Question

The amount of solute (molar mass $60gmo{l}^{-1}$ ) that must be added to $180g$ of water so that the vapour pressure of water is lowered by $10\%$, is

• A. $30g$
• B. $60g$
• C. $120g$
• D. $12g$
• E. $24g$

Answer 1

Answer: (B)

Relative lowering of vapour pressure is given by the formula
$\frac{p{}^{\circ }-p_s}{p{}^{\circ }}=\frac{w_A}{M_A}\times \frac{M_B}{w_B}$
As vapour pressure of water is lowered by 10%.
$\therefore$ $\frac{p{}^{\circ }-p_s}{p{}^{\circ }}=\frac{10}{100}$
$\therefore$ $\frac{10}{100}=\frac{w_A}{60}\times \frac{18}{180}$
or $w_A=60g$