₂O
(c) CH₂O₂
(d) CH₂O
Answer:
(d) CH₂O

Question 16.
The number of moles of hydrogen required to produce 20 moles of ammonia is
(a) 10
(b) 15
(c) 30
(d) 20
Answer:
(c) 30

Question 17.
Which of the following is/are redox reactions?
(i) 4Fe + 3O₂ → 2Fe₂O₃
(ii) H₂S + Cl₂ → 2HCl + S
(iii) CuO + C → Cu + CO
(iv) S +H₂ → H₂S
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv)
Answer:
(d) (iii) and (iv)

Question 18.
The oxidation number of sulphur on H₂SO₄ is
(a) -2
(b) -6
(c) +2
(d) +6
Answer:
(d) +6

Question 19.
Choose the correct pair:

Compound	Oxidation number
A. Cr in Cr2O7	(i) +4
B. C in CO2	(ii) +2
C. C in CH2F2	(iii) +6
D. Mn in MnSO4	(iv) 0

(a) A – i, B – iv, C- iii, D – ii
(b) A – iii, B – i, C – iv, D – ii
(c) A – i, B – ii, C – iv, D – iii
(d) A – iii, B – iv, C – i, D – ii
Answer:
(b) A – iii, B – i, C – iv, D – ii

Question 20.
The change in the oxidation number of Manganese in the following reaction 2 KMnO₄ + 10 FeSO₄ + 8 H₂SO₄ → K₂SO₄ + 5 Fe₂(SO₄)₃ + 8 H₂O
(a) +2 to +7
(b) +2 to +5
(c) +7 to +2
(d) +5 to +2
Answer:
(c) +7 to +2

Question 21.
Which of the following statements are correct about oxidation?
(i) Removal of oxygen
(ii) Addition of hydrogen
(iii) Loss of electron
(iv) Gain of electron
(v) Addition of oxygen
(vi) Removal of hydrogen
(a) i, ii, iv
(b) iii, v, vi
(c) i, iii, v
(d) iii, iv, v
Answer:
(b) iii, v, vi

Question 22.
Choose the correct oxidation number of oxygen in the following compounds.

(A) KO2	(i) -2
(B) H2O	(ii) +2
(C) H2O2	(iii) -1/2
(D) OF2	(iv) -1

(a) A -iii, B- iv, C – i, D – ii.
(b) A – iv, B – i, C – iii, D – ii
(c) A – iii, B – i, C – iv, D – ii
(d) A – iv, B – iii, C – i, D – ii
Answer:
(c) A – iii, B – i, C – iv, D – ii

Question 23.
The correct order of electron releasing tendency of the following elements is
(a) Zn > Cu > Ag
(b) Cu > Zn > Ag
(c) Ag > Zn > Cu
(d) Ag > Cu > Zn
Answer:
(a) Zn > Cu > Ag

Question 24.
H₂O₂ → 2 H₂O + O₂ is a
(a) Displacement reaction
(b) Combinationreaction
(c) Decomposition reaction
(d) Disproportionate reaction
Answer:
(d) Disproportionate reaction

Question 25.
The molar mass and empirical formula mass of a compound are 78 and 13 respectively. The molecular formula of the compound is (Empirical formula is CH)
(a) C₂H₂O₂
(b) C₂H₄
(c) C₆H₆
(d) C₃H₈
Answer:
(c) C₆H₆

II. Very Short Question and Answers (2 Marks):

Question 1.
State Avogadro’s Hypothesis.
Answer:
It states that ‘Equal volume of all gases under the same conditions of temperature and pressure contain the same number of molecules’.

Question 2.
How is matter classified physically?
Answer:
Matter can be classified as solids, liquids, and gases based on their physical state. The physical state of matter can be converted into one another by modifying the temperature and pressure suitably.

Question 3.
How is matter classified chemically?
Answer:
Matter can be classified into mixtures and pure substances based on chemical compositions.

Question 4.
Calculate a number of moles of carbon atoms ¡n three moles of ethane.
Answer:
Ethane – Molecular formula = C₂H₆
1 mole of ethane contains 2 atoms of carbon (6.023 x 10²³ C)
∴ 3 moles of ethane contains 6 atoms of Carbon.
∴ No. of moles of Carbon atoms = 3 x 6.023 x 10²³ Carbon atoms.
= 18.069 x 10²³ Carbon atoms.

Question 5.
What are pure substances? How are they classified?
Answer:
Pure substances are composed of simple atoms or molecules. They are further classified as elements and compounds.

Question 6.
Mass of one atom of an element ¡s 6.66 x 10²³ g. How many moles of the element are there in 0.320 kg?
Answer:
Mass of one atom of an element = 6.66 x 10²³g
No. of moles = \(\frac {Mass}{Molecular mass}\) 3
Molecular mass = Mass of 1 atom x Avogadro number
6.66 x 10²³ x 6.023 x 10²³
= 6.66 x 6.023 = 40.11318
Number of moles = \(\frac {Mass}{Molecular mass}\) = \(\frac{0.320 \mathrm{kg} \times 10^{3}}{40}\) = 8 moles.

Question 7.
What are compounds? Give examples.
Answer:
Compounds are made up of molecules which contain two or more atoms of different elements.
Example: Carbon dioxide (CO₂), Sodium Chloride (NaCl).

Question 8.
Calculate the weight of 0.2 moles of sodium carbonate.
Answer:
Sodium carbonate = Na₂CO₃
Molecular mass of Na₂CO₃ = (23 x 2)+(12 x 1)+(16 x 3)
= 46 + 12 + 48 = 106 g
Mass of 1 mole of Na₂CO₃ = \(\frac{106 \times 0.2}{1}\) = 21.2 g

Question 9.
Define relative atomic mass.
Answer:
The relative atomic mass is defined as the ratio of the average atomic mass factor to the unified atomic mass unit.
Relative atomic mass
(A_r) = Average mass of the atom / Unified atomic mass

Question 10.
What is the average atomic mass?
Answer:
Average atomic mass is defined as the average of the atomic masses of all atoms in their naturally occurring isotopes.

Question 11.
Calculate the equivalent mass of barium hydroxide.
Answer:
Barium hydroxide = Ba(OH)₂
Molecular mass of Ba(OH)₂ = 137 + (16 x 2) + (1 x 2) = 171.0 g / mol.
Acidity = 2
Equivalent mass of Ba(OH)₂ = \(\frac {17 1.0}{2}\) = 85.5

Question 12.
What is a mole?
Answer:
One mole is the amount of substance of a system, which contains as many elementary particles as there are atoms in 12 g of carbon – 12 isotope. The elementary particles can be molecules, atoms, ions, electrons or any other specified particles.

Question 13.
What do you understand by the terms empirical formula and molecular formula?
Answer:
Empirical Formula:

• It is the simplest formula.
• It shows the ratio of the number of atoms of different elements in one molecule of the compound.

Molecular Formula:

• It is the actual formula.
• It shows the actual number of different types of atoms present in one molecule of the compound.

Question 14.
What is Avogadro’s number?
Answer:
The total number of entities present in one mole of any substance is equal to 6.022 × 10²³. This number is called the Avogadro number.

Question 15.
State Avogadro hypothesis.
Answer:
Equal volume of all gases under the same conditions of temperature and pressure contain equal number of molecules.

Question 16.
Calculate the mass of sodium (in kg) present in 95 kg of a crude sample of sodium nitrate whose percentage purity is 70%.
Answer:
Sodium Nitrate = NaNO₃
Molecular mass of Sodium Nitrate = 23 + 14 + 48 = 85
100% pure 85 g of NaNO₃ contains 23 g of Sodium.
100% pure 95 x 103 g of NaNO₃ will contains \(\frac {23}{85}\) x 95 x 10³
= 25.70 x 10³ g of Sodium.
100% pure NaNO₃  contains 25.70 x 10³ g of Sodium.
∴ 70% pure NaNO₃  will contains = 17990 g (or) 17.99 Kg of Na.

Question 17.
Define molar volume.
Answer:
The volume occupied by one mole of any substance in the gaseous state at a given temperature and pressure is called molar volume.

Question 18.
What is gram equivalent mass?
Answer:
Gram equivalent mass of an element, compound or ion is the mass that combines or displaces 1.008 g hydrogen or 8 g oxygen or 35.5 g chlorine.
Gram equivalent mass = (Molar mass(g mol^-1)) / Equivalence factor (eq mol^-1)

Question 19.
What is meant by Plasma state? Give an example.
Answer:
The gaseous state of matter at a very high temperature containing gaseous ions and free-electron is referred to as the Plasma state. e.g. Lightning.

Question 20.
What is the acidity of a base? Give an example.
Answer:
The acidity of a base is the number of moles of ionizable OH- ions present in 1 mole of the base. The acidity of potassium hydroxide (KOH) is 1.

Question 21.
What is empirical formula ola compound?
Answer:
The empirical formula of a compound is the formula written with the simplest ratio of the number of different atoms present in one molecule of the compound as a subscript to the atomic symbol.

Question 22.
Chlorine has a fractional average atomic mass. Justify this statement.
Answer:
Chlorine molecule has two isotopes as in ₁₇Cl³⁵, ₁₇ Cl³⁷ in the ratio of 77 : 23, so when we are calculating the average atomic mass, it becomes fractional.
The average relative atomic mass of Chlorine = \(\frac {(35 x 77) + (37 x 23)}{100}\) = 35.46 amu

Question 23.
What is meant by Stoichiometry?
Answer:
Stoichiometry is the quantitative relationship between reactants and products in a balanced chemical equation in moles. The quantity of reactants and products can be expressed in moles or in terms of mass unit or as volume.

Question 24.
What are limiting and excess reagents?
Answer:
When a reaction is carried out using non-stoichiometric quantifies of the reactants, the product yield will be determined by the reactant that is completely consumed and is called the limiting reagent. It limits the further reaction to take place. The other reagent which is in excess is called the excess reagent.

Question 25.
Define Avogadro Number.
Answer:
Avogadro number is the number of atoms present in one mole of an element or number of molecules present in one mole of a compound. The value of Avogadro number (N) = 6.023 x 10²³

Question 26.
Define oxidation number.
Answer:
The oxidation number is defined as the imaginary charge left on the atom when all other atoms of the compound have been removed in their usual oxidation states that are assigned according to a set of rules.

Question 27.
Calculate the equivalent mass of Copper. (Atomic mass of copper = 63.5)
Answer:
Equivalent mass = \(\frac {Atomic mass}{Valency}\)
Equivalent mass of Copper = \(\frac {63.5}{2}\) = 31.75 g eq^-1.

Question 28.
Mention the types of redox reactions?
Answer:
The types of redox reactions are combination reaction, decomposition reaction, displacement reaction, disproportionate reaction and competitive electron transfer reactions.

III. Short Question and Answers (3 Marks):

Question 1.
Describe the chemical classification of matter.
Answer:
Matter can be classified into mixtures and pure substances based on chemical compositions. Mixtures consists of more than one chemical entity present without any chemical interactions. They can be further classified as homogeneous or heterogeneous mixtures based on their physical appearance. Pure substances are composed of simple atoms or molecules. They are further classified as elements and compounds.

Question 2.
Distinguish between element and compound.
Answer:
An element consists of only one type of atom. Element can exists as monoatomic or polyatomic units. The polyatomic elements are called molecules.
Example: Monoatomic unit – Gold (Au), Copper (Cu);
Polyatomic unit: Hydrogen (H₂), Phosphorous (P₄).

Compounds are made up of molecules which contain two or more atoms of different elements.
Example: Carbon dioxide (CO₂), Glucose (C₆ H₁₂ O₆)

Properties of compounds are different from those of their constituent elements. For example, sodium is a shiny metal, and chlorine is an irritating gas. But the compound formed from these two elements, sodium chloride shows different characteristics as it is crystalline solid, vital for biological functions.

Question 3.
What is average atomic mass? How is average atomic mass of chlorine calculated?
Answer:
Average atomic mass is defined as the average of the atomic masses of all atoms in their naturally occurring isotopes. Chlorine consists of two naturally occurring isotopes ₁₇Cl³⁷ and ₁₇Cl³⁵ in the ratio 77 : 23, the average relative atomic mass of chlorine is
= ((35 × 77) + (37 × 23)) / 100
= 35.46 u

Question 4.
What will be the mass of one ¹²C atom in g?
Answer:
Molar mass of ¹²C = 12.00 g mol^-1.
∴ Mass of 6.023 x 10²³ carbon atom = 12.0 g
∴ Mass of 1 carbon atom = \(\frac{12}{6.023 \times 10^{23}}\) = 1.992 x 10 g.

Question 5.
Discuss the role of antacids.
Answer:
Gastric acid is a digestive fluid formed in the stomach and it contains hydrochloric acid. The typical concentration of the acid in gastric acid is 0.082 M. When the concentration exceeds 0.1M it causes heartburn and acidity. Antacids used to treat acidity contain mostly magnesium hydroxide or aluminium hydroxide that neutralizes the excess acid. The chemical reactions are as follows.

3 HCl + Al(OH)₃ → AlCl₃ + 3H₂O
2HCl + Mg(OH)₂ → MgCl₂ + 2H₂O.

From the above reactions, we know that 1 mole of aluminium hydroxide neutralizes 3 moles of HCl while 1 mole of magnesium hydroxide neutralizes 2 moles of HCl.

Question 6.
Explain gram equivalent mass.
Answer:
Gram equivalent mass of an element, compound, or ion is the mass that combines or displaces 1.008 g hydrogen or 8 g oxygen or 35.5 g chlorine.
Consider the following reaction:
Zn + H₂ SO₄ → ZnSO₄ + H₂
In this reaction, 1 mole of zinc (65.38 g) displaces one mole of hydrogen molecule (2.016 g). Mass of zinc required to displace 1.008 g hydrogen is
= \(\frac{65.38}{2.016}\) × 1.008
= \(\frac{65.38}{2}\)
= 32.69
The equivalent mass of zinc = 32.69
The gram equivalent mass of zinc = 32.69 g eq^-1.
The expression used to calculate gram equivalent mass is
Gram equivalent mass = Molar mass(g mol^-1) / Equivalence factor (eq mol^-1)

Question 7.
Calculate the gram equivalent mass of sulphuric acid.
Answer:
Basicity of sulphuric acid (H₂SO₄) = 2eq mol^-1
Molar mass of H₂SO₄ = (2 × 1) +(1 × 32) +(4 × 16) = 56 g mol^-1
Gram equivalent mass of H₂SO₄ = \(\frac{98}{2}\)
= 49 g eq^-1

Question 8.
Calculate the gram equivalent mass of potassium hydroxide.
Answer:
Acidity of potassium hydroxide (KOH) = 1 eq mol^-1
Molar mass of KOH = (1 × 39) + (1 × 16) + (1 × 1) = 56 g mol^-1
Gram equivalent mass of KOH = \(\frac{56}{1}\) = 56 g eq^-1.

Question 9.
Calculate the gram equivalent mass of Potassium permanganate.
Answer:
Potassium permanganate is an oxidizing agent. Molar mass of
KMnO₄ = (1 × 39)+ (1 × 55)+ (4 × 16)
= 158 g mol^-1

In an acidic medium, permanganate is reduced during oxidation and is given by the following equation,
MnO₄^– + 8H⁺ + 5e → Mn² + 4H₂O
Therefore, n = 5.
Gram equivalent mass of KMnO₄ = \(\frac{158}{5}\) = 31.6 g mol^-1

Question 10.
An acid found in tamarind on analysis shows the following percentage composition: 32 % Carbon; 4 % Hydrogen; 64 % Oxygen. Find the empirical formula of the compound.
Answer:

The ratio of C : H : O is 2 : 3 : 3 and hence, the empirical formula of the compound is CH₂O

Question 11.
An organic compound present in vinegar has 40 % Carbon, 6.6 % Hydrogen: and 53.4 % Oxygen. Find the empirical formula of the compound.
Answer:

The ratio of C: H : O is 1 : 2 : 1 and hence, the empirical formula of the compound is CH₂O.

Question 12.
How much copper can be obtained from 100 g of anhydrous copper sulphate?
Answer:
Anhydrous copper sulphate = CuSO₄
Molecular mass of CuSO₄ = 63.5 + 32 + (16 x 4)
= 63.5 + 32 + 64
= 159.5 g
159.5 g of CuSO₄ contains 63.5 g of copper.
∴ 100 g of CuSO₄ contains \(\frac{6.35}{159.5}\) x 100 = 0.39811 x 100 = 39.81 g of Copper.

Question 13.
Explain the classical concept of oxidation and reduction.
Answer:
According to classical concept, the addition of oxygen or removal of hydrogen is called oxidation.
Consider the following reactions,
4Fe + 3O₂ → 2Fe₂O₃
H₂S +Cl₂ → 2 HCl + S

In the first reaction, which is responsible for the rusting of iron, the oxygen adds on to the metal, iron. In the second reaction, hydrogen is removed from Hydrogen sulphide.
According to classical concept, addition of hydrogen or removal of oxygen is called reduction.
Consider the following reactions,
CuO + C → Cu + CO
S + H₂ → H₂S
In the first reaction, oxygen is removed from cupric oxide and in the second reaction, hydrogen is added to sulphur.

Question 14.
Describe the electron concept of oxidation and reduction.
Answer:
The reaction involving loss of electron is termed as oxidation and gain of electron is termed as reduction.
Fe²⁺ → Fe³⁺ + e^– (loss of electron – oxidation)
Cu²⁺ + 2e^– → Cu ( gain of electron – reduction)

Question 15.
Describe the oxidation number concept of oxidation and reduction.
Answer:
During redox reactions, the oxidation number of elements changes. A reaction in which oxidation number of the element increases is called oxidation whereas the oxidation number of the element 3 decreases is called reduction.

In this reaction, manganese in potassium permanganate favours the oxidation of ferrous sulphate into ferric sulphate by gets reduced.

Question 16.
Write notes on displacement reaction.
Answer:
Redox reactions in which an ion or an atom in a compound is replaced by an ion or an atom of another element are called displacement reactions. They are further classified into
(i) metal displacement reactions
(ii) non-metal displacement reactions.

(i) Metal displacement reactions:
Place a zinc metal strip in an aqueous copper sulfate solution taken in a beaker. The intensity of blue colour of the solution slowly reduced and finally disappeared. The zinc metal strip became coated with brownish metallic copper. This is due to the following metal displacement reaction.
CuSO₄(aq)+ Zn(s) → Cu + ZnSO₄.

(ii) Non-metal displacement reaction:
Zn + 2HCl → ZnCl₂ + H₂

IV. Long Question and Answers:

Question 1.
What is a matter? Explain its classification.
Answer:
Matter is defined as anything that has mass and occupies space. All matter is composed of atoms.

Physical Classification:
Matter can be classified as solids, liquids and gases based on their physical state. The physical state of matter can be converted into one another by modifying the temperature and pressure suitably.

Chemical Classification:

Classification of Matter

Matter can be classified into mixtures and pure substances based on chemical compositions. Mixtures consists of more than one chemical entity present without any chemical interactions. They can be further classified as homogeneous or heterogeneous mixtures based on their physical appearance. Pure substances are composed of simple atoms or molecules. They are further classified as elements and compounds.

An element consists of only one type of atom. Element can exists as monoatomic or polyatomic units. The polyatomic elements are called molecules.
Example: Monoatomic unit – Gold (Au), Copper (Cu);
Polyatomic unit: Hydrogen (H₂), Phosphorous (P₄).

Compounds are made up of molecules which contain two or more atoms of different elements.
Example: Carbon dioxide (CO₂), Glucose (C₆H₁₂O₆)
Properties of compounds are different from those of their constituent elements. For example, sodium is a shiny metal, and chlorine is an irritating gas. But the compound formed from these two elements, sodium chloride shows different characteristics as it is a crystalline solid, vital for biological function.

Question 2.
How is empirical formula of a compound determined from the elemental analysis?
Answer:
The empirical formula of a compound determined from the elemental analysis by the following steps.
(i) Since the composition is expressed in percentage, we can consider the total mass of the compound as 100 g and the percentage values of individual elements as a mass in grams.
(ii) Divide the mass of each element by its atomic mass. This gives the relative number of moles of various elements in the compound.
(iii) Divide the value of a relative number of moles obtained in the step – (ii) by the smallest number of them to get the simplest ratio.
(iv) In case the simplest ratios obtained in step – (iii) are not whole numbers then they may be converted into the whole numbers by multiplying a suitable smallest number.

Question 3.
An organic compound present in vinegar has 40% carbon, 6.6 % hydrogen, and 53.4 % oxygen, Find the empirical formula and molecular formula of the compound. (Given, Molar mass: 60 g mol^-1).
Answer:

The ratio of C: H : O is 1 : 2 : 1 and hence, the empirical formula of the compound is CH₂O.
Empirical Formula mass = (1 × 12 + 1 × 2 + 1 × 16) = 12 + 2 + 16 = 30.
Whole number η =  Molar mass / Empirical formula mas = \(\frac{60}{30}\) = 2
Therefore, Molecular formula = (CH₂O)₂ = C₂H₄O₂

Question 4.
An organic compound present in vinegar has 40% carbon, 6.6 % hydrogen, and 53.4 % oxygen, Find the empirical formula and molecular formula of the compound. (Given, Molar mass: 90 g mol^-1).
Answer:

The ratio of C : H : O is 1 : 2 : 1 and hence, the empirical formula of the compound is CH₂O.
Empirical Formula mass = (1 × 12 + 1 × 2 × 16) = 12 + 2 + 16 = 30.
Whole number, n = \(\frac{\text { Molar mass }}{\text { Empirical formula mass }}\)
= \(\frac{90}{30}\) = 3
Therefore, Molecular formula = (CH₂O)₃ = C₃H₆O₃

Question 5.
What is a redox reaction? Explain the different concepts of redox reaction.
Answer:
The reaction involving loss of electron is oxidation and gain of electron is reduction. Both these reactions take place simultaneously and are called as redox reactions.

Classical concept of oxidation and reduction:
According to classical concept, addition of oxygen or removal of hydrogen is called oxidation.
Consider the following reactions,

4Fe + 3O₂ → 2Fe₂O₃
H₂S + Cl₂ → 2HCl + S

In the first reaction, which is responsible for the rusting of iron, the oxygen adds on to the metal, iron. In the second reaction, hydrogen is removed from Hydrogen sulphide.
According to classical concept, addition of hydrogen or removal of oxygen is called reduction.
Consider the following reactions,
CuO + C → Cu + CO
S + H₂ → H₂S

In the first reaction, oxygen is removed from cupric oxide, and in the second reaction, hydrogen is added to sulphur.

Electron concept of oxidation and reduction.
The reaction involving loss of electron is termed as oxidation and gain of an electron is termed as reduction.
Fe²⁺ → Fe³⁺ + e^– (loss of electron – oxidation)
Cu²⁺ + 2e^– → Cu (gain of electron – reduction)

Oxidation number concept of oxidation and reduction:
During redox reactions, the oxidation number of elements changes. A reaction in which oxidation number of the element increases is called oxidation whereas the oxidation number of the element decreases is called reduction.

In this reaction, manganese in potassium permanganate favours the oxidation of ferrous sulphate into ferric sulphate by gets reduced.

Question 6.
What is an oxidation number? State the rules to find the oxidation number.
Answer:
Oxidation number is defined as the imaginary charge left on the atom when all other atoms of the compound have been removed in their usual oxidation states that are assigned according to set of rules. A term that is often used interchangeably with oxidation number is oxidation state.

1. The oxidation state of a free element (i.e., uncombined state) is zero.
   Example: H₂ Cl₂, Na, and S8 have the oxidation number of zero.
2. For a monoatomic ion, the oxidation state is equal to the net charge on the ion.
   Example: The oxidation number of sodium in Na⁺ is +1.
   The oxidation number of chlorine in Cl^– is -1.
3. The algebraic sum of oxidation states of all atoms in a molecule is equal to zero, while in ions, it is equal to the net charge on the ion.
   Example: In H₂SO₄, 2 × (oxidation number of hydrogen) + 1 × (oxidation number of sulphur) + 4 × (oxidation number of oxygen) = 0.
4. Hydrogen has an oxidation number of +1 in all its compounds except in metal hydrides where it has -1 value.
   Example: Oxidation number of hydrogen in hydrogen chloride (HCl) is + 1.
   Oxidation number of hydrogen in sodium hydride (NaH) is -1.
5. Fluorine has an oxidation state of -1 in all its compounds.
6. The oxidation state of oxygen in most compounds is -2. Exceptions are peroxides, superoxides, and compounds with fluorine.
   Example: Oxidation number of oxygen
   (i) in water is -2,
   (ii) in hydrogen peroxide is -1,
   (iii) in superoxides such as KO₂ is 4, and
   (iv) in oxygen difluoride (OF₂) is +2.
7. Alkali metals have an oxidation state of +1 and alkaline earth metals have an oxidation state of +2 in all their compounds.

Question 7.
Balance the following chemical equation by oxidation number method.
KMnO₄ + FeSO₄ + H₂SO₄ → K₂SO₄ + Fe(SO₄)₃ + 8H₂O
Answer:
Using oxidation number concept, the reactants which undergoes oxidation and reduction are as follows:

The oxidation number of Mn in KMnO₄ changes from +7 to +2 by gaining five electrons and the oxidation number of Fe FeSO₄ changes from +2 to +3 by loosing one electron.
Since the total number of electrons lost is equal to the total number of electrons gained, the number of electrons, by cross multiplication of the respective formula with suitable integers on reactant side.

Based on the reactant side, the products are balanced.

2KMnO₄ + 10 FeSO₄ + H₂SO₄ → K₂SO₄ + MnSO₄ + Fe( SO₄)₃ + H₂O

Balance the other elements except H and O atoms. K and S are balanced as follows

2KMnO₄ + 10 FeSO₄ + H₂SO₄ → K₂SO₄ + MnSO₄ + Fe( SO₄)₃ + H₂O
The difference of 8 – S atoms in reactant side, has to be balanced by multiplying H₂SO₄ by ‘8’. The equation now becomes,

2KMnO₄ + 10 FeSO₄ + 8 H₂SO₄ → K₂SO₄ + MnSO₄ + Fe( SO₄)₃ + H₂O

‘H’ and ‘O’ atoms are balanced by multiplying H₂O molecules in the product side by ‘8’.

2KMnO₄ + 10 FeSO₄ + 8 H₂SO₄ → K₂SO₄ + 2 MnSO₄ + 5 Fe( SO₄)₃ + 8 H₂O
The above equation is a balanced equation.

Question 8.
How is the following equation ¡s balanced by Ion electron method?
MnO₄^– + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O.
Answer:
Using the oxidation number concept, the reactants which undergoes oxidation and reduction are as follows;

The two half reactions are,
Fe²⁺ → Fe³⁺ + 1e ……….(1)
MnO₄^– + 5e + 8H⁺ → Mn²⁺ + 4H₂O ……….(2)
Balancing the atoms and charges on both sides of the half reactions.
There is no change in the equation (1) whereas in equation (2), there are four ‘O’ atoms on the reractant side .
Therefore, four H₂ is added on the product side, to balance ‘H’ – add, 8 H+ in the reactant side.
MnO₄^– + 5e + 8H⁺ → Mn²⁺ + 4H₂O ………..(3)
The two half reactions are equated in such a way that the number of electrons lost is equal to number of electrons gained.
Adding the two half reactions as follows:
(1) × 5 5 Fe²⁺ → 5 Fe³⁺ + 5e ……………(4)
(3) × 1 MnO₄^– + 5e + 8H⁺ → Mn²⁺ + 4H₂O ………(5)

(4) + (5)
MnO₄^– + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5 Fe³⁺ + 4H₂O ………..(6)
The equation (6) is a balanced equation.

Question 9.
Calculate the percentage composition of the elements present in lead nitrate. How many Kg of 02 can be obtained from 50 kg of 70% pure lead nitrate?
Answer:
Lead nitrate = Pb (NO₃)₂
Molecular mass of lead nitrate = 207 + (14 x 2) + (16 x 6)
= 207 + 28 + 96 = 331 g / mol.
331 g of lead nitrate contains 96 g of oxygen.
∴ 50 x 10³ g of lead nitrate will contain \(\frac {96}{331}\) x 50 x 10³
= 14501.5 g
= 14.501 Kg of oxygen.
100 % pure lead nitrate contains 14.501 Kg of oxygen.
70 % pure lead nitrate will contain = \(\frac {14.501}{100}\) x 70 = 10.15 Kg of oxygen.
.’. 70 % pure lead nitrate will contain 10.15 Kg of oxygen.

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Chemistry Model Question Paper 4 English Medium

Time: 3 Hours
Maximum Marks: 70

Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the parts An internal choice of questions is provided wherever applicable
3. All questions of Part I, II, III, and IV are to be attempted separately
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are lo be answered in about one or two sentences
6. Question numbers 25 to 33 in Part III are three-marks questions. These are to be answered in about three to five short sentences
7. Question numbers 34 to 38 in Part IV are five-mark Questions These are to answered in detail. Draw diagrams wherever necessary

Part – 1

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
Depressing agents used to separate ZnS from PbS is …………..
(a) NaCN
(b) NaCl
(c) NaNO₃
(d) NaNO₂
Answer:
(a) NaCN

Question 2.
The basic structural unit of silicates is
(a)(SiO₃)^2-
(b) (SiO₄)^2-
(c) (SiO)^–
(d) (SiO₄)^4-
Answer:
(d) (SiO₄)^4-

Question 3.
………………… is a pungent smelling gas.
(a) Ammonia
(b) Nitric acid
(c) Fluorite
(d) Sodium chloride
Answer:
(a) Ammonia

Question 4.
Consider the following statement.
(i) All the actinoids are non radioactive.
(ii) Neptunium and other heavier elements are produced by artificial transformation of naturally occurring elements by nuclear reactions.
(iii) Most.of the actinoids have long half lives.
Which of the above statements is/are not correct.
(a) i only
(b) i and ii
(c) ii and iii
(d) i and iii
Answer:
(d) i and iii

Question 5.
How many geometrical isomers are possible for [Pt (Py) (NH₃) (Br) (Cl)]?
(a) 3
(b) 4
(c) 0 3
(d) 15
Answer:
(a) 3

Question 6.
Metal excess defect is possible in……………….
(a) AgCl
(b) AgBr
(c) KCl
(d) Fes
Answer:
(c) KCl

Question 7.
The correct difference between first and second order reactions is that ……………………
(a) A first order reaction can be catalysed; a second order reaction cannot be catalysed.
(b) The half life of a first order reaction does not depend on [A₀]; the half-life of a second order reaction does depend on [A₀].
(c) The rate of a first order reaction does not depend on reactant concentrations; the rate of a second order reaction does depend on reactant concentrations.
(d) The rate of a first order reaction does depend on reactant concentrations; the rate of a second order reaction does not depend on reactant concentrations.
Answer:
(b) The half life of a first order reaction does not depend on [A₀]; the half-life of a second order reaction does depend on [A₀].
Solution:
For a first order reaction

Question 8.
Concentration of the Ag⁺ ions in a saturated solution of Ag₂C₂O₄ is 2.24 × 10^-4 mol L^-1 solubility product of Ag₂C₂O₄ is
(a) 2.42 × 10^-8 mol³ L^-3
(b) 2.66 × 10^-12mol³ L^-3
(c) 4.5 × 10^-11mol³ L^-3
(d) 5.619 × 10^-12mol³ L^-3
Answer:
(d) 5.619 × 10^-12mol³ L^-3
Solution:

Question 9.
Assertion(A): Copper Sulphate can be stored in a Zinc vessel.
Reason (R): Zinc is less reactive than Copper.
(a) Both A and R are correct
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(b) Both A and R are wrong

Question 10.
Fog is colloidal solution of……………..
(a) solid in gas
(b) gas in gas
(c) liquid in gas
(d) gas in liquid
Answer:
(c) liquid in gas

dispersion medium-gas
dispersed phase-liquid

Question 11.
Match the following Column-I with Column-II using the code given below.
c
Answer:
(a) A – 2,B – 3, C – 4, D – 1

Question 12.

(a) anilinium chloride
(b) O – nitro aniline
(c) benzene diazonium chloride
(d) m – nitro benzoic acid

Answer:
(c) benzene diazonium chloride

Question 13.
Which one of the following is the IUPAC name of CH₃ – CH₂ – CH₂ CN?
(a) Propiono nitrile
(b) Butane cyanide
(c) Isobutyro nitrile
(d) Butane nitrile
Answer:
(d) Butane nitrile

Question 14.
The correct corresponding order of names of four aldoses with configuration given below Respectively is ……………
(a) L-Erythrose, L-Threose, L-Eiythrose, D-Threose
(b) D-Threose, D-Erythrose, L-Threose, L-Erythrose
(c) L-Erythrose, L-Threose, D-Erythrose, D-Threose
(d) D-Erythrose, D-Threose, L-Erythrose, L-Threose
Answer:
(d) D-Erythrose, D-Threose, L-Erythrose, L-Threose

Question 15.
Tranquilisers are substances used for the treatment of ………………
(a) cancer
(b) AIDS.
(c) mental diseases
(d) blood infection
Answer:
(c) mental diseases

Part – II

Answer any six questions. Question No. 22 is compulsory. [6 × 2 = 12]

Question 16.
What are leaching process?
Answer:
This method is based on the solubility of the ore in a suitable solvent and the reactions in aqueous solution. In this method, the crushed ore is allowed to dissolve in a suitable solvent, the metal present in the ore is converted to its soluble salt or complex while the gangue remains insoluble. This process is also called chemical method.

Question 17.
What happens when PCI5 is heated?
Answer:
On heating phosphorous pentachloride, it decomposes into phosphorus trichloride and chlorine.

Question 18.
Write the biologically importance of coordination compounds.
Answer:
Biological important of coordination compounds:

(i) Ared blood corpuscles (RBC) is composed of heme group, which is Fe²⁺– Porphyrin complex, it plays an important role in carrying oxygen from lungs to tissues and carbon dioxide from tissues to lungs.

(ii) Chlorophyll, a green pigment present in green plants and algae, is a coordination complex containing Mg²⁺ as central metal ion surrounded by a modified Porphyrin ligand called corrin ring. It plays an important role in photosynthesis, by which plants converts CO₂ and water into carbohydrates and oxygen.

(iii) Vitamin B₁₂(cyanocobalamine) is the only vitamin consist of metal ion. it is a coordination complex in which the central metal ion is CO⁺ surrounded by Porphyrin like ligand.

(iv) Many enzymes are known to be metal complexes, they regulate biological processes. For example, Carboxypeptidase is a protease enzyme that hydrolytic enzyme important in digestion, contains a zinc ion coordinated to the protein.

Question 19.
Mention the factors affecting the reaction rate.
Answer:
The rate of the reaction is affected by the following factors.

• Nature and state of the reactant .
• Concentration of the reactant
• Surface area of the reactant
• Temperature of the reaction
• Presence of a catalyst

Question 20.
What is meant by standard reduction potential? What is its application?
Answer:

• The standard reduction potential (E°) is a measure of the oxidising tendency of the species.
• The greater the E° value means greater is the tendency shown by the species to accept electrons and undergo reduction.
• So higher the E° values, lesser is the tendency to undergo corrosion.

Question 21.
Give two important characteristics of physisorption.
Answer:
Important characteristics of physisorption:

• It is reversible
• It has low heat of adsorption
• It has weak van der Waals forces of attraction with adsorbent.
• It increases with increase in pressure.
• It forms multimolecular layer.

Question 22.
Draw the major product formed when 1-ethoxyprop-l-ene is heated with one equivalent
Answer:

This reaction follows SN¹ mechanism because in this reaction the more stable carbocation is formed that is double bonded carbocation. Therefore, the given molecule reacts with HI to form ethanol and 1-iodo prop-1-ene.

Question 23.
Define Tautomerism. Give example. Why tertiary nitro alkanes do not exhibit tautomerism?
Answer:
(i) Tautomerism is an isomerism in which the isomers change into one another with great ease of shifting of proton so that they exist together in equilibrium.

(ii) Tertiary nitro alkanes do not exhibit tautomerism due to absence of a-H atom.

Question 24.
Why do soaps not work in hard water?
Answer:
Soaps are sodium or potassium salts of long-chain falty acids. Hard water contains calcium and magnesium ions. When soaps are dissolved in hard water, these ions displace sodium or potassium from insoluble calcium or magnesium salts of fatty acids. These insoluble salts separate as scum.

This is the reason why soaps do not work in hard water.

Part – III

Answer any six questions. Question No. 31 is compulsory. [6 × 3 = 18]

Question 25.
Write the application of Iron (Fe).
Answer:

• Iron is one of the most useful metals and its alloys are used everywhere including bridges, electricity pylons, bicycle chains, cutting tools and rifle barrels.
• Cast iron is used to make pipes, valves and pumps stoves etc.
• Magnets can be made of iron and its alloys and compounds.

An important alloy of iron is stainless steel, and it is very resistant to corrosion. It is used in architecture, bearings, cutlery, surgical instruments and jewellery. Nickel steel is used for making cables, automobiles and aeroplane parts. Chrome steels are used for manufacturing cutting tools and crushing machines.

Question 26.
Give a reason to support that sulphuric acid is a dehydrating agent.
Answer:
Sulphuric acid is highly soluble in water and has strong affinity towards water and hence it can be used as a dehydrating agent. When dissolved in water it forms mono (H₂SO_4.H₂O) and di (H₂SO_4.H₂O) hydrates and the reaction is exothermic.
The dehydration property can also be illustrated by its reaction with organic compounds such as sugar, oxalic acid and formic acid.

Question 27.
Give evidence that [CO(NH₃)gCl]SO₄ and [CO(NH₃)₅SO₄]Cl are ionisation isomers.
Answer:
When they are dissolved in water, they will give different ions in the solution which can be
tested by adding AgNO₃ solution and BaCl₂ solution. When Cl^– ions are the counter ions, a white precipitate will be obtained with AgNO₃ solution. If SO₄^2- ions are the counter ions, a white precipitate will be obtained with BaCl₂ solution.

Question 28.
For a reaction, X + Y → Product; quadrupling [x] , increases the rate by a factor of 8. Quadrupling both [x] and [y], increases the rate by a factor of 16. Find the order of the reaction with respect to x and y. what is the overall .order of the reaction?
Answer:
z = k [x]^m [y]ⁿ …… (1)
8z = k [4x]^m [y]ⁿ…… (2)
I 6z = k [4x]^m [4y]ⁿ…… (3)
Dividing Eq (2) by Eq (1) we get,

1.5 order with respect to x.
Dividing Eq (3) by Eq (1) we get,

16 = 4^m . 4ⁿ
16 = 4².4ⁿ
16/16 = 4ⁿ
1 = 4n
∴ n = 0 [Zero order with respect toy]
Overall order of the reaction, ‘
k[x]^m[y]ⁿ
k [x]^1.5 [y]⁰
Order = (1.5 + 0) = 1.5

Question 29.
Identify the conjugate acid base pair for the following reaction in aqueous solution
Answer:

• HF and F^– , HS^– and H₂S are two conjugate acid – base pairs.
• F^– is the conjugate base of the acid HF (or) HF is the conjugate acid of F^– .
• H₂S is the conjugate acid of HS^– (or) HS^– is the conjugate base of H₂S.

(ii)

• HPO₄^2- and PO₄^3- , SO₃^-2 and HSO₃^– are two conjugate acid – base pairs.
• PO₄^3- is the conjugate base of the acid HPO₄^2- (or)
  HPO₄^2- is the conjugate acid of PO₄^3-.
• HSO₃^– is the conjugate acid of SO₃^-2 (or) SO₃^-2 is the conjugate base of HSO₃^–

(iii)

• NH₄⁺ and NH₃, CO₃^2- and HCO₃^– are two conjugate acid – base pairs.
• HCO₃^– is the conjugate of acid CO₃^2- (or) CO₃^2- is the conjugate bases of HCO₃^–
• NH₃ is the conjugate base of NH₄⁺ (or) NH₄⁺ is the conjugate acid of NH₃.

Question 30.
In the following fields, how adsorption is applied?
Answer:
(i) Medicine (ii) Metallurgy (iii) Mordant & Dyes (iv) indicators
Answer:
(i) Medicine: – Drugs cure diseases by adsorption on body tissues.

(ii) Metallurgy:- Sulphide ores are concentrated by a process called froth floation in which lighter ore particles are adsorbed by pine oil.

(iii) Mordants and Dyes:- Most of the dyes are adsorbed on the surface of the fabric. Mordants are the substances used for fixing dyes onto the fabric.

(iv) In the precipitation titrations, the end point is indicated by an external indicator which changes its colour after getting absorbed on precipitate. It is used to indicate the end point of filtration.

Question 31.
Identify A, B, C, and D

Answer:

Question 32.
Draw the structural formula and write the IUPAC name of
(i) N, N-dimethyl aniline (ii) Benzyl amine (iii) N-methyl benzylamine
Answer:

Question 33.
Differentiate soap and detergents?
Soap :

1. Soaps are sodium or potassium salt of long chain fatty acid.
2. Soaps are made from animal (or) plant fats and oils.
3. Soaps have lesser cleansing action.
4. Soaps are bio degradable.
5. Soaps are less effective in hard water.
6. They have a tendency to form a scum in hard water.
7. Example: Sodium palmitate.

Detergent :

1. Detergent is sodium salt of alkyl hydrogen sulphate or alkyl benzene sulphonic acid.
2. Detergents are made from petrochemicals.
3. Detergents have more cleansing action.
4. Detergents are non-bio degradable.
5. Detergents are more effective even in hard water.
6. They do not form scum with hard water.
7. Example: Sodium lauryl sulphate.

Part – IV

Answer all the questions. [5 × 5 = 25]

Question 34.
(a) (i) Explain the role of carbon monoxide in the purification of nickel? (2)
(ii) Describe the structure of diborane. (3)
[OR]
(b) (i) What type of hybridisation occur in 1. BrF₅. 2. BrF₃ (2)
(ii) Most of the transition metals act as catalyst. Justify this statement. (3)
Answer:
(a) (i) During the purification of Nickel by Mond’s process, carbon monoxide (CO) is used to convert impure nickel to nickel carbonyl.
Nickel carbonyl is an unstable compound. Heating to higher temperature decomposes it to give pure nickel.

(ii) In diborane two BH₂ units are linked by two bridged hydrogens. Therefore, it has eight B-H bonds. However, diborane has only 12 valance electrons and are not sufficient to
form normal covalent bonds. The four terminal B-H bonds are normal covalent bonds (two centre – two electron bond or 2c-2e bond). The remaining four electrons have to used for the bridged bonds.
1. e. two three centred B-H-B bonds utilise two electrons each. Hence, these bonds are three centre- two electron bonds. The bridging hydrogen atoms are in a plane as shown in the figure.

In dibome, the boron is sp³ hybridised. Three of the four sp³ hybridised orbitals contains single electron and the fourth orbital is empty. Two of the half filled hybridised orbitals of each boron overlap with the two hydrogens to form four terminal 2c-2e bonds, leaving one empty and one half filled hybridised orbitals on each boron. The Three centre – two electron bonds, B-H-B bond formation involves overlapping the half filled hybridised orbital of one boron, the empty hybridised orbital of the other boron and the half filled Is orbital of hydrogen.

[OR]

(b) (i) 1. BrF₅. is a AX₅ type. Therefore is has sp³ d² hybridisation. Hence, BrF₅ molecule has square pyramidal shape.
2. BrF₃ is a AX₃ type. Therefore it has sp³ d hybridisation. Hence, BrF₃ molecule has T-shape.

(ii) Many industrial processes use transition metals or their compounds as catalysts. Transition metal has energetically available d orbitals that can accept electrons from reactant molecule or metal can form bond with reactant molecule using its ‘d’ electrons.

For example, in the catalytic hydrogenation of an alkene, the alkene bonds to an active site by using its n electrons with an empty d orbital of the catalyst.

Question 35.
(a) (i) Why tetrahedral complexes do not exhibit geometrical isomerism. (2)
(ii) Explain about the importance and application of coordination complexes. (3)
[OR]
(b) Explain AAAA and ABABA and ABCABC type of three dimensional packing with the help of neat diagram.
Answer:
(a) (i) In tetrahedral geometry

• AH the four ligands are adjacent or equidistant to one another.
• The relative positions of donor atoms of ligands attached to the central metal atom are same with respect to each other.
• It has plane of symmetry

Therefore, tetrahedral complexes do not exhibit geometrical isomerism.

(ii) 1. Phthalo blue – a bright blue pigment is a complex of copper (II) ion and it is used in printing ink and packaging industry.

2. Purification of Nickel by Mond’s process involves formation of [Ni (CO)₄ ] which yields 99.5% pure on decomposition.

3 . EDTA is used as a chelating ligand for the separation of lanthanides, in softening of hard water and also in removing lead poisoning.

4. Coordination complexes are used in the extraction of silver and gold from their ores
by forming soluble cyano complex. These cyano complexes are reduced by zinc to
yield metals. This process is called Mac – Arthur Forrest cyanide process.

5. Some metal ions are estimated more accurately by complex fonnation. For eg., Ni present in Nickel chloride solution is estimated accurately forming an insoluble, complex called [Ni (DMG)₂].

6. Many of the complexes are used as catalyst in organic and inorganic reactions. For e.g.,
(i) Wilkinson’s Catalyst – [(PPh₃)₃, Rh Cl] is used for hydrogenation of alkenes.
(ii) Ziegler – Natta Catalyst [TiCl₄ + Al (C₂H₅) ]₃ is used in the polymerisation of ethene.

7. In photography, when the developed film is washed with sodium thio sulphate solution (hypo), the negative film gets fixed. Undecomposed AgBr forms a soluble complex called sodium dithio sulphate argentate (I) which can be removed easily by washing the film with water.

AgBr + 2 Na₂ S₂O₃ → Na₃ [Ag (S₂O₃)₂] + 2NaBr

[OR]

(b) (i) AAAA type of three dimensional packing:
This type of three dimensional packing arrangements can be obtained by repeating the AAAA type two dimensional arrangements in three dimensions, i.e., spheres in one layer sitting directly on the top of those in the previous layer so that all layers are identical. All spheres of different layers of crystal are perfectly aligned horizontally and also vertically, so that any unit cell of such arrangement as simple cubic structure as shown in fig.

In simple cubic packing, each sphere is in contact with 6 neighbouring spheres – Four in its own layer, one above and one below and hence the coordination number of the sphere in simple cubic arrangement is 6.

(ii) ABABA type of three dimensional packing:
In this arrangement, the spheres in the first layer (A type) are slightly separated and the second layer is formed by arranging the spheres in the depressions between the spheres in layer A as shown in figure. The third layer is a repeat of the first. This pattern ABABAB is repeated throughout the crystal. In this arrangement, each sphere has a coordination number of 8, four neighbors in the layer above and four in the layer below.

(iii) ABCABC type of three dimensional packing:
In this arrangement (FCC) second layer spheres are arranged at the dips of first layer. Third layer spheres are arranged in a manner such that it cover the octahedral void. Then no longer third layer is similar to first or second layer. Third layer gives different arrangement. Fourth Layer spheres are similar to first layer. If the first, second and third layer are represented as A,B,C then this type of packing gives the arrangement of layers as ABCABC… (Le.,), the first three layers do not resemble first, second and third layers respectively and the sequence is repeated with the addition of more layers.

In this arrangement atoms occupy 74% of the available space and thus has 26% vacant space. The coordination number is 12.

Voids: The empty spaces between the three dimensional layers are known as voids. There are
two types of common voids possible. They are tetrahedral and octahedral voids.

Tetrahedral void:
A void formed by three spheres of a layer in contact with each other and also with a sphere on the top or bottom r Layer a layer is a hole between four spheres. The spheres are arranged at the vertices Layerc of a regular tetrahedron such a hole or void is called tetrahedral void.

Octahedral void:
A hole or void formed by three spheres of a hexagonal layer and another three spheres of the adjacent layer is a hole between six spheres. The spheres are arranged at the vertices of a regular octahedron. Such a hole or void is abc arrangement – ccp structure called octahedral void.

The buffer capacity or index is defined as the concentration of acid or base to add in order to modify the pH of an aqueous solution.

Question 36.
(a) (i) The rate constant for a first order reaction is 1.54 × 10^-3s^-1 (2)
Calculate its half life time.
(ii) Explain about protective action of colloid. (3)
[OR]
(b) (i) What is buffer solution? Give an example for an acidic buffer and a basic buffer. (2)
(ii The value of k_spof two sparingly soluble salts Ni(OH)₂ and AgCN are 2.0 × 10^-15 and 6 × 10^-17 respectively. Which salt is more soluble? Explain. (3)
Answer:
(a) (i) We know that, t_1/2 = 0.693/ k
t_1/2= O.693/1.54 ×10^-3s ^-1 = 450s

(ii) 1. Lyophobic sols are precipitated readily even with small amount of electrolytes. But they are stabilised by the addition of small amount of lyophillic colloid.

2. A smaLl amount of gelatine sol is added to gold sol to protect gold sol.

3. Gold number is a measure of protecting power of a colloid. Gold number is defined as the number of milligrams of hydrophillic colloid that will just prevent the precipitation of 10 ml of gold sol on the addition of ImI of 10% NaCI solution. Smaller the gold number, greater the protective power.

[OR]

(b) (i) 1. Buffer is a solution which consists of a mixture of weak acid and its conjugate base (or) a weak base and its conjugate acid.

2. This buffer solution resists drastic changes in its pH upon addition of a small quantities of acids (or) bases and this ability is called buffer action.

3. Acidic buffer solution : Solution containing acetic acid and sodium acetate.
Basic buffer solution : Solution containing NH₄O and NH₄Cl.

(ii)

Ni(OH)₂ is more soluble than AgCN.

Question 37.
(a) Describe about lead storage battery construction and its uses. (5)
[OR]
(b) (i) What happens when m-cresol is treated with acidic solution of sodium dichromate? (3)
(ii) Formic acid is more stronger than acetic acid. Justify this statement. (2)
(a) Lead storage battery :
1. Anode : Spongy lead
Cathode : Lead plate bearing PbO₂
Electrolyte : 38% by mass of H₂SO₄ with density 1.2 g/ml

2. Oxidation occurs at the anode

Pb_(s) → Pb²⁺ _(aq) + 2e^– …………(1)

The Pb²⁺ ions combine with SO₄^2- to form PbSO₄ precipitate

Pb²⁺_(aq) + SO₄^2- → PbS_4(s) …………….(2)

3. Reduction occurs at the cathode

PbO_2(s) + 4H⁺_(aq) + 2e^– – → Pb²⁺_(aq) + 2H₂O ………….(3)

The Pb²⁺ ions also combines with SO₄^2- ions to form sulphuric acid to form PbSO₄ Precipitate

Pb²⁺_(aq) + SO₄^2-_(aq) → PbSO₄ …………..(4)

4. The overall reaction is,
(1) + (2) + (3) + (4)
pb_(s) + PbO_2(s) + 4H⁺ + 2SO₄^2-_(aq) → 2PbSO_4(s) + 2H₂O(l)

5. The emf of a single cell is about 2V. Usually six such cells are combined in series to produce 12 volts.

6. The emf of the cell depends on the concentration of H₂SO₄. As the cell reaction uses SO₄^2- ions, the concentration H₂SO₄ decreases. When the cell potential falls to about 1,8V, the cell has to be recharged.

7. Recharge of the cell: During recharge process, the role of anode and cathode is reversed and H₂SO₄ is regenerated.

The above reaction is exactly the reverse of redox reaction which takes place while discharging.

8. Uses: Lead storage battery is used in automobiles, trains, inverters.

[OR]

(b) (i) When m-cresol is treated with acidic solution of sodium dichromate it gives 4-hydroxy benzoic acid.

(ii) The electron releasing groups (+1 groups) increase the relative charge on the carboxylate ion and destabilise it and hence the loss of proton becomes difficult.
+I groups are CH₃, – C₂H₅, – C₃H₇

Question 38.
(i) An aromatic compound ‘A’ on treatment with aqueous ammonia and heating
forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’
of molecular formula C₆H₇N. Write the structures and IUPAC names of compound
A,BandC. (3)
(ii) Lactose act as reducing sugar. Justify this statement. (2)
[OR]
(b) (i) Explain the mechanism of enzyme action? (3)
(ii) Which chemical is responsible for the antiseptic properties of dettol? (2)
Answer:
(a) (i) Step-1: To find out the structure of compounds‘B’and‘C’.
1. Since compound ‘C’ with molecular formula C₆H₇N is formed from compound ‘B’ on treatment with Br₂ + KOH (i.e, ., Hoffmann bromamide reaction). Therefore, compound ‘B’ must be an amide and ‘C’ must be an amine. The only amine having the molecular formula C₆H₇N is C₆H₅NH₂ (i.e., aniline or benzenamine).

2. Since ‘C’ is aniline, therefore, the amide from which it is formed must be benzamide (C₆H₅CONH₂). Thus, compound ‘B’ is benzamide:
The chemical equation showing the conversion of ‘B’ to ‘C’ is

Step-2: To find out the structure of compound ‘A’. Since compound ‘B’ is formed from compound ‘A’ by treatment with aqueous ammonia and heating. Therefore, compound ‘A’ must be benzoic acid or benzenecarboxylic acid.

(ii) Lactose is a disaccharide and contains one galactose unit and one glucose unit.
In lactose, the β-D galactose and β-D glucose are linked by β-1, 4 – glycosidic bond.
The aldehyde carbon is not involved in the glycosidic bond hence it retains its reducing property and is called a reducing sugar.

[OR]

(b) (i) Enzymes are bio catalysts that catalyse a specific bio chemical reaction. They generally activate the reaction by reducing the activation energy by stabilising the transition state.

In a typical reaction, enzyme E binds with the substrate molecule leversity to produce an enzyme substrate complex. During this stage the substrate is converted into product and the enzyme becomes free and is ready to bind to another substrate molecule.

(ii) (a) Chloroxylenol and (b) Terpineol are the chemicals responsible for the antiseptic properties of dettol. But among these two, chloroxylenol plays more important role. Chloroxylenol is an antiseptic and disinfectant which is used for skin disinfection and cleaning surgical instruments.