Question

The freezing point of benzene decreases by ${0.45}^{0}C$ when $0.2$ g of acetic acid is added to $20$ g of benzene. If acetic acid is associates to form a dimer in benzene, percentage association of acetic acid in benzene will be:
($K_f$ for benzene = 5.12 K kg $mo{l}^{-1}$)

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

Answer 1

The correct option is D $94.6$ %

We know that,

Degree of dissolution, $\alpha =\frac{i-1}{1/n-1}=\frac{i-1}{1/2-1}$

$(\because$ dimer is formed n = 2)

$\Rightarrow \alpha =2(1-i)$

To calculate i (can't off factor ), we will use

$△T_f=i{K}_{f}m$

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

$\Rightarrow i=0.527$

$\therefore \alpha =2(1-i)=2(1-0.527)=0.945$

Thus, % association $=\alpha \times 100\%=94.5\%$

Answer is someone close to 94.6%

$\therefore$ correct Answer : (D)