Question

The freezing point of a solution containing $0.2g$ of acetic acid in $20.0g$ 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 $=5.12Kkgmo{l}^{-1}$ )
$M_{\text{observed }}$ of acetic acid $=113.78$

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

Answer 1

Answer: (A)

Given, $w_2=0.2g,w_1=20g,\Delta T_f=0.45^{\circ }C$
$\Delta T_f=\frac{1000\times K_f\times W_2}{w_1\times M}$
$\Rightarrow 0.45=\frac{1000\times 5.12\times 0.2}{20\times M}$
$\therefore M_{\text{(observed) }}=113.78$ (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 $94.5\%$