Question

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

• A. $10\, g$
• B. $11.4\, g$
• C. $9.8\, g$
• D. $12.8\, g$

Answer 1

Answer: (A)

$10\, g$

Answer 2

$\because \frac{p^{\circ}-p}{p}=\frac{w}{m} \times \frac{M}{W}$
$\therefore \frac{100-80}{80}=\frac{w}{40} \times \frac{114}{114}$

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

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