Question

A dry air is passed through the solution, containing $15 \,g$ of solute and $85 \,g$ of water and then it is passed through pure water. There is the depression in weight of solution by $2.5\, g$ and in weight of pure solution by $0.05\, g .$ The molecular weight of solute is______.

Answer 1

$\because$ Lowering in weight of solution $\propto$ solution pressure $\left(p_{s}\right)$ and lowering in weight of solvent $\propto p ^{\circ}- p _{ s }$

$\left(\because p^{\circ}=\right.$ vapour pressure of pure solvent)

Thus, $ \frac{p^{\circ}-p_{s}}{p_{s}}=\frac{\text { lowering in weight of solvent }}{\text { lowering in weight of solution }}$

$=\frac{0.05}{2.5}$

But according to Raoult's law,

$\frac{p^{\circ}-p_{s}}{p_{s}}=\frac{w_{2} \times M_{1}}{M_{2} \times w_{1}}$

$\frac{0.05}{2.5}=\frac{15 \times 18}{85 \times M_{1}} $

$M_{1}=\frac{15 \times 18 \times 2.5}{85 \times 0.05}=158.8 \,g$