SolutionHard
Question
The vapour pressure of a pure liquid A is 40 mmHg at 310 K. The vapour pressure of this liquid in a solution with liquid B is 32 mmHg. Mole fraction of A in the solution, if it obeys Raoult′s law is :
Options
A.0.8
B.0.5
C.0.2
D.0.4
Solution
pA = XA poA
32 = XA 40
∴ XA =
= 0.8.
32 = XA 40
∴ XA =
= 0.8.Create a free account to view solution
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