Class 12 Chemistry Case Study Worksheet Chapter 1 Solutions | Detailed Solutions & Marking Scheme
VEDANT WORKSHEET SERIES
Class: XII
Subject: Chemistry
Chapter: Solutions
Time: 1 Hour
Maximum Marks: 20
Section A: Case Study Based Questions (5 × 4 = 20 Marks)
Q1.
A solution is defined as a homogeneous mixture of two or more components. The component present in larger quantity is called solvent and the other is solute. Solutions can exist in solid, liquid or gaseous states. For example, alloys are solid solutions, while air is a gaseous solution. The concentration of a solution can be expressed in different ways such as molarity, molality, mole fraction, etc.
(a) What type of solution is an alloy? (1)
(b) Define molality. (1)
(c) Calculate the mole fraction of solute when 2 moles of solute are dissolved in 8 moles of solvent. (2)
Q2.
The solubility of gases in liquids depends upon pressure and temperature. According to Henry’s Law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. This principle is widely used in carbonated beverages. At higher temperatures, gases tend to escape from solutions.
(a) State Henry’s Law. (1)
(b) What happens to gas solubility with increase in temperature? (1)
(c) A gas has Henry’s constant kH = 1.25 × 10^5 Pa. Calculate its solubility when pressure is 2.5 × 10^5 Pa. (2)
Q3.
Raoult’s Law states that the partial vapour pressure of each component of an ideal solution is directly proportional to its mole fraction. It is applicable to ideal solutions where intermolecular interactions between components are similar. Non-ideal solutions show deviation from Raoult’s law.
(a) State Raoult’s Law. (1)
(b) What type of deviation occurs when A–B interactions are weaker than A–A and B–B? (1)
(c) Calculate vapour pressure of solution if mole fraction of solvent is 0.8 and vapour pressure of pure solvent is 100 mm Hg. (2)
Q4.
Colligative properties depend only on the number of solute particles, not their nature. These include relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. These properties help in determining molar mass of solutes.
(a) Name any two colligative properties. (1)
(b) Define osmotic pressure. (1)
(c) Calculate elevation in boiling point if Kb = 0.52 K kg mol^-1 and molality = 0.5 mol kg^-1. (2)
Q5.
Sometimes solutes undergo association or dissociation in solution, leading to abnormal molar mass. This is explained using Van’t Hoff factor (i), which is the ratio of actual number of particles to expected number. For example, NaCl dissociates into Na⁺ and Cl⁻ ions in solution.
(a) Define Van’t Hoff factor. (1)
(b) What is the value of i for complete dissociation of NaCl? (1)
(c) Calculate the Van’t Hoff factor if observed molar mass is half of normal molar mass. (2)
VEDANT WORKSHEET SERIES - ACADEMIC YEAR 2026-27
Class: XII | Subject: Chemistry | Chapter: Solutions
DETAILED ANSWER KEY WITH MARKING SCHEME
Section A: Case Study Based Questions (5 × 4 = 20 Marks)
Q1. Solution
(a)
An alloy is a solid solution.
Marks: 1
(b)
Molality (m) is defined as the number of moles of solute dissolved in 1 kg of solvent.
m = moles of solute / mass of solvent in kg
Marks: 1
(c)
Given:
Moles of solute = 2
Moles of solvent = 8
Mole fraction of solute (Xsolute) = moles of solute / total moles
Xsolute = 2 / (2 + 8) = 2 / 10 = 0.2
Final Answer: 0.2
Marks: 2 (1 for formula, 1 for calculation)
Q2. Solution
(a)
Henry’s Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
p = kH × x
Marks: 1
(b)
With increase in temperature, solubility of gas decreases.
Marks: 1
(c)
Given:
kH = 1.25 × 10^5 Pa
p = 2.5 × 10^5 Pa
Using formula:
p = kH × x
x = p / kH
x = (2.5 × 10^5) / (1.25 × 10^5) = 2
Final Answer: x = 2 (theoretical value; indicates high solubility)
Marks: 2 (1 for formula, 1 for calculation)
Q3. Solution
(a)
Raoult’s Law states that the partial vapour pressure of a component in a solution is equal to the product of its mole fraction and vapour pressure in pure state.
p = x × p⁰
Marks: 1
(b)
If A–B interactions are weaker → Positive deviation from Raoult’s Law.
Marks: 1
(c)
Given:
x = 0.8
p⁰ = 100 mm Hg
p = x × p⁰
p = 0.8 × 100 = 80 mm Hg
Final Answer: 80 mm Hg
Marks: 2 (1 for formula, 1 for calculation)
Q4. Solution
(a)
Any two colligative properties:
• Elevation of boiling point
• Depression of freezing point
• Osmotic pressure
• Relative lowering of vapour pressure
(Any two)
Marks: 1
(b)
Osmotic pressure is the pressure required to stop the flow of solvent through a semipermeable membrane.
Marks: 1
(c)
Formula:
ΔTb = Kb × m
Given:
Kb = 0.52 K kg mol^-1
m = 0.5 mol kg^-1
ΔTb = 0.52 × 0.5 = 0.26 K
Final Answer: 0.26 K
Marks: 2 (1 for formula, 1 for calculation)
Q5. Solution
(a)
Van’t Hoff factor (i) is the ratio of actual number of particles in solution to the expected number of particles.
i = observed colligative property / calculated value
Marks: 1
(b)
For complete dissociation of NaCl:
NaCl → Na⁺ + Cl⁻
Number of particles = 2
i = 2
Marks: 1
(c)
Given:
Observed molar mass = (1/2) × normal molar mass
Formula:
i = normal molar mass / observed molar mass
i = M / (M/2) = 2
Final Answer: i = 2
Marks: 2 (1 for formula, 1 for calculation)
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