ChemistryCBSE Class 12 · Board exam 2024, set 56/1/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2024: question 31(a)

Set in the CBSE Class 12 board exam of 2024, paper 56/1/1, section E, question 31(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q31(a) · 5 marks · Long answerSolutions

(i) At the same temperature, $C O_{2}$ gas is more soluble in water than $O_{2}$ gas. Which one of them will have higher value of $K_{H}$ and why ?
(ii) How does the size of blood cells change when placed in an aqueous solution containing more than 0·9% (mass/volume) sodium chloride ?
(iii) 1 molal aqueous solution of an electrolyte $A_{2} B_{3}$ is 60% ionized. Calculate the boiling point of the solution. (Given : $K_{b}$ for $H_{2} O$ = 0·52 K kg mol $^{- 1}$ )

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 56/1/1.

Hints

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

Hint 1

In Henry's law, K_H and solubility are inversely related.

Hint 2

Count the ions given by A₂B₃ to get n, then find i from the degree of ionisation before using ΔT_b = iK_bm.

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) $O_{2}$ has the higher $K_{H}$ . By Henry's law $p = K_{H} x$ , so at a given pressure, the higher the $K_{H}$ , the lower the solubility; O₂ is less soluble than CO₂.
1 mark
(ii) The solution (> 0.9% NaCl) is hypertonic, so water moves out of the blood cells by osmosis and the cells shrink.
1 mark
(iii) $A_{2} B_{3} \to 2 A^{3 +} + 3 B^{2 -}$ , so n = 5 ions. $α = \frac{i - 1}{n - 1} \Rightarrow 0.6 = \frac{i - 1}{4}$
½ mark
$i = 1 + 2.4 = 3.4$
1 mark
$Δ T_{b} = i K_{b} m = 3.4 \times 0.52 K kg mo l^{- 1} \times 1 mol k g^{- 1} = 1.768 K$
1 mark
$T_{b} = 373.15 + 1.768 = 374.918$ K (or 374.768 K taking the boiling point of water as 373 K).
½ mark

Answer: (i) O₂ (higher K_H → lower solubility); (ii) cells shrink; (iii) T_b = 374.918 K (374.768 K if T_b° = 373 K)

Where the marks are

O₂ has higher K_H (½ mark) with reason via Henry's law (½ mark)
Cells shrink in hypertonic solution (1 mark)
n = 5 and use of α = (i–1)/(n–1) (½ mark)
i = 3.4 (1 mark)
ΔT_b = 1.768 K (1 mark)
T_b = 374.918 K / 374.768 K (½ mark)

Revise the chapter

Chapter: Solutions (notes, formulas, where marks are lost and every past question)
Topic: Elevation of Boiling Point and Depression of Freezing Point

Practise this chapterBoard-style questions from Solutions, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 56/1/1: every question, by section.

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CBSE Class 12 Chemistry 2024: question 31(a)

Hints

Step-by-step solution

Revise the chapter

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