JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Solutions

Raoult's law gives the solution's vapour pressure as $P_s=P_A^{\circ }{x}_A+P_B^{\circ }{x}_B$. With one mole of A and three of B, $x_A=1/4$ and $x_B=3/4$, so $500=200\times \frac{1}{4}+P_B^{\circ }\times \frac{3}{4}=50+\frac{3}{4}{P}_B^{\circ }$, giving $P_B^{\circ }=600\text{mm Hg}$. A higher pure vapour pressure marks the more volatile liquid, so against $P_B^{\circ }=600$ the lower $P_A^{\circ }=200$ identifies A as the least volatile; the pure vapour pressure of B is $600\text{mm Hg}$ and the least volatile component is A.