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  4. 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

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Q. 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

KCETKCET 2012Solutions

Solution:

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$\because \frac{p^{\circ}-p}{p}=\frac{w}{m} \times \frac{M}{W}$

$\therefore \frac{100-80}{80}=\frac{w}{40} \times \frac{114}{114}$

or $\frac{20}{80}=\frac{w}{40}$

or $w=\frac{20 \times 40}{80}=10\, g$