Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and 15 kN respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A

Question

Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and 15 kN $m^{- 2}$ respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8? [2026]

Smart Achievers · Accepted Answer

(1)

$\frac{Y_{A}}{Y_{B}} = \frac{P_{A}^{0}}{P_{B}^{0}} \cdot \frac{X_{A}}{X_{B}}$

$\frac{0.8}{0.2} = \frac{55}{15} \times \frac{X_{A}}{X_{B}}$

$\frac{X_{A}}{X_{B}} = \frac{60}{55} = \frac{12}{11}$

$X_{A} = \frac{12}{23} = 0.5217$

Solution

Q.
Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and 15 kN $m^{- 2}$ respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8 [2026]

1 0.5217

2 0.340

3 0.663

4 0.480

Ans.
(1)

$\frac{Y_{A}}{Y_{B}} = \frac{P_{A}^{0}}{P_{B}^{0}} \cdot \frac{X_{A}}{X_{B}}$

$\frac{0.8}{0.2} = \frac{55}{15} \times \frac{X_{A}}{X_{B}}$

$\frac{X_{A}}{X_{B}} = \frac{60}{55} = \frac{12}{11}$

$X_{A} = \frac{12}{23} = 0.5217$

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