Question

Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and pure B are 55 and 15 kPa 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?

• A. 0.340
• B. 0.663
• C. 0.480
• D. 0.5217

Hint:

For ideal solutions, always combine Raoult’s law with Dalton’s law to relate liquid and vapour compositions.

Answer 1

The Correct Option is D

Step 1: Applying Raoult’s law.
For an ideal solution, partial vapour pressure of A is given by:
$p_A = x_A P_A^0$
Step 2: Using Dalton’s law of partial pressures.
Mole fraction of A in vapour phase is given as:
$y_A = \dfrac{p_A}{p_A + p_B} = 0.8$
Step 3: Substituting values.
$p_A = 55x_A$ and $p_B = 15(1 - x_A)$
\[ 0.8 = \frac{55x_A}{55x_A + 15(1 - x_A)} \]
Step 4: Solving the equation.
\[ 0.8(55x_A + 15 - 15x_A) = 55x_A \]
\[ 0.8(40x_A + 15) = 55x_A \]
\[ 32x_A + 12 = 55x_A \]
\[ 23x_A = 12 \]
\[ x_A = 0.5217 \]
Step 5: Conclusion.
The mole fraction of A in the liquid phase is 0.5217.