Solutions and Colligative Properties 2 Question 13
13. The freezing point of benzene decreases by $0.45^{\circ} \mathrm{C}$ when $0.2 \mathrm{~g}$ of acetic acid is added to $20 \mathrm{~g}$ of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be $\left(K_{f}\right.$ for benzene $\left.=5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$
(2017 Main)
(a) $64.6 %$
(b) $80.4 %$
(c) $74.6 %$
(d) $94.6 %$
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Solution:
- Let the degree of association of acetic acid $\left(\mathrm{CH}_{3} \mathrm{COOH}\right)$ in benzene is $\alpha$, then
| $2 \mathrm{CH}_{3} \mathrm{COOH}$ | $\rightleftharpoons\left(\mathrm{CH}{3} \mathrm{COOH}\right){2}$ | |
|---|---|---|
| Initial moles | 1 | 0 |
| Moles at equilibrium | $1-\alpha$ | $\frac{\alpha}{2}$ |
$\therefore$ Total moles $=1-\alpha+\frac{\alpha}{2}=1-\frac{\alpha}{2}$ or $i=1-\frac{\alpha}{2}$
Now, depression in freezing point $\left(\Delta T_{f}\right)$ is given as
$$ \Delta T_{f}=i K_{f} m $$
where, $K_{f}=$ molal depression constant or
cryoscopic constant.
$m=$ molality
Molality $=\frac{\text { number of moles of solute }}{\text { weight of solvent (in kg) }}=\frac{0.2}{60} \times \frac{1000}{20}$
Putting the values in Eq. (i)
$$ \begin{array}{ll} \therefore & 0.45=\left1-\frac{\alpha}{2}\right\left[\frac{0.2}{60} \times \frac{1000}{20}\right] \ \Rightarrow & 1-\frac{\alpha}{2}=\frac{0.45 \times 60 \times 20}{5.12 \times 0.2 \times 1000} \ \therefore & 1-\frac{\alpha}{2}=0.527 \Rightarrow \frac{\alpha}{2}=1-0.527 \ \therefore & \quad \alpha=0.946 \end{array} $$
Thus, percentage of association $=94.6 %$
