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(C) The loss in weight of the solution is proportional to the vapour pressure of the solution $(p)$.The loss in weight of the solvent is proportional

Vedclass · Accepted Answer

$100$

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(C) The loss in weight of the solution is proportional to the vapour pressure of the solution $(p)$.The loss in weight of the solvent is proportional to the lowering of vapour pressure $(p^0 - p)$.According to the Ostwald-Walker method,the relative lowering of vapour pressure is given by:$\frac{p^0 - p}{p} = \frac{	ext{Loss in weight of solvent}}{	ext{Loss in weight of solution}}$Also,$\frac{p^0 - p}{p} = \frac{w 	imes M_0}{m 	imes W_0}$,where $w$ is the mass of solute,$m$ is the molar mass of solute,$W_0$ is the mass of solvent,and $M_0$ is the molar mass of water $(18 \ g/mol)$.Substituting the values:$\frac{0.05}{2.5} = \frac{10 	imes 18}{m 	imes 90}$$0.02 = \frac{180}{90m}$$0.02 = \frac{2}{m}$$m = \frac{2}{0.02} = 100 \ g/mol$.

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

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