Question

A solution of urea (mol. mass $56{\text{g mol}}^{-1}$) boils at $100$. ${18}^0\text{C}$ at the atmospheric pressure. If $K_f\text{and}{K}_b$ for water are $1.86$ and $0.512{\text{K kg mol}}^{-1}$ respectively, the above solution will freeze at :

• A. ${0.654}^0\text{C}$
• B. $-{0.654}^0\text{C}$
• C. ${6.54}^0\text{C}$
• D. $-{6.54}^0\text{C}$

Answer 1

The correct option is B $-{0.654}^0\text{C}$

$\begin{array}{c}\Delta T_f=K_{f}m.......\left(1\right)\\ \Delta T_b=K_{b}m.......\left(2\right)\end{array}\}\Rightarrow \frac{\Delta T_f}{\Delta T_b}=\frac{K_f}{K_b}...\left(3\right)$

$\Delta T_f\to$ depression in freezing point
$\Delta T_b\to$ elevation in b.p.
B.P. of water = ${100}^0\text{C};K_f=1.86{\text{kg mol}}^{-1}$
B.P. of urea in water = ${100.18}^0\text{C}$
$K_b=0.512{\text{kg mol}}^{-1}\Rightarrow \Delta T_b=0.18$
F.P. of water = $0^0\text{C}$
F.P. of urea in water = $-T^0\text{C}$
$\Rightarrow \Delta T_f=T$
$\Rightarrow \text{from eq. (3),}\frac{T}{0.18}=\frac{1.86}{0.512}\Rightarrow T=0.6539$
$\Rightarrow \text{F.P. of urea in water}=-{0.654}^0\text{C}$