Two components A and B have their pure vapor pressures in the ratio 1 ∶ 4 and respective mole fractions in solution in ratio 1 ∶ 2. What is the mole fraction of component B in vapor phase?(a) 0.8889(b) 0.1250(c) 0.8000(d) 0.2000This question was addressed to me during an internship interview.The query is from Colligative Properties and Determination of Molar Mass topic in portion Solutions of Chemistry – Class 12

Two components A and B have their pure vapor pressures in the ratio 1

1.	Two components A and B have their pure vapor pressures in the ratio 1 ∶ 4 and respective mole fractions in solution in ratio 1 ∶ 2. What is the mole fraction of component B in vapor phase?(a) 0.8889(b) 0.1250(c) 0.8000(d) 0.2000This question was addressed to me during an internship interview.The query is from Colligative Properties and Determination of Molar Mass topic in portion Solutions of Chemistry – Class 12
Answer» The correct option is (a) 0.8889 Explanation: Given, 4P^0A = P^0A ———- (1) 2XA = XB ———— (2) Using law of relative lowering of vapor pressure: PA = P^0A x XA ——- (3) PB = P^0B x XB ——- (4) Using values of (1) and (2) in (4) PA = P^0A x XA PB = 4P^0Ax 2XA =8(P^0A x XA) Therefore, PB = 8PA Mole fraction of component B in vapor phase, YB = PB/(PA + PB) On simplifying, YB = 8PA/(PA + 8PA) = 8/9 = 0.8889.

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