Question

17. Relation between relative lowering of vapour pressure and molar mass.

Answer 1

Consider a solution in which $W_{2g}$ of solute of molar mass $M_2$ is dissolved in $W_{1g}$ of solvent of molar mass $M_1$.

The relation between relative lowering of vapour pressure and molar mass is

$\frac{P_1^0-P}{P_1^0}=\frac{W_2{M}_1}{W_1{M}_2}$

$\therefore \frac{\Delta P}{P_1^0}=\frac{n_2}{n_1}=\frac{\frac{W_2}{M_2}}{\frac{W_1}{M_1}}=\frac{W_2{M}_1}{W_1{M}_2}$

Hence, by determining the vapour pressure of a pure solvent and a solution, the molar mass of a non-volatile solute can be measured.