Two liquids A and B are miscible over the whole range of composition and may be treated as ideal (obeying Raoult’s law.) At 350 K the vapour pressure of pure A is 24.0 kPa and of pure B is 12.0 kPa. A mixture of 60% A and 40% B is distilled at this temperature. A small amount of the distillate is collected and and redistilled at 350 K; what is the mole percent of B in the second distillate ?
Text Solution
Verified by Experts14
(14)
P T = 
× 24 + 
× 12 = 
= 
= 19.2 kPa
P A ′ = X A P A = 
× 24 = 
= 14.4 = X A ′ × 19.2
so X A ′ = 
= 75% X B ′ = 25%
so P T = 
× 24 + 
× 12 = 18 + 3 = 21 k Pa
& P A = 
× 24 = 18 = X A ′ (21)
so X A ′ = 
= 
= 85.7% of A ⇒ X B ′ = 14.3%
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