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Standard XII

Chemistry

Relative Lowering of Vapour Pressure

Dry air was p...

Question

Dry air was passed successively through a solution of 5 g of the solute in 180 g of water and then through pure water. The loss in the weight of solution was 2.50 g and that of pure solvent 0.04 g. The molecular weight of the solute is

31.25

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3.125

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312.5

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None

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Solution

The correct option is A 31.25
$P^{\circ} - P_{S} \propto$ loss in weight of water chamber and $P_{S} \propto$ loss in weight of solution chamber.
$\frac{p^{\circ} - p_{S}}{p_{S}} = \frac{n_{2}}{n_{1}} = \frac{W_{2} \times M w_{1}}{M w_{2} \times W_{1}}$
or $\frac{0.04}{2.50} = \frac{5 \times 18}{M w_{2} \times 180}$
$∴ M w_{2} = 31.25$

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Dry air was passed successively through a solution of 5 g of the solute in 180 g of water and then through pure water. The loss in the weight of solution was 2.50 g and that of pure solvent 0.04 g. The molecular weight of the solute is

The correct option is A 31.25P∘−PS∝ loss in weight of water chamber and PS∝ loss in weight of solution chamber. p∘−pSpS=n2n1=W2×Mw1Mw2×W1 or 0.042.50=5×18Mw2×180 ∴Mw2=31.25