An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molecular mass of the solute?

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20.8g $m o l^{- 1}$

10 g $m o l^{- 1}$

23.4g $m o l^{- 1}$

41.35g $m o l^{- 1}$

answer is B.

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Detailed Solution

Vapour pressure of pure water at boiling point = 1 atm = 1.013 bar

Vapour pressure of solution $(p_{S})$ =1.004 bar

Let mass of solution = 100g

Mass of solute = 2g

Mass of solvent = 100-2=98 g

$\frac{p^{o} - p_{s}}{p^{o}} = \frac{n_{2}}{n_{2} + n_{2}} = \frac{n_{2}}{n_{1}} = \frac{W_{2} / M_{2}}{W_{1} / M_{1}}$

$(∵ n_{2} < < < n_{1})$

$\frac{1.013 - 1.004}{1.013} = \frac{2}{M_{2}} X \frac{18}{98}$ or $M_{2} = \frac{2 X 18}{98} X \frac{1.013}{0.009} = 41.35 g$ $m o l^{- 1}$

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