Question

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

• A. $30\, g$
• B. $60\, g$
• C. $120\, g$
• D. $12\, g$

Answer 1

Answer: (B)

$60\, g$

Answer 2

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 ${{w}_{A}}=60\text{ }g$