Chapter 2: Solutions Class 12 Chemistry NCERT Solutions

Class 12 Chemistry – Chapter 2: Solutions (Short Questions and Answers)

Question 2.1: Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Answer: Total mass of solution = 22 g + 122 g = 144 g
Mass % of benzene = (22 / 144) × 100 = 15.28%
Mass % of CCl₄ = (122 / 144) × 100 = 84.72%

Question 2.2: Calculate the mole fraction of benzene in a solution containing 30% by mass in carbon tetrachloride.

Answer: Given: 30 g of benzene, 70 g of CCl₄
Molar mass of benzene = 78 g/mol
Molar mass of CCl₄ = 154 g/mol
Moles of benzene = 30 / 78 = 0.385 mol
Moles of CCl₄ = 70 / 154 = 0.454 mol
Total moles = 0.385 + 0.454 = 0.839 mol
Mole fraction of benzene = 0.385 / 0.839 = 0.459

Question 2.3: Calculate the molarity of the following solutions:

Answer: (a) 30 g of Co(NO₃)₂·6H₂O in 4.3 L of solution:
Molar mass = 291 g/mol
Moles = 30 / 291 = 0.103 mol
Molarity = 0.103 / 4.3 = 0.0239 M

(b) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL:
Using M₁V₁ = M₂V₂
M₂ = (0.5 × 30) / 500 = 0.03 M

Question 2.4: Calculate the mass of urea (NH₂CONH₂) required to prepare 2.5 kg of 0.25 molal aqueous solution.

Answer: Molality (m) = 0.25 mol/kg
Molar mass of urea = 60 g/mol
Mass of urea for 1 kg solvent = 0.25 × 60 = 15 g
Total mass of solution = 1000 + 15 = 1015 g
For 2.5 kg solution:
Mass of urea = (15 / 1.015) × 2.5 = 37 g (approximately)

Question 2.5: A 20% (mass/mass) aqueous KI solution has a density of 1.202 g/mL. Calculate: (a) Molality (b) Molarity (c) Mole fraction of KI

Answer: Mass of KI = 20 g
Mass of water = 80 g = 0.08 kg
Molar mass of KI = 166 g/mol

(a) Molality = (20 / 166) / 0.08 = 1.51 mol/kg
(b) Molarity = (20 / 166) / (100/1.202 × 1/1000) = 0.1205 / 0.0832 = 1.45 M
(c) Moles of water = 80 / 18 = 4.44 mol
Mole fraction of KI = 0.1205 / (0.1205 + 4.44) = 0.0264

Question 2.6: If the solubility of H₂S in water at STP is 0.195 m, calculate Henry’s law constant.

Answer: At STP, p = 1 atm = 101325 Pa
Mole fraction, x = 0.195 / (0.195 + 55.5) ≈ 0.0035
Henry’s law constant, KH = p / x = 101325 / 0.0035 = 2.90 × 10⁷ Pa

Question 2.7: Henry’s law constant for CO₂ in water at 298 K is 1.67 × 10⁸ Pa. Calculate the amount of CO₂ dissolved in 500 mL soda water packed under 2.5 atm CO₂ pressure at 298 K.

Answer: p = 2.5 atm = 253312.5 Pa
x = p / KH = 253312.5 / (1.67 × 10⁸) = 1.516 × 10⁻³
Moles of water = 500 / 18 = 27.78 mol
Moles of CO₂ = 1.516 × 10⁻³ × 27.78 = 0.0421 mol
Mass of CO₂ = 0.0421 × 44 = 1.85 g

Question 2.8: Vapour pressures of pure liquids A and B at 350 K are 450 mmHg and 700 mmHg, respectively. If total vapour pressure of mixture is 600 mmHg, find the liquid and vapour phase compositions.

Answer: Let mole fraction of A = x
450x + 700(1−x) = 600
Solving, x = 0.4
Mole fraction of A in liquid = 0.4, B = 0.6
In vapour phase:
pA = 0.4 × 450 = 180 mmHg
pB = 0.6 × 700 = 420 mmHg
Mole fraction of A in vapour = 180/600 = 0.3
Mole fraction of B in vapour = 420/600 = 0.7

Question 2.9: The vapour pressure of pure water at 298 K is 23.8 mmHg. 50 g of urea is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and relative lowering.

Answer: Moles of urea = 50 / 60 = 0.833 mol
Moles of water = 850 / 18 = 47.22 mol
Mole fraction of water = 47.22 / (47.22 + 0.833) = 0.9827
Vapour pressure = 0.9827 × 23.8 = 23.38 mmHg
Relative lowering = (23.8 − 23.38) / 23.8 = 0.0176 or 1.76%

Question 2.10: Boiling point of water at 750 mmHg is 99.63°C. How much sucrose is added to 500 g of water so that it boils at 100°C? (Kb for water = 0.52 K kg/mol)

Answer: ΔTb = 0.37°C
Molality = 0.37 / 0.52 = 0.711 mol/kg
Moles of sucrose = 0.711 × 0.5 = 0.3555 mol
Mass = 0.3555 × 342 = 121.5 g

Question 2.11: Calculate the mass of ascorbic acid (C₆H₈O₆) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. (Kf for CH₃COOH = 3.9 K kg/mol)

Answer: m = 1.5 / 3.9 = 0.3846 mol/kg
Moles of solute = 0.3846 × 0.075 = 0.0288 mol
Mass = 0.0288 × 176 = 5.07 g

Question 2.12: Calculate the osmotic pressure exerted by a solution made by dissolving 1.0 g of polymer (molar mass = 185,000 g/mol) in 450 mL water at 37°C.

Answer: Moles of solute = 1 / 185000 = 5.41 × 10⁻⁶ mol
Volume = 0.45 L
T = 310 K
Using: Π = (n/V)RT
Π = (5.41 × 10⁻⁶ / 0.45) × 0.0821 × 310 = 3.06 × 10⁻⁴ atm
In Pascals: 3.06 × 10⁻⁴ × 101325 = 31 Pa

NCERT Class 12 Chemistry Chapter 2 – Solutions: Exercise Questions with Answers

Question 2.1: Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Answer: A solution is a homogeneous mixture of two or more chemically non-reacting substances. Based on the physical states of solute and solvent, solutions are classified into nine types:

Type	Solute in Solvent	Example
Gaseous Solution	Gas in Gas	Air (O₂ and N₂ mixture)
	Liquid in Gas	Water vapour in air
	Solid in Gas	Camphor vapour in nitrogen, smoke
Liquid Solution	Gas in Liquid	CO₂ in water (aerated drinks)
	Liquid in Liquid	Ethanol in water
	Solid in Liquid	Sugar in water, saline water
Solid Solution	Gas in Solid	Hydrogen in palladium
	Liquid in Solid	Amalgam (e.g., Na-Hg)
	Solid in Solid	Alloys like gold with copper or silver

Question 2.2: Suppose a solid solution is formed between two substances, one whose particles are very large and the other whose particles are very small. What type of solid solution is this likely to be?

Answer: Such a solution is likely to be an interstitial solid solution, where smaller particles fit into the interstitial spaces between the larger atoms of the host crystal structure.

Question 2.3: Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage

Answer:
(i) Mole Fraction – Ratio of the number of moles of a component to the total moles of all components.
(ii) Molality (m) – Number of moles of solute present in 1 kg of solvent.
(iii) Molarity (M) – Number of moles of solute present in 1 litre of solution.
(iv) Mass Percentage – Mass of solute in 100 g of solution.

Question 2.4: Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of acid if the density of the solution is 1.504 g mL⁻¹?

Answer: Mass of solution = 1504 g
Mass of HNO₃ = 68% of 1504 = 1022.72 g
Molar mass of HNO₃ = 63 g/mol
Moles of HNO₃ = 1022.72 / 63 = 16.23 mol
Molarity = 16.23 M

Question 2.5: 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: Mass of glucose = 10 g
Mass of water = 90 g = 0.09 kg
Moles of glucose = 10 / 180 = 0.0556 mol
Molality = 0.0556 / 0.09 = 0.617 mol/kg
Moles of water = 90 / 18 = 5 mol
Mole fraction of glucose = 0.0556 / (0.0556 + 5) ≈ 0.011
Density = 1.2 g/mL ⇒ Volume = 100 / 1.2 = 83.33 mL = 0.0833 L
Molarity = 0.0556 / 0.0833 = 0.667 M

Question 2.6: How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both?

Answer: Moles of Na₂CO₃ = 0.00472 mol
Moles of NaHCO₃ = 0.00595 mol
Total HCl required = (2 × 0.00472) + 0.00595 = 0.01539 mol
Volume = 0.01539 / 0.1 = 0.1539 L = 153.9 mL

Question 2.7: Calculate the percentage composition in terms of mass of a solution obtained by mixing 300 g of a 25% and 400 g of a 40% solution by mass.

Answer: Total solute = 75 + 160 = 235 g
Total mass = 700 g
% Composition = (235 / 700) × 100 = 33.57%

Question 2.8: An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL⁻¹, then what shall be the molarity of the solution?

Answer: Molality = 3.59 / 0.2 = 17.95 mol/kg
Volume = 422.6 / 1.072 ≈ 394 mL = 0.394 L
Molarity = 3.59 / 0.394 = 9.11 M

Question 2.9: A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃), supposed to be a carcinogen. The level of contamination was 15 ppm (by mass).

Answer: (i) % by mass = 0.0015%
(ii) Molality = 1.255 × 10⁻⁴ mol/kg

Question 2.10: What role does the molecular interaction play in solution of alcohol in water?

Answer: Alcohol and water form hydrogen bonds, but new interactions are weaker than original, leading to positive deviation from Raoult’s law and lowering boiling point.

Question 2.11: Why do gases always tend to be less soluble in liquids as the temperature is raised?

Answer: Gas solubility decreases with temperature because dissolution is exothermic, and heating favours gas phase (Le Chatelier’s principle).

Question 2.12: State Henry’s law and mention some of its important applications.

Answer: Henry’s law states that the solubility of a gas in a liquid is proportional to its pressure. Applications: Carbonated drinks, scuba diving, oxygen transport in blood.

Question 2.13: 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, what shall be the partial pressure of the gas?

Answer: Ratio = 5.00 × 10⁻² / 6.56 × 10⁻³ ≈ 7.62
New pressure = 1 × 7.62 = 7.62 bar

Question 2.14: According to Raoult’s law, what is meant by positive and negative deviations and how is the sign of ∆solH related to positive and negative deviations?

Answer: Positive deviation (∆solH > 0): Weaker A–B interactions. Negative deviation (∆solH < 0): Stronger A–B interactions.

Question 2.15: An aqueous solution of 2 percent non-volatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molecular mass of the solute?

Answer: Using Raoult’s law: Molecular mass ≈ 41.35 g/mol

Question 2.16: 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 35.0 g of octane?

Answer: Total vapour pressure = 73.56 kPa

Question 2.17: 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: Vapour pressure of solution = 12.083 kPa