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Question
The vapour pressures of pure liquids A and B are 0.600 bar and 0.933 bar respectively, at a certain temperature.
What is the mole fraction of liquid B in the solution when the total vapour pressure of their mixture is 0.8 bar?
Solution
Given: Vapour pressure of pure liquid A `("P"_1^0)` = 0.600 bar
Vapour pressure of pure liquid B `("P"_2^0)` = 0.933 bar
Total pressure (P) = 0.8 bar
To find: Mole fraction of liquid B in the solution
Formula: P = `("P"_2^0 - "P"_1^0)x_2 + "P"_1^0`
0.8 bar = (0.933 bar − 0.600 bar) x2 + 0.600 bar
∴ 0.8 − 0.600 = 0.333 x2
∴ `x_2 = 0.2/0.333 = 0.6`
Mole fraction of liquid B in the solution is 0.6.
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