Question

When a $0.374m$ of tetrachloroethane is added to benzene, the solution freezes at ${3.55}^{\circ }C$. Find the $K_f$ of benzene?
[Pure benzene freezes at ${5.45}^{o}C$]

• A. $505Kkgmo{l}^{-1}$
• B. $5.08Kkgmo{l}^{-1}$
• C. $0.508Kkgmo{l}^{-1}$
• D. ${50.8}^{\circ }Ckgmo{l}^{-1}$

Answer 1

The correct option is B $5.08Kkgmo{l}^{-1}$

Freezing point depression of a solution is given by,
$△T_f=K_f\times m$
where,
$△T_f$ is the depression in freezing point.
$K_f$ is molal depression constant.
m is molality of the solution.

$△T_f=5.45-3.55$
$△T_f=1.9K$
Given,
$\text{Molality of tetrachloroethane}=0.374m$
$K_f=\frac{△T_f}{m}$

$K_f=\frac{1.9}{0.374}=5.087Kkg/mol$
$\therefore$
Option (B) is correct.