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A solution contains 10 g of a solute and 116 g of acetone at 30 °C. Its vapour pressure is 290 Torr. If the vapour pressure of pure acetone is 300 Torr at 30 °C, the molar mass of solute is

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$75 g {mol}^{- 1}$

$100 g {mol}^{- 1}$

$125 g {mol}^{- 1}$

$150 g {mol}^{- 1}$

answer is D.

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Detailed Solution

$x_{2} = - \frac{Δ p}{p_{1}^{*}} = \frac{10 Torr}{300 Torr} = \frac{1}{30}$ Now $x_{2} ≃ \frac{(m_{2} / M_{2})}{(m_{1} / M_{1})}$

Hence, $\frac{1}{30} = \frac{(10 g / M_{2})}{(116 g / 58 g {mol}^{- 1})}$ This gives $M_{2} = 150 g {mol}^{- 1}$

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