Question

The weight in grams of a non-volatile solute (mol. wt. 60) to be dissolved in $90g$ 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.)
$\therefore 0.02=\frac{\frac{x}{60}}{\frac{x}{60}+\frac{90}{18}}$
$0.02=\frac{\frac{x}{60}}{\frac{x}{60}+5}$
$\frac{1}{0.02}=\frac{\frac{x}{60}+5}{\frac{x}{60}}$
$50=1+\frac{5\times 60}{x}$
$49=\frac{300}{x}$
$x=\frac{300}{49}=6g$