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 = $5.12Kkgmo{l}^{-1}$)
$M_{observed}$ of acetic acid = 113.78

• A. 94.5%
• B. 54.9%
• C. 78.2%
• D. 100%

Answer 1

Answer: (A)

Given : w${}_2$ = 0.2 g, w${}_1$ = 20 g, $\Delta T_f=0.45^{\circ }C$
$\Delta T_f=\frac{1000\times K_f\times w_2}{w_1\times M}\Rightarrow 0.45=\frac{1000\times 5.12\times 0.2}{20\times M}$
$\therefore M_{\left(observed\right)}=113.78\left(aceticacid\right)$
$2C{H}_{3}COOH\rightleftharpoons {\left(C{H}_{3}COOH\right)}_2$
Before association $10$
After association$1-\alpha \alpha /2$
(where a is degree of association)
Molecular weight of acetic acid = 60
$i=\frac{Normalmolecularmass}{Observedmolecularmass}$
$\therefore \frac{M_{\left(normal\right)}}{M_{\left(observed\right)}}=1-\alpha +\frac{\alpha }{2}$
or,$\frac{60}{113.78}=1-\alpha +\frac{\alpha }{2}\therefore \alpha$ = 0.945 or 94.5%