Question

The freezing point of benzene decrease 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. $64.6\%$
• B. $80.4\%$
• C. $74.6\%$
• D. $94.6\%$

Answer 1

Answer: (D)

$94.6\%$

$△T_f=0.45$ $0.2g$ of acetic acid in $20g$ of benzene.

To form dimer, $2\left(C{H}_{3}COO\right)\rightleftarrows {C{H}_{3}COO}_2$

$i=1+(\frac{1}{n}-1)a=1+\left(\frac{1}{2}\right)a=1-\frac{a}{2}$

$△T_f=i{K}_{f}m$

$0.45=(1-\frac{a}{2})\left(5.12\right)\left(\frac{\frac{0.2}{60}}{\frac{20}{1000}}\right)$

$1-\frac{a}{2}=0.527$

$a=0.945$

Percentage of association is $94.5\%$