NCERT Solutions Class 12 Chemistry Chapter 1 - Solutions

Page No.5 - Intext Questions

Q1.1. Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Ans: 

= 22/(22+122)x x 100%
= 15.28%

= 84.72%
Alternatively,
Mass percentage of CCl₄ = (100 − 15.28)%
= 84.72%

Q1.2: Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride. 
Ans: Benzene

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 (C₆H₆) = (6 × 12 + 6 × 1) g mol⁻¹ = 78 g mol⁻¹
∴ Number of moles of C₆H₆ = 30 /78 mol = 0.3846 mol
Molar mass of carbon tetrachloride (CCl₄) = 1 × 12 + 4 × 355 = 154 g mol⁻¹
∴ Number of moles of CCl₄ = 70/154 mol = 0.4545 mol
Thus, the mole fraction of C₆H₆ is given as:
= 0.458

Q1.3: Calculate the molarity of each of the following solutions:
(a) 30 g of Co(NO₃)₂. 6H₂O in 4.3 L of solution
(b) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL.
Ans: Molarity is given by:
Molarity = Moles of solute/Volume of solution in litre
(a) Molar mass of Co (NO₃)₂.6H₂O = 59 + 2 (14 + 3 × 16) + 6 × 18 = 291 g mol⁻¹
∴ Moles of Co (NO₃)₂.6H₂O = 30 / 291 mol = 0.103 mol
Therefore, molarity = 0.103 mol/4.3 L = 0.023 M
(b) Number of moles present in 1000 mL of 0.5 M H₂SO₄ = 0.5 mol
∴ Number of moles present in 30 mL of 0.5 M H₂SO₄ = 0.5 × 30/1000 mol = 0.015 mol
Therefore, molarity = 0.015/0.5 L mol = 0.03 M

Q1.4: Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.
Ans: 0.25 Molal aqueous solution to urea means that
moles of urea = 0.25 mole
mass of solvent (NH2CONH2) = 60 g mol-1
∴ 0.25 mole of urea = 0.25 x 60=15g
Mass of solution = 1000+15 = 1015g = 1.015 kg
1.015 kg of urea solution contains 15g of urea
∴ 2.5 kg of solution contains urea =15/1.015 x 2.5 = 37 g

Q1.5: 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.
Ans: 
(a) Molar mass of KI = 39 + 127 = 166 g mol⁻¹
20% (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 − 20) g of water = 80 g of water
Therefore,
Molality of the solution

= 1.506 m
= 1.51 m (approximately)
(b) It is given that the density of the solution = 1.202 g mL⁻¹
∴ Volume of 100 g solution = Mass / Density
 
= 83.19 mL
= 83.19 × 10⁻³ L
Therefore, molarity of the solution
 
= 1.45 M

(c) Moles of Kl = 20/166 = 0.12 mol
Moles of water = 80/18 = 4.44
Therefore, mole fraction of Kl


= 0.0263

Page No.9 - Intext Questions

Q1.6: H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry’s law constant.
Ans: H₂SIt is given that the solubility of H₂S in water at STP is 0.195 m, i.e., 0.195 mol of H₂S is dissolved in 1000 g of water.

∴ Mole fraction of H₂S,

At STP, pressure (p) = 0.987 bar
According to Henry’s law:
p = K_Hx
= 0.987 / 0.0035 Bar
= 282 bar

Q1.7: Henry’s law constant for CO₂ in water is 1.67 × 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm CO₂ pressure at 298 K.
Ans: It is given that: K_H = 1.67 × 10⁸ Pa
p_co2 = 2.5 atm = 2.5 × 1.01325 × 10⁵ Pa
= 2.533125 × 10⁵ Pa
According to Henry’s law:
 

= 0.00152
We can write,

[Since,  is negligible as compared to ]
In 500 mL of soda water, the volume of water = 500 mL
[Neglecting the amount of soda present]
We can write:
500 mL of water = 500 g of water

= 27.78 mol of water
Now,

Hence, quantity of CO₂ in 500 mL of soda water = (0.042 × 44)g
= 1.848 g

Page No.15 - Intext Questions

Q1.8: 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 find the composition of the vapour phase.
Ans: It is given that:

from Rault's law, we have:

Therefore, total pressure

⇒ 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,

= 450 × 0.4
= 180 mm of Hg

= 700 × 0.6
= 420 mm of Hg
Now, in the vapour phase:
Mole fraction of liquid

And, mole fraction of liquid B = 1 − 0.30 = 0.70

Page No.23 - Intext Questions

Q1.9: Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering. 
Ans: It is given that vapour pressure of water,  p⁰1 = 23.8 mm of Hg
Weight of water taken, w₁ = 850 g
Weight of urea taken, w₂ = 50 g
Molecular weight of water, M₁ = 18 g mol⁻¹
Molecular weight of urea, M₂ = 60 g mol⁻¹
Now, we have to calculate vapour pressure of water in the solution. We take vapour
pressure as p₁.
Now, from Raoult’s law, we have:

Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173.

Q1.10: 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. Molal elevation constant for water is 0.52 K kg mol⁻¹.
Ans: Here, elevation of boiling point
ΔT_b = (100 273) − (99.63 273)
= 0.37 K
Mass of water, w_l = 500 g
Molar mass of sucrose (C₁₂H₂₂O₁₁), M₂ = 11 × 12 + 22 × 1 + 11 × 16
= 342 g mol⁻¹
Molal elevation constant, K_b = 0.52 K kg mol⁻¹
We know that:

= 121.67 g (approximately)
Hence, 121.67 g of sucrose is to be added.
Note: There is a slight variation in this answer and the one given in the NCERT textbook.

Q1.11: Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) 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.
Ans: Mass of acetic acid, w₁ = 75 g
Molar mass of ascorbic acid (C₆H₈O₆), M₂ = 6 × 12 + 8 × 1 + 6 × 16
= 176 g mol⁻¹
Lowering of melting point, ΔT_f = 1.5 K
We know that:

Hence, 5.08 g of ascorbic acid is needed to be dissolved.

Q1.12: 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.
Ans: It is given that:
Volume of water, V = 450 mL = 0.45 L
Temperature, T = (37 + 273)K = 310 K
Number of moles of the polymer, n = 1/18500 mol
We know that:

= 30.98 Pa
= 31 Pa(approximately)

Exercises

Q1.1: Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Ans: Homogeneous mixtures of two or more than two components are known as solutions.
There are three types of solutions.
(i) Gaseous solution: The solution in which the solvent is a gas is called a gaseous solution. In these solutions, the solute may be liquid, solid, or gas.
Example: a mixture of oxygen and nitrogen gas is a gaseous solution.
(ii) Liquid solution: The solution in which the solvent is a liquid is known as a liquid solution. The solute in these solutions may be gas, liquid, or solid.
Example: a solution of ethanol in water is a liquid solution.
(iii) Solid solution: The solution in which the solvent is a solid is known as a solid solution. The solute may be gas, liquid or solid. For example, a solution of copper in gold is a solid solution.

Q1.2: Give an example of solid solution in which the solute is a gas.
Ans: In case a solid solution is formed between two substances (one having very large particles and the other having very small particles), an interstitial solid solution will be formed.
Example: a solution of hydrogen in palladium is a solid solution in which the solute is a gas.