Class 12 Chemistry QP Worksheet Chapter 1 Solutions | Detailed Solutions & Marking Scheme
VEDANT WORKSHEET SERIES
Class: XII
Subject: Chemistry
Chapter: 1 – Solutions
Time: 1 Hour
Maximum Marks: 30
Section A (1 × 7 = 7 Marks)
Q1. The molarity of a solution containing 5 g NaOH in 500 mL solution is:
A. 0.1 M
B. 0.25 M
C. 0.5 M
D. 1.0 M
Q2. Which of the following is an example of a solid solution?
A. Sugar in water
B. Alloy of copper and zinc
C. CO₂ in water
D. Ethanol in water
Q3. According to Henry’s law, solubility of a gas in liquid increases with:
A. Increase in temperature
B. Decrease in pressure
C. Increase in pressure
D. Decrease in concentration
Q4. The unit of molality is:
A. mol L⁻¹
B. mol kg⁻¹
C. g L⁻¹
D. mol m⁻³
Q5. Which property depends only on number of solute particles?
A. Surface tension
B. Viscosity
C. Colligative property
D. Density
Q6. Assertion (A): Relative lowering of vapour pressure is a colligative property.
Reason (R): It depends only on the number of solute particles present.
A. Both A and R are true, R is correct explanation of A
B. Both A and R are true, R is not correct explanation of A
C. A is true, R is false
D. A is false, R is true
Q7. Assertion (A): Van’t Hoff factor can be greater than 1.
Reason (R): It is due to association of solute particles.
A. Both A and R are true, R is correct explanation of A
B. Both A and R are true, R is not correct explanation of A
C. A is true, R is false
D. A is false, R is true
Section B (2 × 4 = 8 Marks)
Q8. Define molality. Calculate molality of a solution containing 10 g NaCl in 200 g water.
Q9. State Henry’s law and write its mathematical expression.
Q10. What are colligative properties? Name any two.
Q11. Calculate the mole fraction of ethanol in a solution containing 46 g ethanol and 54 g water.
Section C (3 × 2 = 6 Marks)
Q12. Derive the relation for elevation in boiling point:
ΔTb = Kb × m
Q13. Explain abnormal molecular mass. How is it related to Van’t Hoff factor?
Section D (5 × 1 = 5 Marks)
Q14. A solution contains 1.8 g of glucose dissolved in 100 g of water. Calculate:
(i) Molality
(ii) Elevation in boiling point (Kb for water = 0.52 K kg mol⁻¹)
(iii) Final boiling point of solution
Section E (Case Study) (4 Marks)
Q15. Case Study: Osmosis in Daily Life
A student prepared a sugar solution and placed a semipermeable membrane between pure water and the solution. He observed movement of water into the solution side. This phenomenon is widely used in biological systems and industrial processes.
(i) What is the phenomenon called? (1 mark)
(ii) Write the formula for osmotic pressure. (1 mark)
(iii) Explain how osmotic pressure can be used to determine molar mass of solute. (2 marks)
End of Worksheet
VEDANT WORKSHEET SERIES – ANSWER KEY WITH DETAILED SOLUTIONS & MARKING SCHEME
Class XII – Chemistry | Chapter: Solutions | Total Marks: 30
Section A (1 × 7 = 7 Marks)
Q1. Answer: B (0.25 M)
Moles of NaOH = 5 / 40 = 0.125 mol
Volume = 500 mL = 0.5 L
Molarity = 0.125 / 0.5 = 0.25 M
Q2. Answer: B (Alloy of copper and zinc)
Solid solution example: Brass (Cu + Zn)
Q3. Answer: C (Increase in pressure)
Henry’s law: Solubility ∝ Pressure
Q4. Answer: B (mol kg⁻¹)
Q5. Answer: C (Colligative property)
Q6. Answer: A
Both true, and R correctly explains A
Q7. Answer: C
A is true, R is false (association decreases particles → i < 1)
Section B (2 × 4 = 8 Marks)
Q8. (2 Marks)
Molality = moles of solute / mass of solvent (kg)
Moles NaCl = 10 / 58.5 = 0.171 mol
Mass of water = 200 g = 0.2 kg
Molality = 0.171 / 0.2 = 0.855 m
Answer: 0.855 m
(1 mark formula, 1 mark calculation)
Q9. (2 Marks)
Henry’s Law: Solubility of a gas in a liquid is directly proportional to pressure at constant temperature.
Expression:
p = KH × x
(1 mark statement, 1 mark formula)
Q10. (2 Marks)
Colligative properties depend only on number of solute particles.
Examples:
• Elevation in boiling point
• Depression in freezing point
(1 mark definition, 1 mark examples)
Q11. (2 Marks)
Moles ethanol = 46 / 46 = 1 mol
Moles water = 54 / 18 = 3 mol
Mole fraction ethanol = 1 / (1 + 3) = 1/4 = 0.25
Answer: 0.25
(1 mark moles, 1 mark final answer)
Section C (3 × 2 = 6 Marks)
Q12. (3 Marks)
Elevation in boiling point:
ΔTb = Tb(solution) – Tb(pure solvent)
Directly proportional to molality:
ΔTb ∝ m
So,
ΔTb = Kb × m
Where Kb = molal elevation constant
(1 mark concept, 1 mark relation, 1 mark final formula)
Q13. (3 Marks)
Abnormal molar mass occurs due to:
• Association (i < 1)
• Dissociation (i > 1)
Relation:
i = Normal molar mass / Observed molar mass
Thus, molar mass = (normal value) / i
(1 mark definition, 1 mark causes, 1 mark relation)
Section D (5 × 1 = 5 Marks)
Q14. Solution
Given:
Mass glucose = 1.8 g
Molar mass = 180 g/mol
Mass solvent = 100 g = 0.1 kg
Kb = 0.52 K kg mol⁻¹
(i) Molality
Moles glucose = 1.8 / 180 = 0.01 mol
Molality = 0.01 / 0.1 = 0.1 m
(ii) Elevation in boiling point
ΔTb = Kb × m = 0.52 × 0.1 = 0.052 K
(iii) Final boiling point
Tb = 100 + 0.052 = 100.052°C
Answer:
Molality = 0.1 m
ΔTb = 0.052 K
Boiling point = 100.052°C
(Marking: 2 marks molality, 2 marks ΔTb, 1 mark final Tb)
Section E (Case Study) (4 Marks)
Q15.
(i) Answer (1 Mark):
Osmosis
(ii) Answer (1 Mark):
π = CRT
(iii) Answer (2 Marks):
Osmotic pressure is measured experimentally.
Using formula:
π = (n/V)RT
So,
Molar mass = (mass × RT) / (π × V)
Thus, molar mass can be determined from osmotic pressure.
(1 mark explanation, 1 mark formula)
Final Remarks (Teacher Use)
• Covers CBSE competency-based + PYQ pattern
• Includes numericals, theory, and application-based questions
• Balanced difficulty level (Moderate to High)
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