Question

Calculate the vapour pressure lowering caused by addition of $50\text{g}$ of sucrose $\left(\text{molecular mass = 342}\right)$ to $500\text{g}$ water, if the vapour pressure of pure water at ${25}^{o}C$ is $23.8\text{mm Hg}$.

Answer 1

According to Raoult's law,
$\frac{P_0-P_s}{P_0}=\frac{n}{n+N}$
Where,
$P_0=\text{Vapour pressure of water}{P}_s=\text{Vapour pressure of solution}n=\text{number of moles of sucrose}N=\text{number of moles of water}$
$\Rightarrow \Delta P=\frac{n}{n+N}\cdot P_0$
Given: $n=\frac{50}{342}=0.146;N=\frac{500}{18}=27.78\text{and}{p}_0=23.8$
Substituting the values in the above equation,
$\Delta P=\frac{0.146}{0.146+27.78}\times 23.8=0.124\text{mm Hg}$