solution of 2 volatile liquids A and B obey raoult's law at a certain temp. Its found that the total V.P of above two is 400mm of hg.The mole fraction of A in vapour phase is 0.45 and in liquid phase is 0.65 Then calculate V.P of pure liqiud A and B at the same temp.
Asked by banga71 | 17th May, 2017, 09:43: PM
Expert Answer:
According to Dalton's Law of Partial Pressures, partial Pressure of each component gas is proportional to their mole fractions in the vapour.
Hence,
Partial Pressure of A = Pa = 400×0.45 = 180
Pb = 400×(1-0.45) = 220
Also, Partial Pressure of the gas in the vapour = (Vapour Pressure) × (Mole Fraction of Liquid in solution)
Hence,
Vapour Pressure of A = Va = 180/0.65 ≈ 277
Vb = 220/(1-0.65) ≈ 629
Pb = 400×(1-0.45) = 220
Also, Partial Pressure of the gas in the vapour = (Vapour Pressure) × (Mole Fraction of Liquid in solution)
Vb = 220/(1-0.65) ≈ 629
Answered by Prachi Sawant | 18th May, 2017, 10:30: AM
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