Class 12 Chemistry Test Paper Chapter 1 Solutions | GSEB Board | Detailed Solutions & Marking Scheme
VEDANT IGNITE TEST SERIES
Class 12 – Chemistry
Chapter 1: Solutions (Theory + Numericals)
Time: 1 Hour | Maximum Marks: 30
Section A – MCQs (1 × 7 = 7 Marks)
The unit of molality is
(a) mol L⁻¹ (b) mol kg⁻¹ (c) g L⁻¹ (d) mol %Which of the following is an ideal solution?
(a) Ethanol + Water (b) Benzene + Toluene (c) Acetone + Water (d) HCl + WaterThe value of van’t Hoff factor (i) for K₂SO₄ in dilute solution is approximately
(a) 1 (b) 2 (c) 3 (d) 4Colligative properties depend on
(a) Nature of solute (b) Number of particles (c) Nature of solvent (d) PressureHenry’s law constant (KH) increases with
(a) Increase in temperature (b) Decrease in temperature (c) Increase in pressure (d) No changeWhich property is most suitable for determining molar mass of proteins?
(a) Boiling point elevation (b) Osmotic pressure (c) Vapour pressure (d) Freezing pointIf solute undergoes association, the van’t Hoff factor
(a) increases (b) decreases (c) remains same (d) becomes zero
Section B – Short Answer (2 × 3 = 6 Marks)
(Attempt any 3)
Define van’t Hoff factor and write its expression.
State Raoult’s law for non-volatile solute.
Define molarity and molality.
What is reverse osmosis?
Define azeotropes with one example.
Section C – Medium Answer (3 × 3 = 9 Marks)
(Attempt any 3)
Mixed Concept Numerical (Molality + i)
18 g glucose is dissolved in 180 g water. Calculate:
(a) Molality
(b) Expected elevation in boiling point (Kb = 0.52 K kg mol⁻¹)
Abnormal Molar Mass (Dissociation)
0.5 g KCl is dissolved in 100 g water and shows a freezing point depression of 0.186 K.
Calculate van’t Hoff factor. (Kf = 1.86 K kg mol⁻¹)
Henry’s Law Application
The solubility of a gas in water at 1 atm pressure is 0.02 mol fraction.
Calculate solubility at 5 atm assuming Henry’s law is valid.
Osmotic Pressure Numerical (Multi-step)
2 g of a solute dissolved in 1 L solution exerts osmotic pressure of 4.92 atm at 27°C.
Calculate molar mass of solute. (R = 0.0821 L atm mol⁻¹ K⁻¹)
Explain positive and negative deviations from Raoult’s law with graphs (theory-based).
Section D – Long Answer (4 × 2 = 8 Marks)
(Attempt any 2)
1. Association Case (Advanced Numerical)
1.5 g benzoic acid (C₆H₅COOH) is dissolved in 50 g benzene.
The observed depression in freezing point is 0.63 K.
Calculate:
(a) van’t Hoff factor (i)
(b) Degree of association
(Given: Kf for benzene = 5.12 K kg mol⁻¹, Molar mass = 122 g mol⁻¹)
2. Combined Concept Numerical (Molality + ΔTb + i)
5 g NaCl is dissolved in 200 g water.
Calculate:
(a) Molality
(b) Elevation in boiling point
(c) If NaCl is 80% dissociated, calculate actual ΔTb
(Kb = 0.52 K kg mol⁻¹)
3. Multi-Concept (Osmotic Pressure + Density)
A solution contains 10 g glucose in 100 mL solution.
Density of solution = 1.2 g mL⁻¹.
Calculate osmotic pressure at 27°C.
(R = 0.0821 L atm mol⁻¹ K⁻¹)
— End of Question Paper —
VEDANT IGNITE TEST SERIES – ANSWER KEY
Class 12 – Chemistry | Chapter: Solutions | Marks: 30
Section A – MCQs (1 × 7 = 7 Marks)
(b) mol kg⁻¹ ✔
(b) Benzene + Toluene ✔
(c) 3 ✔
(b) Number of particles ✔
(a) Increase in temperature ✔
(b) Osmotic pressure ✔
(b) decreases ✔
Marking Scheme:
+1 for each correct answer
Section B – Short Answer (2 × 3 = 6 Marks)
(Any 3, each 2 marks)
1. Van’t Hoff Factor
i = (Observed colligative property) / (Calculated value)
OR
i = (Actual number of particles) / (Expected number)
2. Raoult’s Law
For a non-volatile solute:
P(solution) = X(solvent) × P⁰(solvent)
3. Molarity and Molality
Molarity (M) = moles of solute / volume of solution (L)
Molality (m) = moles of solute / mass of solvent (kg)
4. Reverse Osmosis
Flow of solvent from solution to pure solvent by applying pressure greater than osmotic pressure.
5. Azeotropes
Mixture having constant boiling point and composition.
Example: Ethanol + Water
Marking Scheme:
Definition: 1 mark
Formula/example: 1 mark
Section C – Medium Answer (3 × 3 = 9 Marks)
1. Molality + Boiling Point Elevation
Given:
Mass of glucose = 18 g
Molar mass = 180 g mol⁻¹
Moles = 18 / 180 = 0.1 mol
Mass of solvent = 180 g = 0.18 kg
(a) Molality:
m = 0.1 / 0.18 = 0.56 mol kg⁻¹
(b) ΔTb:
ΔTb = Kb × m × i
= 0.52 × 0.56 × 1
= 0.29 K
2. Van’t Hoff Factor (KCl)
Given:
ΔTf = 0.186 K
Formula:
ΔTf = i × Kf × m
Moles KCl = 0.5 / 74.5 = 0.0067 mol
Mass solvent = 0.1 kg
m = 0.0067 / 0.1 = 0.067
i = ΔTf / (Kf × m)
= 0.186 / (1.86 × 0.067)
= 1.5
3. Henry’s Law
P₁ = 1 atm, X₁ = 0.02
P₂ = 5 atm
X₂ = X₁ × (P₂ / P₁)
= 0.02 × 5
= 0.10
4. Osmotic Pressure
Given:
π = 4.92 atm
V = 1 L
T = 300 K
Formula:
π = (n/V)RT
n = πV / RT
= (4.92 × 1) / (0.0821 × 300)
= 0.2 mol
Molar mass = mass / moles
= 2 / 0.2
= 10 g mol⁻¹
5. Positive & Negative Deviation
Positive Deviation:
A–B interaction weaker
Vapour pressure increases
Negative Deviation:
A–B interaction stronger
Vapour pressure decreases
Marking Scheme (each question 3 marks):
Formula: 1 mark
Substitution: 1 mark
Final answer: 1 mark
Section D – Long Answer (4 × 2 = 8 Marks)
1. Benzoic Acid (Association)
Given:
ΔTf = 0.63 K
Formula:
ΔTf = i × Kf × m
Moles = 1.5 / 122 = 0.0123
Mass solvent = 0.05 kg
m = 0.0123 / 0.05 = 0.246
i = 0.63 / (5.12 × 0.246)
= 0.5
Degree of association (α):
i = 1 − α/2
0.5 = 1 − α/2
α = 1 (100% association)
2. NaCl (Dissociation)
Given:
Mass = 5 g
Molar mass = 58.5
Moles = 5 / 58.5 = 0.0855
Mass solvent = 0.2 kg
(a) Molality:
m = 0.0855 / 0.2 = 0.43 mol kg⁻¹
(b) ΔTb (ideal):
ΔTb = 0.52 × 0.43 × 1
= 0.22 K
(c) With dissociation (i):
i = 1 + α(n−1)
= 1 + 0.8(2−1)
= 1.8
ΔTb = 0.52 × 0.43 × 1.8
= 0.40 K
3. Osmotic Pressure (Density)
Given:
Mass = 10 g
Molar mass = 180
Moles = 10 / 180 = 0.0556
Volume = 100 mL = 0.1 L
π = (n/V)RT
= (0.0556 / 0.1) × 0.0821 × 300
= 13.7 atm
Marking Scheme (each question 4 marks):
Formula: 1 mark
Calculation steps: 2 marks
Final answer: 1 mark
✅ Final Notes for Evaluation
Accept answers within rounding error
Units must be correct
Step marking allowed
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