Solutions Part 2

Concept:Raoult's law for an ideal solution: $P_{\text{total}} = x_A P_A^0 + x_B P_B^0$, where $x$ is the mole fraction and $P^0$ is the vapour pressure of the pure component.Explanation:First mixture: 2 moles of A and 3 moles of B. Total moles = 5. Mole fractions: $x_A = \frac{2}{5}$, $x_B = \frac{3}{5}$. Given $P_{\text{total}} = 320$ mm Hg: $\frac{2}{5}P_A^0 + \frac{3}{5}P_B^0 = 320$. Multiply by 5: $2P_A^0 + 3P_B^0 = 1600$ ...(1).After adding 1 mole each of A and B: A = 3 moles, B = 4 moles, total = 7 moles. New mole fractions: $x_A = \frac{3}{7}$, $x_B = \frac{4}{7}$. New $P_{\text{total}} = 328.6$ mm Hg: $\frac{3}{7}P_A^0 + \frac{4}{7}P_B^0 = 328.6$. Multiply by 7: $3P_A^0 + 4P_B^0 = 2300.2$ ...(2).Solve the equations: Multiply (1) by 3: $6P_A^0 + 9P_B^0 = 4800$. Multiply (2) by 2: $6P_A^0 + 8P_B^0 = 4600.4$. Subtract: $P_B^0 = 4800 - 4600.4 = 199.6 \approx 200$ mm Hg. Substitute $P_B^0$ in (1): $2P_A^0 + 3(199.6) = 1600$ → $2P_A^0 + 598.8 = 1600$ → $2P_A^0 = 1001.2$ → $P_A^0 = 500.6 \approx 500$ mm Hg.Answer:Vapour pressures: $P_A^0 = 500$ mm Hg and $P_B^0 = 200$ mm Hg. Correct option: C.