Dry air was passed successively through a solution of 5 g of a solute in 80 g of water and then thro

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Dry air was passed successively through a solution of 5 g of a solute in 80 g of water and then through pure water. The loss in mass of solution was 2.5 g and that of pure solvent was 0.04 g. What is the molecular mass of the solute?

Accepted Answer

Loss in mass of solution$\propto$P and loss in mass of water$\propto$ (P° – P).

$\therefore\frac{\text{Loss in mass of water}}{\text{Loss in mass of solution}} = \frac{P^{o} - P}{P} = \frac{n_{1}}{n_{2}}$

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