At T(K), 2 moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of ideal solution fonned is 320 mm Hg. At this stage, one mole of A and one mole of B are added to the solution. The vapour pressure is now measmed as 328.6 mm Hg. The vapom pressure (in mm Hg) of A and B are respectively:

Let $$P_A^0$$ and $$P_B^0$$ be vapor pressures of pure A and B.

Mixture 1: 2 mol A, 3 mol B. $$x_A = 2/5, x_B = 3/5$$. $$P = 0.4P_A^0 + 0.6P_B^0 = 320 \quad (1)$$

Mixture 2: 3 mol A, 4 mol B. $$x_A = 3/7, x_B = 4/7$$. $$P = \frac{3}{7}P_A^0 + \frac{4}{7}P_B^0 = 328.6 \quad (2)$$

From (2): $$3P_A^0 + 4P_B^0 = 2300.2 \quad (2')$$

From (1): $$2P_A^0 + 3P_B^0 = 1600 \quad (1')$$

Multiply (1') by 4/3: $$8P_A^0/3 + 4P_B^0 = 6400/3$$

Subtract from (2'): $$3P_A^0 - 8P_A^0/3 = 2300.2 - 2133.3$$

$$P_A^0/3 = 166.9 \implies P_A^0 = 500.7 \approx 500$$

From (1'): $$P_B^0 = (1600 - 1000)/3 = 200$$

The answer is Option 2: 500, 200.