JEE Chemistry Trick: Vapour Phase Composition from Raoult’s Law Explained 💡
❓ Question Liquid A and B form an ideal solution. The vapour pressures of pure liquids A and B are 350 and 750 mm Hg respectively at the same temperature. If xA and xB are the mole fraction of A and B in solution while yA and yB are the mole fraction of A and B in vapour phase then 🖼️ Question Image ✍️ Short Solution For an ideal solution Raoult’s law applies: p A = x A p A ∗ , p B = x B p B ∗ p_A \;=\; x_A\,p_A^*,\qquad p_B \;=\; x_B\,p_B^* Total vapour pressure above the solution: p tot = p A + p B = x A p A ∗ + x B p B ∗ . p_{\text{tot}} \;=\; p_A + p_B \;=\; x_A p_A^* + x_B p_B^*. The mole fractions in the vapour (via Dalton’s law): y A = p A p tot = x A p A ∗ x A p A ∗ + x B p B ∗ , y_A \;=\; \frac{p_A}{p_{\text{tot}}} =\frac{x_A p_A^*}{x_A p_A^* + x_B p_B^*}, y B = p B p tot = x B p B ∗ x A p A ∗ + x B p B ∗ . y_B \;=\; \frac{p_B}{p_{\text{tot}}} =\frac{x_B p_B^*}{x_A p_A^* + x_B p_B^*}. Since x B = 1 − x A x_B=1-x_A ...