ChemistryCBSE Class 12 · Board exam 2023, set 56/5/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2023: question 33(a)

Set in the CBSE Class 12 board exam of 2023, paper 56/5/1, section E, question 33(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(a) · 5 marks · Long answerSolutions

(i) Why is boiling point of 1M NaCl solution more than that of 1M glucose solution ?
(ii) A non-volatile solute ‘X’ (molar mass = 50 g mol⁻¹) when dissolved in 78g of benzene reduced its vapour pressure to 90%. Calculate the mass of X dissolved in the solution.
(iii) Calculate the boiling point elevation for a solution prepared by adding 10g of $M g C l_{2}$ to 200g of water assuming $M g C l_{2}$ is completely dissociated.
( $K_{b}$ for Water = 0.512 K kg mol⁻¹, Molar mass $M g C l_{2}$ = 95g mol⁻¹)

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 56/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Colligative properties depend on the number of particles — think about dissociation.

Hint 2

If vapour pressure falls to 90%, the relative lowering is 0.1; equate this to the mole fraction of solute (benzene, C₆H₆, has M = 78).

Hint 3

For MgCl₂, count the ions formed per formula unit to get i.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Elevation of boiling point is a colligative property depending on the number of solute particles. NaCl dissociates into $N a^{+}$ and $C l^{-}$ ( $i \approx 2$ ), while glucose does not dissociate ( $i = 1$ ). So 1 M NaCl has about twice as many particles, giving larger $Δ T_{b}$ and a higher boiling point.
1 mark
(ii) Raoult's law: $\frac{p ^{\circ} - p}{p ^{\circ}} = x_{2} = \frac{n _{2}}{n _{1} + n _{2}}$
½ mark
$p = 0.9 p^{\circ} \Rightarrow \frac{p ^{\circ} - 0.9 p ^{\circ}}{p ^{\circ}} = 0.1$ ; $n_{1} = \frac{78}{78} = 1$ mol (benzene, M = 78 g mol⁻¹), $n_{2} = \frac{w}{50}$
$0.1 = \frac{w /50}{1 + w /50}$
1 mark
$0.1 + \frac{0.1 w}{50} = \frac{w}{50} \Rightarrow 0.9 \frac{w}{50} = 0.1 \Rightarrow w = \frac{5}{0.9} = 5.55 g$ (≈ 5 g if the dilute-solution approximation $x_{2} \approx n_{2} / n_{1}$ is used)
½ mark
(iii) $MgC l_{2} \to M g^{2 +} + 2C l^{-}$ , so $i = 3$ .
½ mark
$Δ T_{b} = i K_{b} m = i K_{b} \times \frac{w _{2} \times 1000}{M _{2} \times w _{1}}$
½ mark
$Δ T_{b} = 3 \times 0.512 \times \frac{10 \times 1000}{95 \times 200}$
½ mark
$Δ T_{b} = 3 \times 0.512 \times 0.526 = 0.81 K$
½ mark

Answer: (i) NaCl gives more particles (i ≈ 2); (ii) mass of X = 5.55 g; (iii) ΔTb = 0.81 K

Where the marks are

Dissociation of NaCl / larger i gives larger ΔTb (1 mark)
Relative lowering of vapour pressure formula (½ mark)
Correct substitution with n₁ = 1 mol (1 mark)
w = 5.55 g (½ mark)
i = 3 (½ mark)
Formula ΔTb = iKbm (½ mark)
Substitution (½ mark)
ΔTb = 0.81 K (½ mark)

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CBSE Class 12 Chemistry 2023: question 33(a)

Hints

Step-by-step solution

Revise the chapter

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