Question

A solution of urea in water has a boiling point of ${100.18}^{\circ }C.$ Calculate the freezing point of the same solution. Molal depression and elevation constants for water $K_f$ and $K_b$ are ${1.86}^{\circ }Ckgmo{l}^{-1}$ and ${0.512}^{\circ }Ckgmo{l}^{-1}$ respectively.

• A. $-{0.654}^{\circ }C$
• B. $-{0.98}^{\circ }C$
• C. $-{1.54}^{\circ }C$
• D. $+{0.101}^{\circ }C$

Answer 1

The correct option is A $-{0.654}^{\circ }C$

Boiling point elevation of a solution is given by,
$△T_b=K_b\times m$
where,
$△T_b$ is the elevation in boiling point.
$K_b$ is molal elevation constant.
m is molality of solution.

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 solution.

For a same solution of same molality m, we have,
$m=\frac{△T_f}{K_f}=\frac{△T_b}{K_b}$
$\therefore △T_f=△T_b\frac{K_f}{K_b};(△T_b=100.18-100={0.18}^{\circ }C)$
$=0.18\times \frac{1.86}{0.512}={0.654}^{\circ }C$
As the freezing point of pure water is $0^{\circ }C,$
The freezing point of the solution will be $-{0.654}^{\circ }C.$