At T(K), 2 moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of ideal solution formed 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 measured as 328.6 mm Hg. The vapour pressure (in mm Hg) of A and B are respectively:
- 400, 300
- 600, 400
- 500, 200
- 300, 200
Answer
(3) 500, 200
Solution
For an ideal solution: Ptotal = pA0xA + pB0xB
Solution 1: xA = 25, xB = 35
25pA0 + 35pB0 = 320 2pA0 + 3pB0 = 1600 … (i)
Solution 2 (3 mol A, 4 mol B): xA = 37, xB = 47
37pA0 + 47pB0 = 328.6 3pA0 + 4pB0 = 2300.2 … (ii)
3 × (i) − 2 × (ii): 9pB0 - 8pB0 = 4800 - 4600.4
pB0 = 199.6 200 mm Hg
From (i): 2pA0 = 1600 - 598.8 = 1001.2 pA0 = 500.6 500 mm Hg
Answer: (3)