Q70 / 180 · AIIMS 2016 · Chemistry
The freezing point of a solution containing $0.2 \mathrm{~g}$ of acetic acid in $20.0 \mathrm{~g}$ benzene is lowered by $0.45^{\circ} \mathrm{C}$. The degree of association of acetic acid in benzene is (Assume acetic acid dimerises in benzene and $K_f$ for benzene $=5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ ) $M_{\text {observed }}$ of acetic acid $=113.78$
- Correct answerA$94.5 \%$
- B$54.9 \%$
- C$78.2 \%$
- D$100 \%$
Solution
Given : $w_2=0.2 \mathrm{~g}, w_1=20 \mathrm{~g}, \Delta T_f=0.45^{\circ} \mathrm{C}$
$\Delta T_f=\frac{1000 \times K_f \times w_2}{w_1 \times M} \Rightarrow 0.45=\frac{1000 \times 5.12 \times 0.2}{20 \times M}$
$\therefore \quad M_{\text {(observed) }}=113.78$ (acetic acid)
$2 \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons\left(\mathrm{CH}_3 \mathrm{COOH}\right)_2$
Before association
After association
$1-\alpha$
$\alpha / 2$
(where $\alpha$ is degree of association)
Molecular weight of acetic acid $=60$
$\begin{aligned} & \quad i=\frac{\text { Normal molecular mass }}{\text { Observed molecular mass }} \\ & \therefore \quad \frac{M_{(\text {normal })}}{M_{(\text {observed) }}}=1-\alpha+\frac{\alpha}{2} \\ & \text { or, } \frac{60}{113.78}=1-\alpha+\frac{\alpha}{2} \therefore \alpha=0.945 \text { or } 94.5 \%\end{aligned}$
Hence, the correct answer is option: 'A'.