Question

Dry air is passed through a solution containing 10 g of a solvent in 90 g of water and then through pure water. The loss in weight of the solution is 2.5 g and that of the pure solvent is 0.05 g. Calculate the molecular weight of the solvent.

• A. 50 g/mol
• B. 180 g/mol
• C. 100 g/mol
• D. 251 g/mol

Answer 1

The correct option is C 100 g/mol

Loss of weight of the solution $\propto$ V.P. of solution
Loss in weight of solvent $\propto P^0-P_s$

$\therefore \frac{P^0-P_s}{P_s}=\frac{\text{Loss in weight of solvent}}{\text{Loss in weight of solution}}\frac{\frac{w_2}{M_2}}{\frac{w_1}{M_1}}=\frac{w_2{M}_1}{w_1{M}_2}\frac{0.05}{2.5}=\frac{\frac{10}{M_2}}{\frac{90}{18}}=\frac{2}{M_2}$

$M_2$ = 100 g/mol