SolutionHard
Question
A mixture of liquids ‘A’ and ‘B’ in the molar ratio 1: 2 forms a maximum boiling azeotrope. Identify the incorrect statement, if ‘A’ is more volatile. (Molar masses: A = 100, B = 50).
Options
A.A liquid solution of ‘A’ and ‘B’ having mass% of A = 50 will have vapours having mass% of A = 50.
B.A liquid solution of ‘A’ and ‘B’ having mass% of A > 50 will have vapours having mass% of A > 50.
C.A mixture of ‘A’ and ‘B’ in the molar ratio 1: 3 can be separated into azeotropic mixture and pure ‘A’.
D.A mixture of ‘A’ and ‘B’ in the molar ratio 2: 3 can be separated into azeotropic mixture and pure ‘A’.
Solution
(a) Mass percent of A = 50 ⇒ nA: nB = 1: 2 ⇒ Azeotrope
Hence, vapour will have the same composition of liquid.
(b) Mass percent of A > 50 ⇒ nA: nB = 1: 2
In this case, the vapour must be more rich in A than liquid.
(c) $X_{B} = \frac{3}{4} = 0.75 > \frac{2}{3} \Rightarrow$ Pure A cannot be obtained.
(d) $X_{B} = \frac{3}{5} = 0.60 < \frac{2}{3} \Rightarrow$Pure A cannot be obtained in traces.
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