Question

The weight of a non-volatile solute (m. wt. 40). Which should be dissolved in $114g$ octane to reduce its vapour pressure to $80\%$, is-

• A. $20g$
• B. $30g$
• C. $10g$
• D. $40g$

Answer 1

The correct option is C $10g$

According to Roults law of relative lowering of vapour pressure:
$\frac{p_1^0-p_1}{p_1^0}=\frac{w_1/MW{t}_1}{w_1/M.w{t}_1+w_2/M.w{t}_2}$
let relative vapour pressure of solvent octane $p_1^0$ = x
vapour pressure of solute $p_1$ = $\frac{80}{100}\ast x=0.8x$
$w_1,M.w{t}_1$ of solute; $w_2,M.w{t}_2$ of solvent.
$\frac{(x-0.8x)}{x}=\frac{w_1/40}{w_1/40+\left(114/114\right)}$
$\frac{0.8w_1}{40}=0.2$
$\therefore$ the weight of solute required = 10g