Question

The weight in grams of a non-volatile solute (mol. wt. 60) to be dissolved in $90 \,g$ of water to produce a relative lowering of vapour pressure of $0.02$ is

• A. 4
• B. 8
• C. 6
• D. 10

Answer 1

Answer: (C)

Relative lowering of vapour pressure,
$\frac{p^{\circ}-p_{s}}{p^{\circ}} =x_{A} $
$=\frac{\frac{w_{A}}{m_{A}}}{\frac{w_{A}}{m_{A}}+\frac{w_{B}}{m_{B}}}$

(where, $w_{A}$ and $m_{A}$ are the mass and molar mass of solute and $w_{B}$ and $m_{B}$ are the mass and molar mass of water.)
$ \therefore 0.02=\frac{\frac{x}{60}}{\frac{x}{60}+\frac{90}{18}} $
$0.02 =\frac{\frac{x}{60}}{\frac{x}{60}+5} $
$ \frac{1}{0.02} =\frac{\frac{x}{60}+5}{\frac{x}{60}} $
$50=1+\frac{5 \times 60}{x} $
$ 49 =\frac{300}{x} $
$ x=\frac{300}{49}=6\, g $