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Question

# Consider a binary mixture of two ideal volatile liquids A and B having vapour pressure of 300 mm Hg and 600 mm Hg in their pure state. What will be the mole fraction of component A in vapour phase (yA) if the total pressure (PT) of the solution is 400 mm Hg?

A
0.2
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B
0.5
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C
0.8
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D
0.9
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Solution

## The correct option is B 0.5χA=yA×PTp∘A χA is the mole fraction of component A in liquid phase χB=yB×PTp∘B χB is the mole fraction of component B in liquid phase Also χA+χB=1 So, 1PT=yAp∘A+yBp∘B1PT=yAp∘A+1−yAp∘B Solving, PT=p∘A×p∘Bp∘A+(p∘B−p∘A)×yA Here , PT=400 mm Hgp∘A=300 mm Hgp∘B=600 mm Hg 400=300×600300+(600−300)×yA4=61+yA4+4yA=6yA=0.5 (b) is the correct option

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