Question

The freezing point of benzene decreases by $0.45^{\circ }C$ when $0.2g$ of acetic acid is added to $20g$ of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be :
($K_f$ for benzene = $5.12Kkgmo{l}^{-1}$)

• A. $74.6\%$
• B. $94.6\%$
• C. $64.6\%$
• D. $80.4\%$

Answer 1

Answer: (B)

$0.45=i(5.12)\frac{0.2/60}{20}\times 1000$

$\Rightarrow i=0.527$

$\underset{1-\alpha }{2C{H}_{3}COOH}\rightleftharpoons \underset{\frac{\alpha }{2}}{(C{H}_{3}COOH{)}_2}$

$\Rightarrow i=1-\frac{\alpha }{2}$

$\Rightarrow 0.527=1-\frac{\alpha }{2}$

$\Rightarrow \frac{\alpha }{2}=0.473$

$\Rightarrow \alpha =0.946$

$\therefore \%$ association $=94.6\%$