Dry air was passed successively through a solution of 5g of a solute in 80g water and then through pure water. The loss in weight of solution was 2.50g and that of pure solvent was  0.04g. The molecular weight of the solute would be 

Loss of weight of solution  $\propto p_s$
Loss of weight of solvent  $\propto p^0-p_s$
$\therefore \frac{p^0-p_s}{p_s}=\frac{Lossofweightofsolvent}{Lossofweightofsolution}$
$or{X}_B=\frac{Lossofweightofsolvent}{Lossofweightofsolution}$
$\frac{W_B/M_B}{W_A/M_A}=\frac{0.04}{2.5}\therefore \frac{W_B}{M_B}\times \frac{M_A}{W_A}=\frac{0.04}{2.5}$
Or  $M_B=\frac{2.5\times W_B\times M_A}{0.04\times W_A}=\frac{2.5\times 5\times 18}{0.04\times 80}=70.31$