Solutions — Class 12 MCQs with Answers

Solutions — important questions & MCQs with answers (Class 12 Chemistry)

14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 84 are timed and scored when you sign in.

1. Q1Easy

   Molarity is defined as:

   A.Moles of solute per kg of solvent

   B.Moles of solute per litre of solution✓ Correct

   C.Mass of solute per litre of solvent

   D.Mass of solute per kg of solution

   Answer: B. Moles of solute per litre of solution

   Explanation: Molarity = moles of solute ÷ litres of SOLUTION — that volume basis is what separates it from molality (mass of solvent) and rules out a kg-of-solvent style answer.

2. Q2Easy

   Raoult's law states that the partial vapour pressure of each component is proportional to its:

   A.Molar mass

   B.Mass fraction in solution

   C.Mole fraction in solution✓ Correct

   D.Boiling point

   Answer: C. Mole fraction in solution

   Explanation: Raoult's law: P_A = x_A·P°_A — partial vapour pressure scales with MOLE fraction, not mass fraction or molar mass, because vapour composition tracks particle count.

3. Q3Easy

   Which of the following is NOT a colligative property?

   A.Boiling point elevation

   B.Freezing point depression

   C.Osmotic pressure

   D.Viscosity✓ Correct

   Answer: D. Viscosity

   Explanation: Colligative properties depend only on the number of dissolved particles; viscosity depends on intermolecular friction and the solute's identity, unlike b.p. elevation, f.p. depression or osmotic pressure.

4. Q4Easy

   The unit of molality (m) is:

   A.mol/L

   B.g/L

   C.mol/kg✓ Correct

   D.kg/mol

   Answer: C. mol/kg

   Explanation: Molality m = moles of solute ÷ kg of SOLVENT, so its unit is mol/kg — distinct from molarity's mol/L, which uses volume of solution instead of mass of solvent.

5. Q5Easy

   For an ideal solution, the enthalpy of mixing (ΔH_mix) is:

   A.Positive

   B.Negative

   C.Infinite

   D.Zero✓ Correct

   Answer: D. Zero

   Explanation: An ideal solution obeys Raoult's law exactly across all compositions, which requires ΔH_mix = 0 and ΔV_mix = 0 — no heat released or absorbed and no volume change on mixing.

6. Q6Easy

   Colligative properties depend on the:

   A.Nature of the solute

   B.Molar mass of the solute

   C.Mass of the solvent

   D.Number of solute particles✓ Correct

   Answer: D. Number of solute particles

   Explanation: Colligative properties are set purely by the NUMBER of solute particles per unit solvent, regardless of what those particles are — identical concentrations of different solutes give the same effect.

7. Q7Medium

   What is the molarity of a solution containing 9.8 g of H₂SO₄ in 250 mL of solution? (Molar mass = 98 g/mol)

   A.0.4 M✓ Correct

   B.0.2 M

   C.0.5 M

   D.1.0 M

   Answer: A. 0.4 M

   Explanation: Moles H₂SO₄ = 9.8/98 = 0.1 mol; molarity = 0.1 mol ÷ 0.250 L = 0.4 M — molarity divides by the SOLUTION's volume in litres, so forgetting to convert 250 mL would distort the result.

8. Q8Medium

   A solution that shows positive deviation from Raoult's law has:

   A.Stronger A–B interactions than A–A and B–B

   B.Equal interactions

   C.Weaker A–B interactions than A–A and B–B✓ Correct

   D.Zero enthalpy of mixing

   Answer: C. Weaker A–B interactions than A–A and B–B

   Explanation: Weaker A–B interactions than A–A/B–B let molecules escape the liquid more easily, so the vapour pressure runs higher than Raoult's law predicts — the signature of positive deviation.

9. Q9Medium

   The van't Hoff factor (i) for NaCl in dilute aqueous solution (assuming complete dissociation) is approximately:

   A.0.5

   B.1

   C.3

   D.2✓ Correct

   Answer: D. 2

   Explanation: NaCl → Na⁺ + Cl⁻ splits into 2 ions per formula unit, so the van't Hoff factor i ≈ 2 for a dilute solution near-complete dissociation — not i = 1, which would apply to a non-electrolyte like glucose.

10. Q10Medium

    Calculate the molality of a solution containing 18 g of glucose (M = 180 g/mol) in 500 g of water.

    A.0.1 m

    B.0.5 m

    C.0.2 m✓ Correct

    D.1.0 m

    Answer: C. 0.2 m

    Explanation: Moles glucose = 18/180 = 0.1 mol; molality = 0.1 mol ÷ 0.500 kg WATER = 0.2 m — dividing by the solvent's mass, not the total solution mass, is what makes this molality.

11. Q11Medium

    An ideal solution has P°_A = 200 mm Hg and P°_B = 100 mm Hg. If x_A = 0.4 in the liquid phase, the total vapour pressure is:

    A.140 mm Hg✓ Correct

    B.120 mm Hg

    C.160 mm Hg

    D.180 mm Hg

    Answer: A. 140 mm Hg

    Explanation: P_total = x_A·P°_A + x_B·P°_B = 0.4×200 + 0.6×100 = 80 + 60 = 140 mm Hg — each component's contribution is weighted by ITS OWN liquid-phase mole fraction, not the other component's.

12. Q12Hard

    A 1.5 M aqueous solution of H₂SO₄ (M = 98 g/mol) has density 1.10 g/mL. The molality of the solution is approximately:

    A.1.45 m

    B.1.58 m✓ Correct

    C.1.65 m

    D.2.00 m

    Answer: B. 1.58 m

    Explanation: 1 L of solution weighs 1100 g (density 1.10 g/mL) and contains 1.5 mol × 98 g/mol = 147 g of H₂SO₄; solvent mass = 1100 − 147 = 953 g, so molality = 1.5 mol ÷ 0.953 kg ≈ 1.58 m — subtracting solute mass to isolate the SOLVENT is the key step.

13. Q13Hard

    A minimum-boiling azeotrope is formed by a solution showing:

    A.Negative deviation from Raoult's law

    B.Positive deviation from Raoult's law✓ Correct

    C.Ideal behaviour

    D.No vapour-liquid equilibrium

    Answer: B. Positive deviation from Raoult's law

    Explanation: Positive deviation raises the solution's vapour pressure above the ideal line, and a higher vapour pressure means the mixture boils at a LOWER temperature — giving a minimum-boiling azeotrope, e.g. ethanol + water.

14. Q14Hard

    Calculate the freezing point depression of a solution containing 1.8 g glucose (M = 180 g/mol) in 100 g water. K_f for water = 1.86 K·kg/mol.

    A.0.186 K✓ Correct

    B.0.093 K

    C.0.372 K

    D.1.86 K

    Answer: A. 0.186 K

    Explanation: Molality = (1.8/180 mol) ÷ 0.100 kg water = 0.1 mol/kg; ΔTf = Kf·m = 1.86 × 0.1 = 0.186 K — freezing-point depression always uses MOLALITY, never molarity.

Solutions — FAQs

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