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 $15 \text{ kN m}^{-2}$ 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?

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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 $15 \text{ kN m}^{-2}$ 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?

0.34

0.663

0.48

0.5217

The correct answer is

0.48

Solving for Mole Fraction in Ideal Solution Vapour Equilibrium

This problem requires determining the mole fraction of component A in an ideal solution based on the vapour pressures of the pure components and the composition of the vapour phase in equilibrium.

Problem Data Summary

Vapour pressure of pure A, $P_A^\circ = 55 \text{ kN m}^{-2}$
Vapour pressure of pure B, $P_B^\circ = 15 \text{ kN m}^{-2}$
Mole fraction of A in vapour, $y_A = 0.8$
Goal: Find the mole fraction of A in the solution, $x_A$.

Applying Raoult's and Dalton's Laws

For an ideal solution, Raoult's Law states that the partial vapour pressure of each component is proportional to its mole fraction in the liquid phase:

$P_A = x_A P_A^\circ$
$P_B = x_B P_B^\circ = (1 - x_A) P_B^\circ$

Dalton's Law of partial pressures relates the total pressure ($P = P_A + P_B$) to the mole fraction in the vapour phase:

$y_A = \frac{P_A}{P}$

Combining these laws, we get the relationship between the mole fraction in the liquid ($x_A$) and vapour ($y_A$) phases for an ideal solution:

$ y_A = \frac{x_A P_A^\circ}{x_A P_A^\circ + (1 - x_A) P_B^\circ} $

Calculation Steps

Substitute the given values into the combined equation:

$ 0.8 = \frac{x_A \times 55}{x_A \times 55 + (1 - x_A) \times 15} $

Simplify the equation:

$ 0.8 = \frac{55 x_A}{55 x_A + 15 - 15 x_A} $

$ 0.8 = \frac{55 x_A}{40 x_A + 15} $

Rearrange to solve for $x_A$:

$ 0.8 (40 x_A + 15) = 55 x_A $

$ 32 x_A + 12 = 55 x_A $

Group the $x_A$ terms:

$ 12 = 55 x_A - 32 x_A $

$ 12 = 23 x_A $

Solve for $x_A$:

$ x_A = \frac{12}{23} $

Calculating the numerical value:

$ x_A \approx 0.5217 $

The calculated mole fraction of A in the solution is approximately 0.5217.

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Substance	$\Delta G_f^\circ / \text{kJ mol}^{-1}$
$\text{A}_2$	-100.00
A	-50.832