QUESTION 2.15

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 molar mass of the solute?

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Solution

Relative lowering in vapour pressure.
$\frac{p_{A}^{\circ} - p_{S}}{p_{A}^{\circ}} = \frac{n_{B}}{n_{A}} = \frac{W_{B}}{M_{B}} \times \frac{M_{A}}{W_{A}} \dots \dots (i) p_{A}^{\circ} (w a t e r) = 1.013 b a r; W_{B} = 2 g; M_{A} = 18 g m o l^{- 1} p_{s} (w a t e r) = 1.004 b a r; W_{A} = (100 - 2) = 98 g; M_{B} = ?$
Placing the values in equation (i)
$\frac{(1.013 - 1.004) b a r}{(1.013 b a r)} = \frac{(2 g) \times (18 g m o l^{- 1})}{M_{B} \times (98) g} M_{B} = \frac{(2 g) \times (18 g m o l^{- 1}) \times (1.013 b a r)}{(0.009 b a r) \times (98) g}$
$= 41.35 g m o l^{- 1}$

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QUESTION 2.15 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 molar mass of the solute?

QUESTION 2.15

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 molar mass of the solute?