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 solute when the total vapour pressure of their mixture is 0.8 bar?​

Answer 1

Explanation:

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Answer 2

Given:

The vapour pressures of pure liquids A and B are 0.600 bar and 0.933 bar respectively, at a certain temperature.

To find:

The mole fraction of solute when the total vapour pressure of their mixture is 0.8 bar?

Calculation:

Let liquid B be solute:

Let total pressure be P.

$P =( P_{A}^{0} \times \chi_{A}) + (P_{B}^{0} \times \chi_{B})$

$\implies P = \{ P_{A}^{0} \times( 1 - \chi_{A})\} + (P_{B}^{0} \times \chi_{B})$

$\implies P = P_{A}^{0} -(P_{A}^{0} \times \chi_{B})+ (P_{B}^{0} \times \chi_{B})$

$\implies P = P_{A}^{0} - \{(P_{A}^{0} - P_{B}^{0} )\times \chi_{B} \}$

$\implies 0.8 = 0.6 - \{(0.6 - 0.933) \times \chi_{B} \}$

$\implies 0.8 = 0.6 - \{ (- 0.333) \times \chi_{B} \}$

$\implies 0.2 = ( 0.333) \times \chi_{B}$

$\implies \chi_{B} \approx 0.6$

So, mole fraction is solute is 0.6 .