QUESTION 2.18

Calculate the mass of a non-volatile solute (molar mass $40 g m o l^{- 1}$ ) which should be dissolved in 114g octane to reduce its vapour pressure to $80 %$ .

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Solution

According to Raoult's law, relative lowering of vapour pressure,
$\frac{p_{A}^{\circ} - p_{s}}{p_{A}^{\circ}} = x_{B} \dots \dots (i) x_{B} = \frac{n_{B}}{n_{B} + n_{A}} = \frac{\frac{W_{B}}{M_{B}}}{\frac{W_{B}}{M_{B}} + \frac{W_{A}}{M_{A}}} \dots \dots (i i)$
Given Vapour pressure is reduced to $80 %$ when non-volatile solute is dissolved in octane. It means
If $p_{A}^{\circ} = 1 a t m$ then $p_{s} = 0.8 a t m$ ; $p_{A}^{\circ} - p_{s} = 0.2 a t m$ ;

$M_{A} (C_{8} H_{18}) = 114 g m o l^{- 1}; W_{A} = 114 g; M_{B} = 40 g m o l^{- 1}; W_{B} = ?$

Applying equation (ii)
$0.2 = \frac{\frac{W_{B}}{40}}{\frac{W_{B}}{40} + \frac{114}{114}} = \frac{\frac{W_{B}}{40}}{\frac{W_{B}}{40} + 1} 0.2 = \frac{W_{B}}{W_{B} + 40} 0.2 W_{B} + 8 = W_{B}$

$W_{B} - 0.2 W_{B} = 8$

$0.8 W_{B} = 8$

$W_{B} = \frac{8}{0.8}$
$W_{B} = 10$ grams

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