The vapour pressure of a solvent at 283 K is 100 mm Hg. Calculate the vapour pressure of a dilute solution containing 1 mole of a strong electrolyte…

\[\begin{aligned} & \frac{p^{\circ}-p_{\mathrm{s}}}{\mathrm{P}^{\circ}}=i \times \chi \\ & A B \longrightarrow A^{+}+B^{-} \\ & \quad i=2 \end{aligned} \] \(\displaystyle \frac{100-\mathrm{P}_{S}}{100}=2 \times \frac{n_{A B}}{n_{\text {solvent }}}\) (For dilute solution) \[\begin{gathered} \frac{100-\mathrm{P}_{\mathrm{s}}}{100}=2 \frac{1}{50} \\ 100-\mathrm{P}_{\mathrm{s}}=4 \end{gathered} \] \(\displaystyle \mathrm{P}_{\mathrm{s}}=96 \mathrm{~mm} \mathrm{Hg}\)