At 20°C, the vapour pressure of diethyl ether is 442mm. When 6.4g. of a non-volatile solute is dissolved in 50g. of ether, the vapour pressure falls to 410mm. The Molecular weight of the solute is

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150

180

160

130.832

answer is B.

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Detailed Solution

$P^{o} - P = 442 - 410 = 32 W_{s} = 6.4 g, W_{o} = 50 g, M_{s} = ? \frac{P^{o} - P}{P^{o}} = \frac{W_{s}}{M_{s}} \times \frac{M_{o}}{W_{o}} \frac{32}{442} = \frac{6.4}{M_{s}} \times \frac{74}{50} M_{s} = \frac{6.4 \times 442 \times 74}{32 \times 50} = 130.832 g$

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