Question

If pure liquids A and B have vapour pressures of 55 kPa and 15 kPa respectively. If in a solution of A and B, mole fraction of A in vapour is 0.8, then find mole fraction of A in liquid phase.

• A. 0.813
• B. 0.5217
• C. 0.407
• D. 0.363

Hint:

Vapour phase is always richer in the more volatile component.

Answer 1

The Correct Option is B

Step 1: State Raoult’s law and Dalton’s law

According to Raoult’s law:

P_A = P_A⁰ x_A,   P_B = P_B⁰ x_B

According to Dalton’s law:

P_total = P_A + P_B

Mole fraction of A in vapour phase:

y_A = P_A / P_total

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Step 2: Write given data

P_A⁰ = 55 kPa
P_B⁰ = 15 kPa
y_A = 0.8

Also,

x_A + x_B = 1 ⇒ x_B = 1 − x_A

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Step 3: Express vapour-phase mole fraction

y_A =

P_A⁰x_A / (P_A⁰x_A + P_B⁰x_B)

Substituting values:

0.8 = 55x_A / [55x_A + 15(1 − x_A)]

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Step 4: Solve for x_A

0.8(55x_A + 15 − 15x_A) = 55x_A

0.8(40x_A + 15) = 55x_A

32x_A + 12 = 55x_A

55x_A − 32x_A = 12

23x_A = 12

x_A = 12/23 ≈ 0.522

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Final Answer:

The mole fraction of component A in the liquid phase is
0.5217