\(2.5 \mathrm{~g}\) of a non-volatile, non-electrolyte is dissolved in \(100 \mathrm{~g}\) of water at…

\(2.5 \mathrm{~g}\) of a non-volatile, non-electrolyte is dissolved in \(100 \mathrm{~g}\) of water at \(25^{\circ} \mathrm{C}\). The solution showed a boiling point elevation by \(2^{\circ} \mathrm{C}\). Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is \(\qquad\) \(\mathrm{mm}\) of \(\mathrm{Hg}\) (nearest integer) [Given : Molal boiling point elevation constant of water \(\left(\mathrm{K}_{\mathrm{b}}\right)=0.52 \mathrm{~K} . \mathrm{kg} \mathrm{mol}^{-1}\), \(1 \mathrm{~atm}\) pressure \(=760 \mathrm{~mm}\) of \(\mathrm{Hg}\), molar mass of water \(=18 \mathrm{~g} \mathrm{~mol}^{-1}\)]

Solution

$\begin{aligned} & 2=0.52 \times \mathrm{m} \\ & \mathrm{m}=\frac{2}{0.52} \end{aligned}$
According to question, solution is much diluted $\begin{aligned} & \text { so } \frac{\Delta \mathrm{P}}{\mathrm{P}^{\circ}}=\frac{\mathrm{n}_{\text {solute }}}{\mathrm{n}_{\text {solvent }}} \\ & \frac{\Delta \mathrm{P}}{\mathrm{P}^{\circ}}=\frac{\mathrm{m}}{1000} \times \mathrm{M}_{\text {solvent }} \\ & \Delta \mathrm{P}=\mathrm{P}^{\circ} \times \frac{\mathrm{m}}{1000} \times \mathrm{M}_{\text {solvent }} \\ & \quad=760 \times \frac{\frac{2}{0.52}}{1000} \times 18=52.615 \\ & \mathrm{P}_5=760-52.615=707.385 \mathrm{~mm} \text { of } \mathrm{Hg} \end{aligned}$

Asked in: JEE Main 2024 (04 Apr Shift 1)

Practice more Solutions questions on Aicharya