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

CBSE Class 12 Chemistry 2023: question 33(b)

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

Q33(b) · 5 marks · Long answerSolutions

(i) Why is the value of Van’t Hoff factor for ethanoic acid in benzene close to 0.5 ?
(ii) Determine the osmotic pressure of a solution prepared by dissolving $2.32 \times 1 0^{- 2}$ g of $K_{2} S O_{4}$ in 2L of solution at 25 °C, assuming that $K_{2} S O_{4}$ is completely dissociated.
(R = 0.082 L atm K⁻¹ mol⁻¹, Molar mass $K_{2} S O_{4}$ = 174g mol⁻¹)
(iii) When 25.6g of Sulphur was dissolved in 1000g of benzene, the freezing point lowered by 0.512 K. Calculate the formula of Sulphur ( $S_{x}$ ).
( $K_{f}$ for benzene = 5.12 K kg mol⁻¹, Atomic mass of Sulphur = 32g 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

An i value of 0.5 means the number of particles is halved — what kind of association causes that?

Hint 2

For osmotic pressure, find moles of K₂SO₄, divide by 2 L and remember K₂SO₄ gives 3 ions; convert 25 °C to kelvin.

Hint 3

From ΔTf find the molar mass of sulphur, then divide by the atomic mass 32.

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) In benzene (non-polar), ethanoic acid molecules associate through hydrogen bonding to form dimers, $2C H_{3} COOH ⇌ (C H_{3} COOH)_{2}$ . The number of particles is nearly halved, so $i \approx 0.5$ .
1 mark
(ii) $K_{2} S O_{4} \to 2 K^{+} + S O_{4}^{2 -}$ , so $i = 3$ .
½ mark
$π = i C R T = i \frac{n}{V} R T$
½ mark
$π = 3 \times \frac{2.32 \times 1 0 ^{- 2}}{174 \times 2} \times 0.082 \times 298$
½ mark
$π = 3 \times 6.67 \times 1 0^{- 5} \times 24.44 = 4.89 \times 1 0^{- 3} atm$
½ mark
(iii) $Δ T_{f} = K_{f} \times \frac{w _{2} \times 1000}{M _{2} \times w _{1}}$
½ mark
$0.512 = 5.12 \times \frac{25.6 \times 1000}{M _{2} \times 1000}$
½ mark
$M_{2} = \frac{5.12 \times 25.6}{0.512} = 256 g mol^{- 1}$
½ mark
$32 x = 256 \Rightarrow x = 8$ ; formula of sulphur is $S_{8}$ .
½ mark

Answer: (i) dimerisation of ethanoic acid; (ii) π = 4.88–4.89 × 10⁻³ atm; (iii) S₈

Where the marks are

Dimer formation by H-bonding halves particles (1 mark)
i = 3 for K₂SO₄ (½ mark)
Formula π = iCRT (½ mark)
Correct substitution (½ mark)
π ≈ 4.88 × 10⁻³ atm (½ mark)
Formula for ΔTf (½ mark)
Substitution (½ mark)
M₂ = 256 g mol⁻¹ (½ mark)
x = 8, S₈ (½ mark)

Revise the chapter

Chapter: Solutions (notes, formulas, where marks are lost and every past question)
Topic: Abnormal Molar Mass and van't Hoff Factor

Practise this chapterBoard-style questions from Solutions, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 56/5/1: every question, by section.

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CBSE Class 12 Chemistry 2023: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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