Two liquids A and B form an ideal solution. At 320 K, the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is 500 mm Hg. At the same temperature, if 1 mol of A is further added to this solution, vapour pressure of the solution increases by 20 mm Hg. Vapour pressure (in mm Hg) of B in pure state is . (Nearest integer)
Answer
200
Solution
Raoult's law: p = xA p A + xB p B
Solution 1 (xA = 34, xB = 14): 34 p A + 14 p B = 500
3p A + p B = 2000 ...(i)
Solution 2 (xA = 45, xB = 15): 45 p A + 15 p B = 520
4p A + p B = 2600 ...(ii)
(ii) − (i): p A = 600 mm Hg
p B = 2000 - 3(600) = 200 mm Hg
Answer: 200