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 \(K_{b}=0.76 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) )

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), \(p=724 \mathrm{~mm}\)