The vapour pressure of acetone at 20 °C is 185 Torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20 °C, its vapour pressure was 193 Torr. The molar mass of the substance is - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
$32 g {mol}^{- 1}$
B
$64 g {mol}^{- 1}$
C
$128 g {mol}^{- 1}$
D
$488 g {mol}^{- 1}$

Solution:

According to Raoult's law,

$p_{1} = x_{1} p_{1}^{*} = (1 - x_{2}) p_{1}^{*} i.e. x_{2} = 1 - \frac{p_{1}}{p_{1}^{*}} = 1 - \frac{183 Torr}{185 Torr} = \frac{2}{185}$ Also, $x_{2} = \frac{n_{2}}{n_{1} + n_{2}} = \frac{(1.2 g / M)}{(100 g / 58 g {mol}^{- 1}) + (1.2 g / M)}$

Hence, $\frac{(1.2 g / M)}{(100 g / 58 g {mol}^{- 1}) + (1.2 g / M)} = \frac{2}{185}$

This gives

$\frac{1.2 g}{M} (1 - \frac{2}{185}) = (\frac{2}{185}) (\frac{100 g}{58 g {mol}^{- 1}})$ or $M = (1.2 g) (\frac{183}{185}) (\frac{185}{2}) (\frac{58 g {mol}^{- 1}}{100 g}) = 63.68 g {mol}^{- 1}$

Approixmate value Writing $x_{2} ≃ n_{2} / n_{1}, we get \frac{2}{185} = \frac{1.2 g / M}{(100 g / 58 g {mol}^{- 1})} .$

This gives $M = (1.2 \times \frac{58}{100} \times \frac{185}{2}) g {mol}^{- 1} = 64.38 g {mol}^{- 1}$

Solution:

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