JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Solutions

For an ideal solution Raoult’s law gives $P=x_A{P}_A^{\circ }+x_B{P}_B^{\circ }$. With 2 mol A and 3 mol B, $x_A=\frac{2}{5}$ and $x_B=\frac{3}{5}$, so $\frac{2}{5}{P}_A^{\circ }+\frac{3}{5}{P}_B^{\circ }=320$, that is $2P_A^{\circ }+3P_B^{\circ }=1600$. After adding one mole of each the mixture is 3 mol A and 4 mol B, so $x_A=\frac{3}{7}$, $x_B=\frac{4}{7}$ and $\frac{3}{7}{P}_A^{\circ }+\frac{4}{7}{P}_B^{\circ }=328.6$, that is $3P_A^{\circ }+4P_B^{\circ }=2300.2$. Multiplying the first relation by 3 and the second by 2 and subtracting eliminates $P_A^{\circ }$ and gives $P_B^{\circ }=200$ mm Hg; substituting back gives $P_A^{\circ }=500$ mm Hg. The vapour pressures of pure A and pure B are 500 mm Hg and 200 mm Hg respectively.