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:
D. 500, 200
For an ideal solution obeying Raoult's Law, total pressure is P = P_A^o x_A + P_B^o x_B. Initially with 2 moles A and 3 moles B: x_A = 0.4, x_B = 0.6, and P = 320 = 0.4P_A^o + 0.6P_B^o After adding 1 mole each: 3 moles A and 4 moles B gives x_A = 3 7 , x_B = 4 7 , and P = 328.6 = 3 7 P_A^o + 4 7 P_B^o From the first equation: 3200 = 4P_A^o + 6P_B^o From the second: 2300.2 = 3P_A^o + 4P_B^o Solving: We find P_B^o = 200 mm Hg and P_A^o = 500 mm Hg.
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