The mass of a non-volatile solute of molar ma | vedclass

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(D) According to Raoult's law for relative lowering of vapour pressure:$\frac{p^{\circ} - p}{p^{\circ}} = \frac{n}{n + N} \approx \frac{n}{N} = \frac{

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$10$

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(D) According to Raoult's law for relative lowering of vapour pressure:$\frac{p^{\circ} - p}{p^{\circ}} = \frac{n}{n + N} \approx \frac{n}{N} = \frac{w 	imes M}{m 	imes W}$Given:$p^{\circ} = 100, \quad p = 100 - 20 = 80$$m = 40 \ g \ mol^{-1}, \quad M(C_{8}H_{18}) = 8 	imes 12 + 18 	imes 1 = 114 \ g \ mol^{-1}$$W = 114 \ g, \quad w = ?$Substituting the values:$\frac{100 - 80}{100} = \frac{w 	imes 114}{40 	imes 114}$$\frac{20}{100} = \frac{w}{40}$$w = \frac{0.2 	imes 40} = 8 \ g$Wait,checking the formula $\frac{p^{\circ} - p}{p} = \frac{w 	imes M}{m 	imes W}$:$\frac{20}{80} = \frac{w 	imes 114}{40 	imes 114}$$0.25 = \frac{w}{40}$$w = 0.25 	imes 40 = 10 \ g$

The mass of a non-volatile solute of molar mass $40 \ g \ mol^{-1}$ that should be dissolved in $114 \ g$ of octane to lower its vapour pressure by $20 \%$ is (in $g$)

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