Two liquids, A and B, form an ideal solution. At the specified temperature, the vapour pressure of pure A is 200 mm Hg while that of pure B is 75 mm Hg. If the vapour over the mixture consists of 50 mol percent A, what is the mole percent A in the liquid?

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Solution

Given, in vapour phase mole fraction of liquid A = 50% = 0.5
Therefore, mole fraction of liquid B = 1-0.5 = 0.5
Now,
$0.5 = \frac{P_{A}}{P_{A} + P_{B}} - - - - - - (1) A n d 0.5 = \frac{P_{B}}{P_{A} + P_{B}} - - - - - - (2) D i v i d i n g (1) b y (2), 1 = P_{A} / P_{B} O r P_{A} = P_{B}$
Now,

$P_{A} = P_{A}^{°} x x_{A} A n d P_{B} = P_{B}^{°} x x_{B} S i n c e, P_{A} = P_{B} P_{A}^{°} x x_{A} = P_{B}^{°} x x_{B} 200 x x_{A} = 75 x x_{B} x_{A} = 0.375 x_{B}$

We know,
x_A + x_B = 1
0.375 x_B + xB = 1
x_B = 0.73
x_A= 1- 0.73
= 0.27
Hence, mole percent of A in the liquid = 27%

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