Question

Find the vapour pressure of solution where $0.5g$ of a non-volatile solute (Mol.wt $=65$ g ) is dissolved in $100mLCCl_{4}.$
Given:
Density of $CCl_{4}=1.58g/cm^{3}$
Vapour pressure of $CCl_{4}$ at $25^\circ C$ is $143mmHg$ .

Answer 1

Mass of $CCl _4( W )=$ Volume $\times$ density $=100 \times 1.58=158 g$
Mass of solute $\left(\right.w\left.\right)=0.5g$
$P_{0}=143mmHg;P_{s}=?$
As we know $\frac{P_{0} - P_{S}}{P_{0}}=\frac{w \times M}{m \times W}$
$\frac{143 - P_{s}}{143}=\frac{0 . 5 \times 154}{65 \times 158}$
$143-P_{S}=\frac{0 . 5 \times 154 \times 143}{65 \times 158}$
$143-P_{s}=1.072$
$P_{s}=143-1.072=141.927mmHg$