Question

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

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

Answer 1

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

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