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.

50 g/mol

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180 g/mol

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100 g/mol

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251 g/mol

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Solution

The correct option is C 100 g/mol
Loss of weight of the solution $\propto$ V.P. of solution
Loss in weight of solvent $\propto P^{0} - P_{s}$

$∴ \frac{P^{0} - P_{s}}{P_{s}} = \frac{Loss in weight of solvent}{Loss in weight of solution} \frac{\frac{w_{2}}{M_{2}}}{\frac{w_{1}}{M_{1}}} = \frac{w_{2} M_{1}}{w_{1} M_{2}} \frac{0.05}{2.5} = \frac{\frac{10}{M_{2}}}{\frac{90}{18}} = \frac{2}{M_{2}}$

$M_{2}$ = 100 g/mol

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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.

The correct option is C 100 g/molLoss of weight of the solution ∝ V.P. of solution Loss in weight of solvent ∝P0−Ps ∴P0−PsPs=Loss in weight of solventLoss in weight of solutionw2M2w1M1=w2M1w1M20.052.5=10M29018=2M2 M2 = 100 g/mol