Question

The mass of a non-volatile, non-electrolyte solute (molar mass $= 50\, g\, mol^{-1}$) needed to be dissolved in $114\, g$ octane to reduce its vapour pressure to $75\%,$ is :

• A. 37.5 g
• B. 75 g
• C. 150 g
• D. 50 g

Answer 1

Answer: (C)

150 g

Answer 2

Relative lowering of vapour pressure is given by

$\frac{\Delta p}{p}=x_{\text {solute }}=\frac{w_{ B } / M_{ B }}{w_{ A } / M_{ A }+w_{ a } / M_{ B }}$

where $w_{A}$ and $w_{ B }$ are the masses of solvent and solute taken and $M_{ A }$ and $M_{ B }$ are the molar masses of the solvent and solute.

$\frac{75}{100}=\frac{w_{ B } / 50}{w_{ B } / 50+114 / 114}$
$0.75=\frac{w_{ B } / 50}{w_{ B } / 50+1} \Rightarrow w_{ B }=150\, g$