Dry air is passed through a solution containi | vedclass

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(C) According to the Ostwald-Walker method,the relative lowering of vapour pressure is given by the ratio of the loss in weight of the solvent to the

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$100$

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(C) According to the Ostwald-Walker method,the relative lowering of vapour pressure is given by the ratio of the loss in weight of the solvent to the loss in weight of the solution.$\frac{P^0 - P_s}{P^0} = \frac{	ext{Loss in weight of solvent}}{	ext{Loss in weight of solution}} = \frac{0.05}{2.5}$Also,by Raoult's Law,$\frac{P^0 - P_s}{P^0} = \frac{n}{n + N} \approx \frac{n}{N} = \frac{w 	imes M}{m 	imes W}$Where $w = 10 \ g$ (solute),$W = 90 \ g$ (solvent),$M = 18 \ g/mol$ (molar mass of water),and $m$ is the molar mass of the solute.$\frac{0.05}{2.5} = \frac{10 	imes 18}{m 	imes 90}$$\frac{1}{50} = \frac{180}{90m} = \frac{2}{m}$$m = 2 	imes 50 = 100 \ g/mol$.

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

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