Question

In a mixture of A and B having vapour pressure of pure A and pure B are 400 mm Hg and 600 mm Hg respectively, the mole fraction of B in the liquid phase is 0.5. Calculate total vapour pressure and mole fraction of A and B in the vapour phase.

• A. 500, 0.4, 0.6
• B. 500, 0.5, 0.5
• C. 450, 0.4, 0.6
• D. 450, 0.5, 0.5

Answer 1

The correct option is A

500, 0.4, 0.6

Total vapour pressure = sum of vapour pressures of pure compounds

$P_{total}=P{°}_A+P{°}_B$

Vapour pressure of pure compound is equal to the product of pressure and mole fraction of compound.

$P_A=x_A\times P_A$

$P_B=x_B\times P_B$

where $x_A$ and $x_B$ are mole fraction.

$P_{total}=0.5\times 400+0.5\times 600$

$P_{total}=500mmofHg.$

Mole fraction: $\frac{1}{P_{total}}=[\frac{X_A}{400}+\frac{(1-x_A)}{600}]$

By solving, we get $x_A=0.6$

Now, $x_B=1-x_A$

$x_B=1-0.6=0.4$

$x_B=0.4$

So, the correct option is A(500, 0.4, 0.6).