Question

The freezing point of a solution containing $0.2 \,g$ of acetic acid in $20.0\, g$ benzene is lowered by $0.45^{\circ} C$. The degree of association of acetic acid in benzene is (Assume acetic acid dimerises in benzene and $K_{f}$ for benzene $=5.12\, K\, kg\, mol ^{-1}$ )
$M_{\text {observed }}$ of acetic acid $=113.78$

• A. $94.5 \%$
• B. $54.9 \%$
• C. $78.7 \%$
• D. $100 \%$

Answer 1

Answer: (A)

Given, $w_{2}=0.2\, g , w_{1}=20\, g , \Delta T_{f}=0.45^{\circ} C$
$\Delta T_{f}=\frac{1000 \times K_{f} \times W_{2}}{w_{1} \times M}$
$\Rightarrow 0.45=\frac{1000 \times 5.12 \times 0.2}{20 \times M}$
$\therefore M_{\text {(observed) }}=113.78$ (acetic acid)

(where, $\alpha$ is degree of association)
Molecular weight of acetic acid $=60$
$ i = \frac{\text{Normal molecular mass}}{\text{Observed molecular mass}}$
$\therefore \frac{M_{(\text{normal})}}{M_{(\text{observed})}} = 1 - \alpha + \frac{\alpha}{2}$
or, $\frac{60}{113.78} = 1 - \alpha + \frac{\alpha}{2}$
$\therefore \alpha = 0.945$ or $94 . 5\%$