Question

What will be the vapour pressure lowering caused by addition of $50\text{g}$ of a non volatile solute $\left(\text{molecular mass = 250 g/mol}\right)$ to $504\text{g}$ water, if the vapour pressure of pure water at ${25}^{o}C$ is nearly $24\text{mm Hg}$.

• A. $0.17\text{mm Hg}$
• B. $0.02\text{mm Hg}$
• C. $1.34\text{mm Hg}$
• D. $4.12\text{mm Hg}$

Answer 1

The correct option is A $0.17\text{mm Hg}$

According to Raoult's law,
$\frac{P_0-P_s}{P_0}=\frac{n}{n+N}$
Where,
$P_0=\text{Vapour pressure of water}{P}_s=\text{Vapour pressure of solution}n=\text{number of moles of solute}N=\text{number of moles of water}$
$\Rightarrow \Delta P=\frac{n}{n+N}\cdot P_0$
Given: $n=\frac{50}{250}=0.2;N=\frac{504}{18}=28\text{and}{p}_0=24$
Substituting the values in the above equation,
$\Delta P=\frac{0.2}{0.2+28}\times 24\approx 0.17\text{mm Hg}$