Question

$1$ mole of a non-volatile solid is dissolved in $200$ moles of water. The solution in taken to a temperature $T_k$ (lower than the freezing point of solution) to cause ice formation. After removing ice, the remaining solution is taken to $373K$ where the vapour pressure is observed to be $740\text{mm of Hg}$. Identify the correct options:
Given data: $K_f\left(H_{2}O\right)=2{\text{K kg mol}}^{-1}$ and normal boiling point of $H_{2}O=373K.$

• A. $163$ moles of ice will be formed
• B. $T_k=273-\frac{2000}{37\times 18}K.$
• C. Freezing point of the original solution should be $-\frac{10}{18}$ ${}^{\circ }C.$
• D. Relative lowering of vapour pressure of final solution will be $\frac{1}{201}$

Answer 1

Answer: (A), (B), (C)

Freezing point of the original solution should be $-\frac{10}{18}$ ${}^{\circ }C.$

Given, $n_{H_{2}O}=200$ moles
$T_f=373K$
$P=740\text{mm of Hg}$
$n_1=1\text{mole}$
$\left(a\right)\frac{P_0-P}{P}=\frac{n_1}{n_2}\Rightarrow \frac{760-740}{740}=\frac{1}{n_2}$
$\therefore$ moles of water separated as ice
$=200–37=163\text{moles}$

(B) Let $\Delta T_k=T^{\circ }C-T_k$
We know, $\Delta T=K_{f.}m$
$=2\times \frac{1}{(37\times 18)/1000}=\frac{2000}{37\times 18}$
$T=(273-\frac{2000}{37\times 18})K$
(C) For original solution
$\Delta T_f=K_{f.}m=2\times \frac{1}{(200\times 18)/1000}\therefore$ Freezing point
$=0-T_f=-\frac{10}{18}$ ${}^{\circ }C$
$\left(d\right)\frac{P_0-P}{P_0}=X_1=\frac{1}{(200-X)+1}=\frac{1}{(200-163)+1}$
$=\frac{1}{38}$