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Solve the following.
When of acetic acid is dissolved in of benzene, the freezing point of the solution is lowered by …
When $0\cdot4\text{ g}$ of acetic acid is dissolved in $40\text{ g}$ of benzene, the freezing point of the solution is lowered by $0\cdot45\text{ K}$. Calculate the degree of association of acetic acid. Acetic acid forms dimer when dissolved in benzene.
($K_f\text{ for benzene} = 5\cdot12\text{ K kg mol}^{-1}, \text{at. wt. C} = 12, \text{H} = 1, \text{O} = 16$)
Answer
Answer
AIFormulas used: Depression of freezing point: dTf = i*Kf*m with i = 1, Kf = 5.12 K kg mol^-1, w1 = 40 g, w2 = 0.4 g, dTf = 0.45 K, so M2 = 113.8 g mol^-1 van 't Hoff factor from molar masses: i = M_normal/M_observed with M_normal = 60 g mol^-1, M_observed = 113.8 g mol^-1, so i = 0.527 Degree of association from i: alpha = (1 - i)/(1 - 1/nu) with i = 0.527, nu = 2, so alpha = 0.945
(The paper prints "oxalic acetic acid"; the question is about acetic acid, $\mathrm{CH_3COOH}$, $M = 60\text{ g mol}^{-1}$.)
Observed molar mass: $M_{obs} = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1} = \frac{5.12 \times 0.4 \times 1000}{0.45 \times 40} = 113.8\text{ g mol}^{-1}$
$i = \frac{M_{normal}}{M_{obs}} = \frac{60}{113.8} = 0.527$
For dimerisation $2\mathrm{CH_3COOH} \rightleftharpoons (\mathrm{CH_3COOH})_2$: $i = 1 - \frac{\alpha}{2}$
$\alpha = 2(1 - 0.527) = 0.945$
Final answer: 0.945
From ISC 2019 Chemistry Paper 1, question 9(a).
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