Question
At 80o C, 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 80o C and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg)
-
52 mol percent
-
34 mol percent
-
48 mol percent
-
50 mol percent
Solution
D.
50 mol percent
PT = PoXA + PoXB
760 = 520XA+ PoB(1-XA)
⇒ = 0.5
Thus, mole% of A = 50%
