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Moles of Na₂SO₄ to be dissolved in 12 mole water to lower its vapour pressure by l0 mm Hg at a temperature at which vapour pressure of pure water is 50 mm is:

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2 mole

1.5 mole

3 mole

1 mole

answer is D.

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Detailed Solution

$\frac{P^{\circ} - P}{P^{\circ}} = \frac{n_{s}}{n_{s} + n_{o}}; here, s = solute, o = solvent \frac{10}{50} = \frac{n}{n + 12} ∴ n = 3$

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