NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

NCERT
SOLUTIONS
CLASS - 12th

aglase .co

Class : 12th
Subject : Chemistry
Chapter : 2
Chapter Name : Solutions

( ) ( )
Q2.1 Calculate the mass percentage of benzene ( C 6H 6 ) and carbon tetrachloride CC1 4 ) if 22 g of
benzene is dissolved in 122 g of carbon tetrachloride.

Answer. Mass percentage Of C 6H 6
Mass of C 6H 6
= Total mass of the solution × 100%
Mass of C 6H 6
= Mass of C H + Mass of CCl × 100%
6 6 4
22
= 22 + 122 × 100%
= 15.28%

Mass percentage of CCl 4
Mass of CCl 4
= Total mass of the solution × 100%
Mass of CCl 4
= Mass of C H + Mass of CCl × 100%
6 6 4

122
= 22 + 122 × 100%

= 84.72%
Alternatively,
Mass percentage of CCl 4 = (100 - 15.28) %
= 84.72 %

Page : 39 , Block Name : Intext Questions

Q2.2 Calculate the mole fraction of benzene in solution containing 30% by mass in carbon
tetrachloride.

Answer. Let the total mass of the solution be 100 g and the mass of benzene be 30 g.
∴ Mass of carbon tetrachloride = (100 — 30) g
= 70 g
Molar mass of benzene ( C 6H 6 ) = (6 × 12 + 6 × 1)gmol − 1
= 78gmol − 1

30
∴ Number of moles of C 6H 6 = 78 mol

= 154 g mol − 1
70
.Number of moles of CCl 4 = 154 mol

= 0.4545mol

Thus, the mole fraction of C 6H 6 is given as:

Number of moles of C 6H 6
Number of moles of C 6H 6 + Number of moles of CCl 4

0.3846
=
0.3846 + 0.4545
= 0.458

Page : 39 , Block Name : Intext Questions

[ ] 2. 6H2O in 4.3 L of
Q2.3 Calculate the molarity of each of the following solutions: (a) 30 g of Co NO 3
solution (b) 30 mL of 0.5 M H 2SO 4 diluted to 500 mL.

Answer.
Molarity is given by:
Moles of solute
Molarity = Volume of solution in litre

( ) 2 ⋅ 6H2O = 59 + 2(14 + 3 × 16) + 6 × 18
(a) Co NO 3
= 291gmol − 1

30
∴ Moles of Co NO 3 ( )2 ⋅ 6H2O = 291 mol
= 0.103 mol
0.103mol
Therefore, molarity = 4.3L
= 0.023 M

(b) Number of moles present in 1000 mL of 0.5 M H 2SO 4 = 0.5 mol

0.5 × 30
∴ Number of moles present in 30 mL of 0.5 H 2SO 4 = = 1000 mol
= 0.015 mol
0.015
Therefore, molarity = = 0.5L mol
= 0.03 M

Page : 39 , Block Name : Intext Questions

( )
Q2.4 Calculate the mass of urea NH 2CONH 2 required in making 2.5 kg of 0.25 molal aqueous
solution.

Answer.
Molar mass of urea ( NH 2CONH 2) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16

= 60gmol − 1
0.25 molar aqueous solution of urea means:
1000g of water contains 0.25mol = (0.25 × 60)g of urea
= 15g of urea
That is,
(1000 + 15) g Of solution contains 15 g Of urea
15 × 2500
Therefore, 2.5 kg (2500 g) of solution contains = = 1000 + 15 g
= 36.95 g
= 37 g of urea ( approximately)
Hence, mass of urea required = 37 g

Page : 39 , Block Name : Intext Questions

Q2.5 Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass)
aqueous KI is 1.202 g mL − 1.

Answer. (a) Molar mass of KI = 39 + 127 = 166gmol − 1
(mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.
That is,
20 g of KI is present in ( 100 - 200) g of water = 80 g of water
Moles of KI
Therefore, molality of the solution = = Mass of water in kg
20
166
= 0.08 m
= 1.506 m
= 1.51 m ( approximately)

(b) It is given that the density of the solution = 1.202gmL − 1
Mass
∴ Volume of 100 g solution = = Density
100g
=
1.202gmL − 1
= 83.19mL
= 83.19 × 10 − 3L

20
166
mol
Therefore, molarity of the solution = =
83.19 × 10 − 3L
= 1.45 M

20
(c) Moles of KI = = 166 = 0.12mol
80
Moles of water = = 18 = 4.44mol

Moles of KI
Therefore, mole fraction of KI = Moles of KI + Moles of water
0.12
= 0.12 + 4.44

= 0.0263

Page : 39 , Block Name : Intext Questions

Q2.6 H 2s, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of
H 2S in water at STP is 0.195 m, calculate Henry’s law constant.

Answer. It is given that the solubility of H 2S in water at STP is 0.195 m, i.e., 0.195 mol of H 2S is
dissolved in 1000 g of water.
1000g
Moles of water = =
18gmol − 1
= 55.56 mol
Moles of H 2S
∴ Mole fraction of H 2S , x = = Moles of H S + Moles of water
2
0.195
= 0.195 + 55.56
= 0.0035
At STP, pressure (p) = 0.987 bar
According to Henry's law:
p = K Hx

p
⇒ KH = x
0.987
= bar
0.0035
= 282bar

Page : 43 , Block Name : Intext Questions

Q2.7 Henry’s law constant for CO 2 in water is 1.67 × 10 8 Pa at 298 K. Calculate the quantity of in 500
mL of soda water when packed under 2.5 atm CO 2 pressure at 298 K.

Answer. given that
Pressure of CO 2 = 2.5atm
1 atm = 1.01325 × 10 5 Pa So that
Pressure of CO 2 = 2.5 × 1.01325 × 10 5Pa = 2.533125 × 10 5Pa

K H = 1.67 × 10 8Pa
According to Henry’s law:

p = KH × X
X = p / K H = 2.533125 × 10 5 / 1.67 × 10 8 = 1.52 × 10 − 3
But
We have 500 mL of soda water so that
Volume of water = 500 mL
[Neglecting the amount of soda present]
Density of water = 1 g/ ml
Use formula mass = volume × density we get
500 mL of water = 500 g of water

( )
Molar mass of water H 2O = 18gmol − 1

Use formula
Number of mole of water = 500 / 18 = 27.78 mol of water
Now use the formula of mole fraction
n CO
2
x= n
CO + n H O
2 2

Value of mole fraction is very small so it is negligible as compared to 1
We get
n CO
2
x≈ n
H 2O

1.52 × 10 − 3 = n CO2 / 27.78
After calculation we get
n CO2 = 0.042mol
molar mass of CO 2 = 12 + 2 × 16 = 44gmol − 1
use formula
mass = molar mass \times number of moles
= 1.848g

Page : 43 , Block Name : Intext Questions

Q2.8 The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find
out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also nd the
composition of the vapour phase.

Answer. It is given that:
p 0A = 450 mm of Hg
p 0B = 700 mm of Hg
P total = 600 mm of Hg
From Raoult's law, we have:
P A = P 0A x A

( )
p B = p 0Bx B = p 0B 1 − x A Therefore, total pressure, P total = P A + P B

0 0
(
⇒ P total = P Ax A + P B 1 − x A )
0 0 0
⇒ P total = P Ax A + P B − P Bx A

( )
⇒ P total = P 0A − P 0B x A + P 0B
⇒ 600 = (450 − 700)x A + 700
⇒ − 100 = − 250x A
⇒ x A = 0.4
Therefore, x B = 1 − x A
= 1- 0.4
= 0.6

Now, p A = p 0A x A
= 450 x 0.4
= 180 mm of Hg
p B = p 0Bx B

= 700 × 0.6
= 420mm of Hg

Now, in the vapour phase :
PA
Mole of fraction of liquid A = = P + P
A B
180
= 180 + 420
180
= 600
= 0.30

Page : 49 , Block Name : Intext Questions

Q2.9 Vapour pressure of pure water at 298 K is 23.8 mm Hg 50 g of urea ( NH 2CONH 2 ) is dissolved in
850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

0
Answer. It is given that vapour pressure of water,p 1 = 23.8 mm of Hg
Weight of water taken, w 1 = 850 g
Weight of urea taken, w 2 = 5O g
Molecular weight of water, M 1 = 18 g mol − 1
Molecular weight of urea, M 2 = 60 g mol − 1

Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p 1.
Now, from Raoult's law, we have:

0
p 1 − p1 n2
0 = n +n
p1 1 2

w2
0
p 1 − p1 M2
⇒ 0 = w1 w2
p1
M +M1 2

50
23.8 − p 1 60
⇒ = 850 50
23.8
18
+ 60
23.8 − p 1 0.83
⇒ =
23.8 47.22 + 0.83
23.8 − p 1
⇒ 23.8
= 0.0173
⇒ p 1 = 23.4mm of Hg
Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is
0.0173.

Page : 57 , Block Name : Intext Questions

Q2.10 Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of
water such that it boils at 100°C.

Answer. Here, elevation of boiling point ΔT b = (100 + 273) - (99.63 + 273)
= 0.37 K
Mass of water, w 1 = 500 g

( )
Molar mass of sucrose C 12H 22O 11 , M 2 = 11 x 12 + 22 x 1 + 11 x 16
= 342 g mol − 1
Molal elevation constant, Kb = 0.52 K kg mol-
We know that:
K b × 1000 × w 2
ΔT b = M2 × w1
ΔT b × M 2 × w 1
⇒ w2 = K b × 1000
0.37 × 342 × 500
= 0.52 × 1000
= 121.67 g ( approximately)
Hence, 121.67 g of sucrose s to be added.

Page : 57 , Block Name : Intext Questions

Q2.11 Calculate the mass of ascorbic acid (Vitamin C, C 6H 8O 6) to be dissolved in 75 g of acetic acid to
lower its melting point by 1.5°C. K f = 3.9 K kg mol − 1.

Answer. Mass of acetic acid, w 1 = 75 g

( )
Molar mass of ascorbic acid C 6H 8O 6 , M 2 = 6 × 12 + 8 × 1 + 6 × 16
= 176 g mol − 1
Lowering of melting point, ΔT f = 1.5 K
We know that:
K f × w 2 × 1000
ΔT f =
M2 × wi
ΔT f × M 2 × w 1 1.5 × 176 × 75
⇒ w2 = =
K f × 1000 3.9 × 1000
= 5.08g( approx )
Hence, 5.08 g of ascorbic acid is needed to be dissolved.

Page : 57 , Block Name : Intext Questions

Q2.12 Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of
polymer of molar mass 185,000 in 450 mL of water at 37°C.

Answer. It is given that:
Volume of water, V = 450 mL = 0.45 L
Temperature, T = (37 + 273) K = 310 K

1
Number of moles of the polymer, n = 185000 mol
We know that:
n
Osmotic pressure, π = V RT
1 1
= mol × × 8.314 × 10 3PaLK − 1mol − 1 × 310K
185000 0.45L
= 30.98Pa
= 31pa( approximately )

Page : 57 , Block Name : Intext Questions

Q2.1 De ne the term solution. How many types of solutions are formed? Write brie y about each type
with an example.

Answer. A solution is a homogenous mixture of two or more than two substances whose composition
can change within a certain limits. A solution of two substances is called binary solution.

In binary solution, the component which is present in smaller amount is called solute and other one is
called solvent.
Types of solutions and there example:-
rst component is solute and second one solvent
1.Solid in solid - stainless steel (C and Fe)
2.Liquid in solid - water and sodium carbonate (10H2O+Na2CO3)

3.Gas in solid - H2 in Pd
4.Solid in liquid - salt water
5.Liquid in liquid - alcohol in water
6.Gas in liquid - O2 in water
7.Solid in gas - Camphor in N2 gas
8.Liquid in gas - moisture
9.Gas in gas - O2 and N2 gas in air

Page : 61 , Block Name : Exercise

Q2.2 Give an example of a solid solution in which the solute is a gas.

Answer. Solution of hydrogen and palladium. In this hydrogen which is a gas is solute and palladium
is solvent.

Page : 61 , Block Name : Exercise

Q2.3 De ne the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage

Answer. (i)Mole fraction –Mole fraction is a way to express concentration of a solution.
Mole fraction of a constituent (either of solute or solvent) is the ratio of number of moles of one
component to the total number of moles of all component present in a solution.
Mole fraction is denoted by letter, ‘x’.
(ii) Molality: Molality is a way to express concentration of solution.
Moles of solute
= Mass of solvent in kg
(iii) Molarity:
Moles of solute
= Volume of solution in litre
(iv) Mass percentage
Mass of the component in solution
% of component = Total mass of solution
× 100

Page : 61 , Block Name : Exercise

Q2.4 Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution.
What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL − 1
?

Answer. Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous
solution. This means that 68 g of nitric acid is dissolved in 100 g of the solution.

( )
Molar mass of nitric acid HNO 3 = 1 × 1 + 1 × 14 + 3 × 16 = 63gmol − 1

Then, number of moles of HNO 3 = 68 / 63mol
= 1.08 mol
= 1.504g / mL − 1
Therefore from the formula density = mass / volume,we get
Volume of solution = 1000/1.504 = 66.49 mL
Therefore molarity of nitric acid = (1.08/66.49) x 1000 = 16.24

Page : 61 , Block Name : Exercise

Q2.5 A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole
fraction of each component in the solution? If the density of solution is 1.2 g mL − 1, then what shall be
the molarity of the solution?

Answer. 10% w/w solution of glucose in water means that 10 g of glucose in present in 100 g of the
solution i.e., 10 g of glucose is present in (100 - 10) g = 90 g of water.

( )
Molar mass of glucose C 6H 12O 6 = 6 × 12 + 12 × 1 + 6 × 16 = 180gmol −

1
Then, number of moles of glucose = 10 / 180mol
= 0.056mol
∴ Molality of solution = 0.056mol / 0.09kg = 0.62m
Number of moles of water = 90g / 18g mol 1 = 5mol

( )
Mole fraction of glucose x g = 0.056 / (0.056 + 5) = 0.011

And, mole fraction of water x W = 1 − x g
= 1 − 0.011 = 0.989
If the density of the solution is 1.2gmL − 1 , then the volume of the 100g
solution can be given as:
= 100g / 1.2gmL − 1
= 83.33mL
= 83.33 × 10 − 3L
∴ Molarity of the solution = 0.056mol / 83.33 × 10 − 3L
= 0.67M

Page : 62 , Block Name : Exercise

Q2.6 How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na 2CO 3 and
NaHCO 3 containing equimolar amounts of both?

Answer.

let the amount of Na 2CO 3 be x
& that of NaHCO 3 be 1 − x
Now moles of Na 2CO 3 = × / 106
& moles of NaHCO 3 = 1 − x / 84
Now according to question, number of moles of Na 2Co 3 = number of
moles of NaHCO 3
Therefore × 106 = 1 − x / 84
84x = 106 − 106x
84x + 106x = 106
190x = 106
&
moles of NaHCO 3 = 1 − 0.558 / 84 = 0.0053
Now Hcl reacts with Na 2Co 3&NaHCO 3 as follows:
Na 2Co 3 + 2Hcl ⟶ 2Nacl + H 2o + CO 2
NaHCO 3 + Hcl ⟶ Nacl + H 2o + CO 2

Page : 62 , Block Name : Exercise

Q2.7 A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass.
Calculate the mass percentage of the resulting solution.

Answer. Total amount of solute present in the mixture is given by,
25 40
300 × 100 + 400 × 100
= 75 + 160
= 235g
Total amount of solution = 300 + 400 = 700 g
235
Therefore, mass percentage ( w / w ) of the solute in the resulting solution, = 700 × 100%
= 33.57%
And, mass percentage (w/w) of the solvent in the resulting solution,
= (100 - 33.57)%
= 66.43%

Page : 62 , Block Name : Exercise

Q2.8 An antifreeze solution is prepared from 222.6 g of ethylene glycol ( C 2H 6O 2 ) and 200 g of water.
Calculate the molality of the solution. If the density of the solution is 1.072gmL − 1, then what shall be
the molarity of the solution?

Answer.

Calculation of Molality:
Mass of ethylene glycol = 222.6 (Given)

[
Molar mass of ethylene glycol C 2H 4(OH) 2 ]
= 2 × 12 + 6 × 1 + 2 × 16
= 62
Therefore moles of ethylene glycol
= 222.6g / 62gmol − 1
= 3.59mol
Mass of water = 200g (Given)
Therefore molality of the solution is = (moles of ethylene glycol / mass of water) x 1000
= (3.59 / 200) × 1000
= 17.95m
Calculation of Molarity:
Moles of ethylene glycol = 3.59 mol (already calculated)
Total Mass of solution = 200 + 222.6
= 422.6g
Volume of solution = mass / density volume
= 422.6 / 1.072
= 394.22 ml
now molarity of the solution is = (moles of ethylene glycol / volume of solution) x 1000
= (3.59 / 394.22) x 1000
= 9.11 M

Page : 62 , Block Name : Exercise

Q2.9 A sample of drinking water was found to be severely contaminated with chloroform (CHCl 3)
supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample.

Answer.

( )
1) 15 ppm means : 15 parts per million 10 6 of the solutions

So, Percent by mass = ( mass of chloroform / total mass ) × 100

( )
= 15 / 10 6 × 100

= 1.5 × 10 − 3%

2) Molality Mass of chloroform = 15g

( )
Molar mass of chloroform CHCl 3 = 1 × 12 + 1 × 1 + 3 × 35.5

= 119.5gmol − 1
Moles of chloroform = 15 / 119.5 = 0.1255mol
Mass of water = 10 6

( )
Therefore molality = (moles of chloroform / mass of water ) × 1000 = 0.1255 / 10 6 × 1000 = 1.255 × 10 − 4m
Therefore molality = (moles of chloroform / mass of water ) × 1000

6
= (0.1255/10 ) × 1000

−4
= 1.255 × 10 m

Page : 62 , Block Name : Exercise

Q2.10 What role does the molecular interaction play in a solution of alcohol and water?

Answer. The lower members of alcohols are highly soluble in water but the solubility decreases with
increase in the molecular weight. The solubility of lower alcohols in water is due to formation of
hydrogen bonds(Hydrogen bonding) between alcohols & water molecules.

However, as the size of alcohol molecule increases, the alkyl groups becomes larger & prevents the
formation of hydrogen bonds with water, & hence the solubility goes on decreasing with increase in
the length of carbon chain.Also the interaction between the molecules of alcohol and water is weaker
than alcohol−alcohol and water−water interactions. As a result, when alcohol and water are mixed,
the intermolecular interactions become weaker and the molecules can easily escape. This increases
the vapour pressure of the solution, which in turn lowers the boiling point of the resulting solution.

Page : 62 , Block Name : Exercise

Q2.11 Why do gases always tend to be less soluble in liquids as the temperature is raised?

Answer. The dissolution of a gas in a liquid is exothermic process.Therefore according to Le Chatelier
principle,with the increase in temperature,the equilibrium shifts in the backward direction.
Gas + Liquid --> Solution + Heat
Therefore the solubility of gas in solution decreases with rise in temperature.

Page : 62 , Block Name : Exercise

Q2.12 State Henry’s law and mention some important applications.

Answer. Henry’s law states that the mass of a gas dissolved per unit volume of the solvent at a given
temperature is proportional to the pressure of the gas in equilibrium with the solution.
Or
It also states that the pressure of a gas over a solution in which the gas is dissolved is proportional to
the mole fraction of the gas dissolved in the solution.
The important applications of Henry’s law are as follows:
1) In the production of carbonated beverages-in order to increase the solubility of CO2 in cold
drinks,beer etc,the bottle are sealed under high pressure.When the bottle is opened under normal
atmospheric pressure,the pressure inside the bottle falls to atmospheric pressure & the excess CO2
bubbles out of the bottle causing effervescence.
2) At high altitudes-the partial pressure of oxygen at high altitudes is less than the ground level.this
results in low concentration of oxygen in the blood & tissues of the peoples
3) In scuba diving- during scuba diving,when the diver breaths in compressed air from the supply
tank,more nitrogen dissolves in the blood & other body uids because the pressure at that depth is far
greater than the surface atmospheric pressure.

Page : 62 , Block Name : Exercise

Q2.13 The partial pressure of ethane over a solution containing 6.56 × 10 − 3 g of ethane is 1 bar. If the
solution contains 5.00 × 10 − 2 g of ethane, then what shall be the partial pressure of the gas?

( )
Answer. Molar mass of ethane C 2H 6 = 2 × 12 + 6 × 1
= 30gmol − 1

6.56 × 10 − 2
Number of moles present in 6.56 × 10 − 2g of ethane = 30
= 2.187 x 10 − 3mol
Let the number of moles of the solvent be x.
According to Henry's law,
p = K Hx
2.187 × 10 − 3
⇒ 1 bar = K H ⋅
2.187 × 10 − 3 + x
2.187 × 10 − 3
⇒ 1 bar = K H x
x
⇒ KH = bar
2.187 × 10 − 3

5.00 × 10 − 2
Number of moles present in 5.00 × 10 − 2g of ethane = 30
mol
= = 1.67 × 10 − 3mol
According to Henry's law,
p = K Hx

x 1.67 × 10 − 3
= −3 ×
2.187 × 10
( 1.67 × 10 ) + x
−3

x 1.67 × 10 − 3
= × x
2.187 × 10 − 3
= 0.764 bar
Hence, partial pressure of the gas shall be 0.764 bar.

Page : 62 , Block Name : Exercise

Q2.14 What is meant by positive and negative deviations from Raoult's law and how is the sign of
Δ mixH related to positive and negative deviations from Raoult's law?

Answer. According to Raoult's law, the partial vapour pressure of each volatile component in any
solution is directly proportional to its mole fraction. The solutions which obey Raoult's law over the
entire range of concentration are known as ideal solutions. The solutions that do not obey Raoult's
law (non-ideal solutions) have vapour pressures either higher or lower than that predicted by Raoult's
law. If the vapour pressure is higher, then the solution is said to exhibit positive deviation, and if it is
lower, then the solution is said to exhibit negative deviation from Raoult's law.

Vapour pressure of a two-component solution showing positive deviation from Raoult's law

Vapour pressure of a two—component solution showing negative deviation from
Raoult's law

In the case of an ideal solution, the enthalpy of the mixing of the pure components for forming the
solution is zero.
Δ solH = 0

In the case of solutions showing positive deviations, absorption of heat takes place.
∴ Δ solH = positive
In the case of solutions showing negative deviations, evolution of heat takes place.
∴ Δ solH = negative

Page : 62 , Block Name : Exercise

Q2.15 An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 barat the normal
boiling point of the solvent. What is the molar mass of the solute?

Answer. Here,

( )
Vapour pressure of the solution at normal boiling point p 1 = 1.004bar

( )
0
Vapour pressure of pure water at normal boiling point p 1 = 1.013bar

Mass of solute, w 2 = 2 g ( )
Mass of solvent (water), w 1 = 98 g ( )
( )
Molar mass of solvent (water), M 1 = 18gmol − 1
According to Raoult's law,
0
p 1 − p1 w2 × M1
= M ×w
p 01 2 1

1.013 − 1.004 2 × 18
⇒ 1.013
= M × 98
2
0.009 2 × 18
⇒ 1.013 = M × 98
2
1.013 × 2 × 18
⇒ M2 = 0.009 × 98

= 41.35gmol − 1
Hence, the molar mass of the solute is 41.35 \) \mathrm { g } \mathrm { mol } ^ { - 1 } \).

Page : 62 , Block Name : Exercise

Q2.16 Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid
components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of
26.0 g of heptane and 35 g of octane?

Answer. kPa

( )
Vapour pressure of heptane p 1 = 105.2kPa
0

Vapour pressure of octane (p )
0
2 = 46.8kPa

We know that,

( )
Molar mass of heptane C 7H 16 = 7 × 12 + 16 × 1

= 100gmol − 1

26
∴ Number of moles of heptane = 100 mol
= 0.26 mol

( )
Molar mass of octane C 8H 18 = 8 × 12 + 18 × 1
= 114gmol − 1

35
∴ Number of moles of octane = 114 mol
= 0.31 mol

0.26
Mole fraction of heptane, x 1 = 0.26 + 0.31
= 0.456
And mole fraction of octane, x 2 = 1 − 0.456
= 0.544

Now, partial pressure of heptane, p 1 = x 1p 01
= 0.456 x 105.2
= 47.97 kPa

0
Now, partial pressure of octane, p 2 = x 2p 2
= 0.544 x 46.8
= 25.46 kPa

Hence vapour pressure of solution p total = p 1 + p 2
= 47.97 + 25.46
= 73.43 kPa

Page : 62 , Block Name : Exercise

Q2.17 The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution
of a non-volatile solute in it.

Answer. 1 molal solution means 1 mol of the solute is present in 100 g of the solvent (water).
Molar mass of water = 18gmol − 1
1000
Number of moles present in 1000 g of water = 18
= 55.56 mol

Therefore, mole fraction of the solute in the solution is
1
x 2 = 1 + 55.56 = 0.0177
It is given that,
Vapour pressure of the water, p 01 = 12.3kPa

0
p 1 − p1
Applying the relation, 0 = x2
p1
12.3 − p 1
⇒ = 0.0177
12.3
⇒ 12.3 − p 1 = 0.2177
⇒ p 1 = 12.0823
= 12.08kPa (approximately)

Hence, the vapour pressure of the solution is 12.08 kPa.

Page : 62 , Block Name : Exercise

Q2.18 Calculate the mass of a non-volatile solute (molar mass 40 g mo1 − 1) which should be dissolved
in 114 g octane to reduce its vapour pressure to 80%.

Answer. Let the vapour pressure of pure octane be p 01 .
Then, the vapour pressure of the octane after dissolving the non-volatile solute is 80
80 0
p = 0.8p 01
100 1
Molar mass of solute, M 2 = 40gmol − 1
Mass of octane, w 1 = 114g

(
Molar mass of octane, C 8H 18 , M 1 = 8 × 12 + 18 × 1 )
= 114gmol − 1
Applying the relation,
0
p 1 − p1 w2 × M1
= M ×w
p 01 2 1

0 0
p 1 − 0.8p 1 w 2 × 114
⇒ = 40 × 114
p 01
0
0.2p 1 w2
⇒ = 40
p 01
w2
⇒ 0.2 = 40

⇒ w 2 = 8g
Hence, the required mass of the solute is 8 g.

Page : 62 , Block Name : Exercise

Q2.19 A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure
of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure
becomes 2.9 kPa at 298 K. Calculate:
(i) molar mass of the solute
(ii) vapour pressure of water at 298 K.

Answer. (i) Let, the molar mass of the solute be Mgmol − 1
90g
Now, the no. of moles of solvent (water), n 1 = = 5mol
18gmol − 1
30g 30
And, the no. of moles of solute, n 2 = = M mol
Mmol − 1
p 1 = 2.8kPa
Applying the relation:
p 01 − p 1 n2
0 = n +n
p1 1 2

0 30
p 1 − 2.8 M
⇒ 0 = 30
p1 5+ M
30
2.8 M
⇒1− 0 = 5M + 30
p1
M
2.8 30
⇒1− = 5M + 30
p 01

2.8 30
⇒ = 1 − 5M + 30
p 01

2.8 5M + 30 − 30
⇒ = 5M + 30
p 01
2.8 5M
⇒ = 5M + 30
p 01
p 01 5M + 30
⇒ 2.8 = 5M
(i)
After the addition of 18 g of water:
90 + 18g
n1 = 18
= 6mol
p 1 = 2.9kPa
Again, applying the relation:
0
p 1 − p1 n2
= n +n
p 01 1 2

30
p 01 − 2.9 M
⇒ 0 = 30
p1 6+ M

30
2.9 M
⇒1− 0
= 6M + 30
p1
M
2.9 30
⇒1− 0
=
p1 6M + 30

2.9 30
⇒ 0
=1−
p1 6M + 30
2.9 6M + 30 − 30
⇒ = 6M + 30
p 01

2.9 6M
⇒ = 6M + 30
p 01
0
p1 6M + 30
⇒ 2.9 = 6M (ii)

Dividing equation (i) by (ii), we have :
5M + 30
2.9 5M
2.8
= 6M + 30
6M
2.9 6M + 30 5M + 30
⇒ 2.8 × 6
= 5
⇒ 2.9 × 5 × (6M + 30) = 2.8 × 6 × (5M + 30)
⇒ 87M + 435 = 84M + 504
⇒ 3M = 69
⇒ M = 23u
Therefore, the molar mass of the solute is 23gmol − 1.

(ii) Putting the value of 'M' in equation (i), we have:
p 01 5 × 23 + 30
2.8
= 5 × 23
p 01 145
⇒ 2.8 = 115

⇒ p 01 = 3.53
Hence, the vapour pressure of water at 298 K is 3.53 kPa.

Page : 62 , Block Name : Exercise

Q2.20 A 5% solution (by mass) of cane sugar in water has freezing point of 271K.Calculate the freezing
point of 5% glucose in water if freezing point of pure water is 273.15 K.

Answer. Here, ΔT f = (273.15 - 271) K
= 2.15K

( )
Molar mass of sugar C 12H 22O 11 = 12 × 12 + 22 × 1 + 11 × 16

= 342gmol − 1
5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 - 5)g = 95 g of
water.

5
Now, number of moles of cane sugar = 342 mol
= 0.0146 mol
0.0146mol
Therefore, molality of the solution, m = 0.095kg
= 0.1537molkg − 1

Applying the relation,
ΔT f = K f × m
ΔT f
⇒ Kf =
m
2.15K
=
0.1537molkg − 1
= 13.99Kkgmol − 1

( )
Molar of glucose C 6H 12O 6 = 6 × 12 + 12 × 1 + 6 × 16

= 180gmol − 1
5% glucose in water means 5g of glucose is present in (100 − 5)g = 95g of water.
5
∴ Number of moles of glucose = 180 mol
= 0.0278 mol
0.0278mol
Therefore, molality of the solution, m = 0.095kg
= 0.2926molkg − 1
Applying the relation,
ΔT f = K f × m

= 13.99Kkgmol − 1 × 0.2926molkg − 1
= 4.09K (approximately)
Hence, the freezing point of 5% glucose solution is (273.15 - 4.09) K- 269.06 K.

Page : 62 , Block Name : Exercise

Q2.21 Two elements A and B form compounds having formula AB 2 and AB 4. When dissolved in 20 g of

( )
benzene C 6H 6 , 1 g of AB 2 lowers the freezing point by 2.3 K whereas 1.0 g of AB 4 lowers it by 1.3 K.
The molar depression constant for benzene is 5.1 K kg mol − 1. Calculate atomic masses of A and B.

Answer. We know that,
1000 × w 2 × k f
M2 = ΔT f × w 1
1000 × 1 × 5.1
Then, M AB = 2.3 × 20
2
= 110.87gmol − 1

1000 × 1 × 5.1
M AB = 1.3 × 20
4
= 196.15gmol − 1

Now, we have the molar masses of AB 2 and AB 4 as 110.87gmol − 1 and 196.15gmol − 1 respectively.
Let the atomic masses of A and B be x and y respectively.
Now, we can write:
x + 3y = 110.87 (i)
x + 4y = 196.15 (ii)
Subtracting equation (i) from (ii), we have
2y = 85.28
⇒ y = 42.64
Putting the value of 'y' in equation (i), we have
x +2 x 42.64 = 110.87
⇒ x = 25.59
Hence, the atomic masses of A and 3 are 25.59 u and 42.64 u respectively.

Page : 62 , Block Name : Exercise

Q2.22 At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If
the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its
concentration?

Answer. Here,
T = 300 K
n = 1.52 bar
R = 0.083 bar LK − 1mol − 1
Applying the relation,
n = CRT
π
⇒C=
RT
1.52bar
=
0.083 bar LK − 1mol − 1 × 300K
= 0.061 mol
Since the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.

Page : 63 , Block Name : Exercise

Q2.23 Suggest the most important type of intermolecular attractive interaction in the following pairs.
(i) n-hexane and n-octane
(ii) I 2 and CCl 4
(iii) NaClO 4 and water
(iv) methanol and acetone

( ) {
(v) acetonitrile CH 3CN and acetone C 3H 6O }

Answer. (i) Van der Wall's forces of attraction.
(ii) Van der Wall's forces of attraction.
(iii) Ion-diople interaction.
(iv) Dipole-dipole interaction.
(v) Dipole-dipole interaction.

Page : 63 , Block Name : Exercise

Q2.24 Based on solute-solvent interactions, arrange the following in order of increasing solubility in
n-octane and explain. Cyclohexane, KCl, CH 3OH, CH 3CN.

Answer. n-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than
that of a
polar solute in the n-octane.
The order of increasing polarity is:
Cyclohexane < CH 3CN < CH 3OH < KCL
Therefore, the order of increasing solubility is:
KCl < CH3OH < CH3CN < Cyclohexane

Page : 63 , Block Name : Exercise

Q2.25 Amongst the following compounds, identify which are insoluble, partially soluble and highly
soluble in water?
(i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform
(vi) pentanol.

Answer. (i) Phenol (C 6H 5OH) has the polar group -OH and non-polar group ( − C 6H 5). Thus, phenol is
partially soluble in water.
(ii) Toluene (C 6H 5 − CH 3) has no polar groups. Thus, toluene is insoluble in water.
(iii) Formic acid (HCOOH) has the polar group —OH and can form H-bond with water.
Thus, formic acid is highly soluble in water.
(iv) Ethylene glycol has polar -OH group and can form H—bond. Thus, it is highly soluble in water.
(v) Chloroform is insoluble in water.
(vi) Pentanol (C 5H 11OH) has polar -OH group, but it also contains a very bulky nonpolar (C 5H 11)
group. Thus, pentanol is partially soluble in water.

Page : 63 , Block Name : Exercise

Q2.26 If the density of some lake water is 1.25gmL − 1 and contains 92g of Na + ions per kg of water,
calculate the molality of Na + ions in the lake.

92g
Answer. Number of moles present in 92 g of Na + ions =
23gmol − 1
= 4 mol
4mol
Number of molality of Na + ions in the lake = 1kg
=4m

Page : 63 , Block Name : Exercise

Q2.27 If the solubility product of CuS is 6 × 10 − 16, calculate the maximum molarity of CuS in aqueous
solution.

Answer. Solubility product of CuS, K sp = 6 × 10 − 16
Let S be the solubility of CuS in molL − 1
CuS ↔ Cu 2 + + S 2 −

[
Now, K sp = Cu 2 + ][s ] 2−

=sxs
= s2
Then, we have, K sp = s 2 = 6 × 10 − 16

⇒s= √6 × 10 − 16
= 2.45 × 10 − 8molL − 1
Hence, the maximum molarity of Cus in an aqueous solution is 2.45 x 10 − 8molL − 1.

Page : 63 , Block Name : Exercise

( ) ( )
Q2.28 Calculate the mass percentage of aspirin C 9H 8O 4 in acetonitrile CH 3CN when 6.5 g of
C 2H 8O 4 is dissolved in 450 g of CH 3CN

Answer. 6.5 g of C 9H 8O 4 is dissolved in 450 g of CH 3CN.
Then, total mass of the solution = ( 6.5 + 450 ) g
= 456.5 g

6.5
Therefore, mass percentage of C 9H 8O 4 = = 456.5 × 100%
= 1.424%

Page : 63 , Block Name : Exercise

( )
Q.2.29 Nalorphene C 19H 21NO 3 , similar to morphine, is used to combat withdrawal symptoms in
narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 − 10 − 3 m
aqueous solution required for the above dose.

( )
Answer. The molar mass of Nalorphene C 19H 21NO 3 is given as:

19 × 12 + 21 × 1 + 1 × 14 + 3 × 16 = 311gmol − 1
In 1.5 × 10 − 3m aqueous solution of nalorphene,
1kg(1000g) of water contains 1.5 × 10 − 3mol = 1.5 × 10 − 3 × 311g
= 0.4665g
Therefore total mass of the solution = (1000 + 0.4665) g
= 1000.4665 g

This implies that the mass of the solution containing 0.4665 g of nalorphene is 1000.4665 g.
Therefore, mass of the solution containing 1.5 mg of nalorphene is:
1000.4665 × 1.5 × 10 − 3
0.4665
g
= 3.22 g
Hence, the mass of aqueous solution required is 3.22 g.
Note: There is a slight variation in this answer and the one given in the NCERT textbook.

Page : 63 , Block Name : Exercise

Q2.30 Calculate the amount of benzoic acid (C 6H 5COOH) required for preparing 250 mL of 0.15 M
solution in methanol.

Answer. 0.15 M solution of benzoic acid in methanol means, 1000 mL of solution contains 0.15 mol of
benzoic acid
0.15 × 250
Therefore, 250 mL of solution contains = 1000 mol of benzoic acid
= 0.0375 mol of benzoic acid

( )
Molar mass of benzoic acid C 6H 5COOH = 7 x 12 + 6 x 1 + 2 x 16
= 122 g mol − 1
Hence, required benzoic acid = 0.0375 mol x 122gmol − 1
= 4.575 g

Page : 63 , Block Name : Exercise

Q2.31 The depression in freezing point of water observed for the same amount of acetic acid,
trichloroacetic acid and tri uoroacetic acid increases in the order given above. Explain brie y.

Answer.

Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can withdraw
electrons towards itself more than Cl and H. Thus, tri uoroacetic acid can easily H + ions i.e.,
tri uoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater is the
depression of the freezing point. Hence, the depression in the freezing point increases in the order:

Acetic acid < trichloroacetic acid < tri uoroacetic acid

Page : 63 , Block Name : Exercise

Q2.32 Calculate the depression in the freezing point of water when 10 g of CH 3CH 2CHClCOOH is
added to 250 g of water. K a = 1.4 × 10 − 3, K f = 1.86 Kkgmol − 1.

Answer. Molar mass of CH 3CH 2CHClCOOH = 15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1
= 122.5 g mol − 1

10g
∴ No . of moles present in 10g of CH 3CH 2CHCICOOH =
122.5gmol − 1
= 0.0816 mol

It is given that 10 g of CH 3CH 2CHClCOOH is added to 250 g of water.
0.0186
∴ Molality of the solution, = 250 × 1000
= 0.3264 mol molkg − 1

Let α be the degree of dissociation of CH 3CH 2CHClCOOH
CH 3CH 2CHClCOOH undergoes dissociation according to the following equation:

Cα ⋅ Cα
∴ Kα = C ( 1 − α )
Cα 2
= 1−α
Since a is very small with respect to 1, 1 − α ≈ 1
Cα 2
Now, K α = 1

⇒ K α = Cα 2
Kα
⇒α=
√ C

1.4 × 10 − 3
=
√ 0.3264 ( ∵ K = 1.4 × 10 )
α
−3

= 0.0655
Again,

Total moles of equilibrium = 1 − α + α + α
=1+α
1+α
∴i= 1
=1+α
= 1 + 0.0655
= 1.0655
Hence, the depression in the freezing point of water is given as:
ΔT f = i. K fm
= 1.0655 × 1.86Kkgmol − 1 × 0.3264molkg − 1
= 0.65K

Page : 63 , Block Name : Exercise

Q2.33 19.5 g of CH 2FCOOH is dissolved in 500 g of water. The depression in the freezing point of
water observed is 1.0 ∘ C. Calculate the van’t Hoff factor and dissociation constant of uoroacetic acid.

Answer. It is given that:
w 1 = 500 g
w 2 = 19.5g
K f = 1.86Kkgmol − 1
ΔT f = 1K
We know that:
K f × w 2 × 1000
M2 = ΔT f × w 1

1.86Kkgmol − 1 × 19.5g × 1000gkg − 1
= 500g × 1K

= 72.54gmol − 1

Therefore, observed molar mass of CH 2FCOOH - 72.54 g mol
The calculated molar mass of CH 2FCOOH is:

(M2 )cal = 14 + 19 + 12 + 16 + 16 + 1
= 78 g mol − 1

( M2 ) cal
Therefore, van’t hoff factor, i =
( M2 ) obs

78gmol − 1
=
= 72.54gmol − 1
= 1.0753

Let α be the degree of dissociation of CH 2FCOOH

C(1 + α)
∴i=
C
⇒i=1+α
⇒α=i−1
= 1.0753 − 1
= 0.0753
Now, the value of K α is given as:

[ CH FCOO ] [ H ]
2
− +

Kα =
[ CH2FCOOH ]
Cα ⋅ Cα
= C(1−α)
Cα 2
= 1−α
Taking the volume of the solution as 500 mL, we have the concentration:
19.5
78
C= × 1000M
500
= 0.5M

Cα 2
Therefore, K α = 1 − α
0.5 × ( 0.0753 ) 2
= 1 − 0.0753
0.5 × 0.00567
= 0.9247

= 0.00307 (approximately)
= 3.07 × 10 − 3

Page : 63 , Block Name : Exercise

Q2.34 Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at
293 K when 25 g of glucose is dissolved in 450 g of water.

Answer. Vapour pressure of water, p 01 = 17.535 mm of Hg
Mass of glucose, w 2 = 25 g
Mass Of Water, w 1 = 450 g

We know that,

( )
Molar mass of glucose C 6H 12O 6 , M , = 6 x 12 + 12 x 1+6 x 16
= 180gmol − 1
Molar mass of water, M 1 = 18gmol − 1

25
Then, number of moles of glucose, n 2 =
180gmol − 1
= 0.139 mol

450g
And, number of moles of water, n 1 =
18gmol − 1
= 25 mol
We know that,
0
p 1 − p1 n1
= n +n
p 01 2 1

17.535 − p 1 0.139
⇒ 17.535
= 0.139 + 25
0.139 × 17.535
⇒ 17.535 − p 1 = 25.139
⇒ 17.535 − p 1 = 0.097
⇒ p 1 = 17.44mm of Hg

Hence, the vapour pressure of water is 17.44 mm of Hg.

Page : 63 , Block Name : Exercise

Q2.35 Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 10 5mm Hg.
Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Answer. p = 760 mm Hg
k H = = 4.27 x 10 3 mm Hg

According to Henry's law,
p = k Hx
p
⇒x=
kH
760mmHg
=
4.27 × 10 5mmHg
= 177.99 × 10 − 5
= 178 × 10 − 5( approximately )

Hence, the mole fraction of methane in benzene is 178 x 10 − 5.

Page : 63 , Block Name : Exercise

Q2.36 100 g of liquid A (molar mass 140gmol − 1 ) was dissolved in 1000 g of liquid B (molar mass
180gmol − 1 ). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour
pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the
solution is 475 Torr.

100
Answer. Number of moles of liquid A, n A = 140 mol
= 0.714 mol
1000
Number of moles of liquid B, n B = 180 mol
= 5.556 mol

nA
Then, mole fraction of A, x A = n + n
A B
0.714
= 0.714 + 5.556
= 0.114
And, mole fraction of B, x B = 1 − 0.114
= 0.886

Vapour pressure of pure liquid B, p 0B = 500 torr
Therefore, vapour pressure of liquid B in the solution,
0
p B = p Bx B
= 500 x 0.886
= 443 torr
Total vapour pressure of the solution, p total = 475 torr
Vapour pressure of liquid A in the solution,

p A = p total − p B
= 475 − 443
= 32 torr
Now,
0
p A = p Ax A

A p
⇒ p A∘ =
xA
32
=
0.114
= 280.7 torr
Hence, the vapour pressure of pure liquid A is 280.7 torr.

Page : 63 , Block Name : Exercise

Q2.37 Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg
respectively. Assuming that they form ideal solution over the entire range of composition, plot
P total , P chloroform , and P acetone as a function of x acetone . The experimental data observed for
different compositions of mixture is:

Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative
deviation from the ideal solution.

Answer. From the question we have the following data

It can be observed from the graph that the plot for the p total of the solution curves downwards.
Therefore, the solution shows negative deviation from the ideal behaviour.

Page : 64 , Block Name : Exercise

Q2.38 Benzene and toluene form ideal solution over the entire range of composition. The vapour
pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively.
Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of
toluene.

Answer.

( )
Molar mass of benzene C 6H 6 = 6 × 12 + 6 × 1

= 78gmol − 1

( )
Molar mass of toluene C 6H 5CH 3 = 7 × 12 + 8 × 1

= 92gmol − 1
80
Now, no. of moles present in 80 g of benzene = = 78 mol = 1.026mol
100
And, no. of moles present in 100 g of benzene = = 92 mol = 1.087mol
1.026
∴ Mole fraction of benzene, x b = = 0.486
1.026 + 1.087
And, mole fraction of toluene, x t = 1 − 0.486 = 0.514

∘
It is given that vapour pressure of pure benzene, p b = 50.71mmHg
∘
And, vapour pressure of pure toluene, p t = 32.06mmHg
Therefore, partial vapour pressure of benzene, p b = x b × p b
= 0.486 × 50.71
= 24.645mmHg
And, partial vapour pressure of toluene, p t = x t × p t
= 0.514 × 32.06
= 16.479mmHg
Hence, mole fraction of benzene in vapour phase is given by:
pb
pb + pt
24.645
= 24.645 + 16.479
24.645
= 41.124

= 0.599
= 0.6

Page : 64 , Block Name : Exercise

Q2.39 The air is a mixture of a number of gases. The major components are oxygen and nitrogen with
approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a
pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are
3.30 × 10 7 mm and 6.51 × 10 7 mm respectively, calculate the composition of these gases in water.

Answer.

( )
Percentage of oxygen O 2 in air = 20%

Percentage of nitrogen (N 2 ) in air = 79%

Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is,
(10 × 760)mmHg = 7600mmHg
Therefore,
20
Partial pressure of oxygen, p O = 100 × 7600mmHg
2
= 1520 mm Hg
79
Partial pressure of nitrogen, p N = 100 × 7600mmHg
2
= 6004 mm Hg

Now, according to Henry's law:
p = KH ⋅ x
For oxygen:
pO = KH ⋅ xO
2 2
pO
2
⇒ xO = K
2 H

1520mmHg
= (Given K H = 3.30 × 10 7mmHg)
3.30 × 10 7mmHg
= 4.61 × 10 − 5

For nitrogen:
pN = KH ⋅ xN
2 2

pN 6004mmHg
2
⇒ xN = =
2 KH 6.51 × 10 7mmHg
= 9.22 × 10 − 5
Hence, the mole fractions of oxygen and nitrogen in water are 4.61 × 10 − 5 and 9.22 × 10 − 5 respectively.

Page : 64 , Block Name : Exercise

Q2.40 Determine the amount of CaCl 2(i = 2.47) dissolved in 2.5 litre of water such that its osmotic
pressure is 0.75 atm at 27° C.

Answer.
We know that:

n
π = i V RT
W
⇒ π = i MV RT
πMV
⇒ w = iRT
π = 0.75atm
V = 2.5L
i = 2.47
T = (27 + 273)K = 300K
Here,
R = 0.0821L atm K − 1mol − 1
M = 1 × 40 + 2 × 35.5
= 111gmol − 1
0.75 × 11 × 2.5
Therefore, w = 2.47 × 0.0821 × 300

= 3.42g
Hence, the required amount of CaCl 2 is 3.42g

Page : 64 , Block Name : Exercise

Q2.41 Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K 2SO 4 in 2 litre
of water at 25° C, assuming that it is completely dissociated.

Answer.
2−
When K 2SO 4 is dissolved in water, K + and SO 4 ions are produced.
2−
K 2SO 4 ⟶ 2K + + SO 4

Total number of ions produced = 3
∴i=3
Given,
w = 25mg = 0.025g
V = 2L
T = 25 ∘ C = (25 + 273)K = 298K
Also, we know that:

R = 0.0821L atm K − 1mol − 1
M = (2 × 39) + (1 × 32) + (4 × 16) = 174gmol − 1

Applying the following relation,

n
π = i RT
v
w1
=i RT
Mv
0.025 1
=3× × × 0.0821 × 298
174 2
= 5.27 × 10 − 3atm

Page : 64 , Block Name : Exercise