0.3 g of acetic acid (Molar mass = 60 g mol^-1) dissolved in 30 g of benzene shows a depression in freezing point equal to 0.45°C. Calculate the percentage association of acid if it forms a dimer in the solution. (Given : K_f for benzene = 5.12 K kg mol^-1)

ChemistryCBSE Class 12 · Board exam 2025, set 56/7/1 · 3 marks · Short answer

CBSE Class 12 Chemistry 2025: question 22

Set in the CBSE Class 12 board exam of 2025, paper 56/7/1, section C, question 22, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q22 · 3 marks · Short answerSolutions

0.3 g of acetic acid (Molar mass = 60 g mol $^{- 1}$ ) dissolved in 30 g of benzene shows a depression in freezing point equal to 0.45 $^{\circ}$ C. Calculate the percentage association of acid if it forms a dimer in the solution.
(Given : $K_{f}$ for benzene = 5.12 K kg mol $^{- 1}$ )

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 56/7/1.

Hints

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

Hint 1

Use the freezing-point depression formula with the van't Hoff factor included.

Hint 2

Calculate the molality first, then solve for i.

Hint 3

For a dimer, start with 1 mol of acid: $1 - α$ mol stays as monomer and $α /2$ mol forms dimer.

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.

Formula used: $Δ T_{f} = i K_{f} m$ , where $m = \frac{w _{B} \times 1000}{M _{B} \times w _{A}}$ .
½ mark
Substitution: $0.45 = i \times 5.12 \times \frac{0.3 \times 1000}{60 \times 30}$
$0.45 = i \times 5.12 \times 0.1667 = i \times 0.8533$
So $i = 0.527$ .
1 mark
For dimerisation, $2 C H_{3} COOH ⇌ (C H_{3} COOH)_{2}$ , so $n = 2$ and $i = 1 - α + \frac{α}{2}$ . This gives $α = \frac{i - 1}{\frac{1}{n} - 1}$ .
½ mark
$α = \frac{0.527 - 1}{\frac{1}{2} - 1} = \frac{- 0.473}{- 0.5}$
½ mark
$α = 0.946$ , i.e. percentage association = 94.6%.
½ mark

Answer: i = 0.527; degree of association = 0.946, i.e. 94.6%

Where the marks are

Formula $Δ T_{f} = i K_{f} m$ (½ mark)
Correct substitution giving $i = 0.527$ (1 mark)
Relation $α = (i - 1) / (1/ n - 1)$ (½ mark)
Substitution with $n = 2$ (½ mark)
$α = 0.946$ or 94.6% (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 paper2025 board paper, set 56/7/1: every question, by section.

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CBSE Class 12 Chemistry 2025: question 22

Hints

Step-by-step solution

Revise the chapter

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