For a dilute solution containing $2.5 \mathrm{~g}$ of a non-volatile non-electrolyte solute in $100…
For a dilute solution containing $2.5 \mathrm{~g}$ of a non-volatile non-electrolyte solute in $100 \mathrm{~g}$ of water, the elevation in boiling point at 1 atm pressure is $2^{\circ} \mathrm{C}$. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure ( $\mathrm{mm}$ of $\mathrm{Hg}$ ) of the solution is (take $K_{b}=0.76 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )
724
740
736
718
Solution
From Raoult's law,
$\frac{p^{\circ}-p}{p^{\circ}}=\frac{\text { No. of moles of solute }}{\text { No. of moles of solvent }+\text { No. of moles of solute }}$
When the concentration of solute is much lower than the concentration of solvent,
$\begin{array}{l}
\frac{p^{\circ}-p}{p^{\circ}}=\frac{\text { No. of moles of solute }}{\text { No. of moles of solvent }} \\
\frac{760-p}{760}=\frac{2.5 / m}{100 / 18}...(i)
\end{array}$
From elevation in boiling point, $\Delta T_{b}=K_{b} \times m$
$\begin{array}{l}
2=0.76 \times m \\
m=\frac{2}{0.76}...(ii)
\end{array}$
From(i) and (ii), $p=724 \mathrm{~mm}$