Class 12 Chemistry Test Paper Chapter 1 Solutions | CBSE Board | Detailed Solutions & Marking Scheme
VEDANT IGNITE TEST SERIES
Class 12 – Chemistry (CBSE)
Chapter 1: Solutions (Competency + PYQ Based)
Time: 1 Hour | Maximum Marks: 30
Section A – MCQs + Assertion Reason (1 × 7 = 7 Marks)
MCQs (5 Questions)
Which solution will have the highest boiling point?
(a) 1% glucose (b) 1% urea (c) 1% NaCl (d) 1% CaCl₂
The value of Henry’s constant (KH) is higher for
(a) gases with high solubility (b) gases with low solubility (c) all gases same (d) independent of solubility
A solution shows positive deviation from Raoult’s law when
(a) A–B interactions > A–A, B–B
(b) A–B interactions < A–A, B–B
(c) ΔHmix = 0
(d) vapour pressure decreases
For an associated solute, van’t Hoff factor (i) is
(a) >1 (b) <1 (c) =1 (d) =0
Relative lowering of vapour pressure depends on
(a) nature of solute (b) number of solute particles (c) pressure (d) temperature only
Assertion–Reason (2 Questions)
Directions:
(a) Both A and R are true and R is correct explanation
(b) Both A and R are true but R is not correct explanation
(c) A is true but R is false
(d) A is false but R is true
Assertion (A): Molarity changes with temperature.
Reason (R): Volume of solution changes with temperature.
Assertion (A): Aquatic organisms are more comfortable in cold water.
Reason (R): Solubility of oxygen decreases with increase in temperature.
Section B – Short Answer (2 × 4 = 8 Marks)
Define azeotropes. What type is formed by positive deviation? Give example. (PYQ based)
Explain why osmotic pressure is preferred for determining molar mass of macromolecules. (PYQ concept)
A solution contains 10 g solute in 200 g water. Vapour pressure of solution is 31.84 mm Hg. Pure solvent = 32 mm Hg. Calculate molar mass. (PYQ type)
Explain why solubility of gases increases with pressure using Henry’s law.
Section C – Medium Answer (3 × 2 = 6 Marks)
1 g of non-volatile solute is dissolved in 100 g water. The freezing point depression is 0.372 K.
Calculate molar mass. (Kf = 1.86 K kg mol⁻¹)
A solution of NaCl shows abnormal molar mass.
(a) Explain why
(b) Calculate van’t Hoff factor if degree of dissociation = 0.8
Section D – Long Answer (5 × 1 = 5 Marks)
Mixed Competency Numerical (PYQ Level)
A solution is prepared by dissolving 5 g urea (M = 60 g mol⁻¹) in 250 g water.
(a) Calculate molality
(b) Calculate elevation in boiling point (Kb = 0.52 K kg mol⁻¹)
(c) If urea associates to form dimers (10%), calculate new ΔTb
Section E – Case Study (4 Marks)
Read the passage carefully and answer the questions:
When a non-volatile solute is added to a solvent, vapour pressure decreases, leading to elevation in boiling point and depression in freezing point. These are colligative properties and depend on the number of solute particles.
Why are colligative properties independent of nature of solute? (1 mark)
What happens to boiling point when vapour pressure decreases? (1 mark)
Explain why NaCl shows greater effect than glucose in colligative properties. (2 marks)
— End of Question Paper —
Here is your detailed solved answer key with marking scheme in simple, clean, copy-friendly format (proper symbols & powers):
VEDANT IGNITE TEST SERIES – ANSWER KEY
Class 12 – Chemistry (CBSE)
Chapter 1: Solutions | Marks: 30
Section A – MCQs + Assertion Reason (1 × 7 = 7 Marks)
MCQs
(d) 1% CaCl₂ ✔
→ Highest number of particles (i ≈ 3)(b) gases with low solubility ✔
(b) A–B interactions < A–A, B–B ✔
(b) < 1 ✔
(b) number of solute particles ✔
Assertion–Reason
(a) Both A and R are true and R is correct explanation ✔
(a) Both A and R are true and R is correct explanation ✔
Marking Scheme:
+1 mark each correct answer
Section B – Short Answer (2 × 4 = 8 Marks)
1. Azeotropes
Azeotropes are mixtures having constant boiling point and composition.
Positive deviation forms minimum boiling azeotrope.
Example: Ethanol + Water
Marks:
Definition (1) + Type (1) + Example (1)
2. Osmotic Pressure (Macromolecules)
Measured at room temperature → no decomposition
Large values even for dilute solutions
Very accurate for high molar mass
Marks:
Any 2–3 valid points = 2 marks
3. Vapour Pressure Numerical
Given:
P⁰ = 32 mm Hg
P = 31.84 mm Hg
Relative lowering:
(P⁰ − P) / P⁰ = (32 − 31.84) / 32
= 0.16 / 32
= 0.005
Formula:
(P⁰ − P) / P⁰ = n₂ / (n₁ + n₂)
Moles solvent (water):
= 200 / 18 = 11.11 mol
Assuming dilute solution:
n₂ ≈ 0.005 × 11.11
= 0.0555 mol
Molar mass = 10 / 0.0555
= 180 g mol⁻¹
4. Henry’s Law
p = KH × x
As pressure increases → mole fraction increases → solubility increases
Marking Scheme:
Formula/concept: 1 mark
Calculation/explanation: 1 mark
Section C – Medium Answer (3 × 2 = 6 Marks)
1. Freezing Point Depression
Given:
ΔTf = 0.372 K
Formula:
ΔTf = Kf × m
m = 0.372 / 1.86 = 0.2 mol kg⁻¹
Mass solvent = 100 g = 0.1 kg
Moles solute = 0.2 × 0.1 = 0.02 mol
Molar mass = 1 / 0.02
= 50 g mol⁻¹
2. NaCl Abnormal Molar Mass
(a) Reason:
NaCl dissociates into Na⁺ and Cl⁻ → increases particles
(b) i = 1 + α(n − 1)
= 1 + 0.8(2 − 1)
= 1.8
Marking Scheme (each 3 marks):
Concept: 1 mark
Formula: 1 mark
Answer: 1 mark
Section D – Long Answer (5 Marks)
Given:
Mass urea = 5 g
Molar mass = 60
Moles = 5 / 60 = 0.0833 mol
Mass solvent = 250 g = 0.25 kg
(a) Molality
m = 0.0833 / 0.25
= 0.333 mol kg⁻¹
(b) ΔTb (normal)
ΔTb = Kb × m × i
= 0.52 × 0.333 × 1
= 0.173 K
(c) Association (10%)
For dimerization:
i = 1 − α/2
= 1 − 0.1/2
= 0.95
ΔTb = 0.52 × 0.333 × 0.95
= 0.164 K
Marking Scheme:
Each part: 1–2 marks
Correct formula + answer required
Section E – Case Study (4 Marks)
1. Why independent of nature?
Because they depend only on number of solute particles, not their identity. ✔ (1 mark)
2. Effect on boiling point
Decrease in vapour pressure → boiling point increases ✔ (1 mark)
3. NaCl vs Glucose
NaCl dissociates into 2 ions → more particles
Glucose does not dissociate
Hence NaCl shows greater colligative effect ✔✔ (2 marks)
✅ Final Evaluation Notes
Allow step marking
Minor rounding acceptable
Units compulsory
Concept-based answers acceptable if correct
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