SolutionHard
Question
At 80oC, the vapour pressure of pure liquid ′A′ is 520 mm Hg and that of pure liquid ′B′ is 1000 mm Hg. If a mixture solution of ′A′ and ′B′ boils at 80oC and 1 atm pressure, the amount of ′A′ in the mixture is (1 atm = 760 mm Hg)
Options
A.52 mol percent
B.34 mol percent
C.48 mol percent
D.50 mol percent
Solution
PT = PAoXA + PBoXB
760 = 520XA + PBo (1 - XA)
⇒ XA = 0.5
Thus, mole% of A = 50%
760 = 520XA + PBo (1 - XA)
⇒ XA = 0.5
Thus, mole% of A = 50%
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