5 g of a nonvolatile solute in 100 g of water has a vapour pressure of 2980 Pa. If the vapour pressure of pure water is 3000 Pa, the molar mass of solute is

5 g of a nonvolatile solute in 100 g of water has a vapour pressure of 2980 Pa. If the vapour pressure of pure water is 3000 Pa, the molar mass of solute is

A
$120 g {mol}^{- 1}$
B
$125 g {mol}^{- 1}$
C
$130 g {mol}^{- 1}$
D
$134 g {mol}^{- 1}$

Solution:

Mole fraction of water in solution is

$x = \frac{(100 / 18) mol}{(5 g / M) + (100 / 18) mol}$

Hence, $p = x_{1} p_{1}^{*}$

$2980 Pa = \frac{(100 / 18) mol}{(5 g / M) + (100 / 18) mol} (3000 Pa)$ This gives $M = 134.1 g {mol}^{- 1}$

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