Question

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

Answer 1

Answer: 158.8

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

$(\because p^{\circ }=$ 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,

$\frac{p^{\circ }-p_s}{p_s}=\frac{w_2\times M_1}{M_2\times w_1}$

$\frac{0.05}{2.5}=\frac{15\times 18}{85\times M_1}$

$M_1=\frac{15\times 18\times 2.5}{85\times 0.05}=158.8g$