Q.The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Concept: Raoult’s Law for a Binary Volatile-Liquid Mixture — this problem uses Raoult’s law to find the liquid composition, then Dalton’s law for the vapour composition.

Let $x_A$ and $x_B$ be the mole fractions in the liquid phase, with $x_A + x_B = 1$.

Raoult’s law: $P_A = x_A P_A^0$, $P_B = x_B P_B^0$.

Total vapour pressure: $P_{\text{total}} = P_A + P_B = x_A P_A^0 + (1 - x_A) P_B^0$.

Substitute given values:

$600 = x_A (450) + (1 - x_A)(700)$

$600 = 450x_A + 700 - 700x_A$

$600 - 700 = -250x_A$ → $-100 = -250x_A$ → $x_A = 0.4$, so $x_B = 0.6$. …

Why this approach works

When two volatile liquids mix, each exerts a partial vapour pressure proportional to its mole fraction in the liquid — that’s Raoult’s law. The total vapour pressure above the mixture is simply the sum of these partial pressures. So if we know the pure vapour pressures and the total pressure, we can solve for the liquid composition.

But the vapour phase composition is different: it depends on the partial pressures of each component in the vapour, not on their liquid mole fractions. Once we find the partial pressures, the mole fraction in vapour is just that component’s partial pressure divided by the total pressure.

Let’s work it through.

Given data

• Pure vapour pressure of A: $P_A^0 = 450\ \text{mm Hg}$
• Pure vapour pressure of B: $P_B^0 = 700\ \text{mm Hg}$
• Total vapour pressure of mixture: $P_{\text{total}} = 600\ \text{mm Hg}$
• Temperature: $350\ \text{K}$ (constant, so vapour pressures are fixed)

1. Apply Raoult’s law for each component

   For a liquid mixture, the partial pressure of A in the vapour is:

where $x_A$ is the mole fraction of A in the liquid phase. Similarly,

Since it’s a binary mixture, $x_B = 1 - x_A$.

2. Write the total pressure equation

Substitute the numbers:

3. Solve for $x_A$

   Expand:

Rearrange:

So the liquid mixture contains 40 mol% A and 60 mol% B. …

Concept: Raoult's Law Application

Step 1 – Define variables

Let $x_A$ and $x_B$ be mole fractions in liquid phase.

Given: $P_A^0 = 450$ mm Hg, $P_B^0 = 700$ mm Hg, $P_{\text{total}} = 600$ mm Hg.

Step 2 – Apply Raoult’s law

$P_{\text{total}} = x_A P_A^0 + x_B P_B^0$ and $x_B = 1 - x_A$

So, $600 = 450 x_A + 700 (1 - x_A)$

$\Rightarrow 600 = 700 - 250 x_A$

$\Rightarrow x_A = 0.4$, hence $x_B = 0.6$

Step 3 – Vapour phase composition …

Common Mistakes in Finding Liquid & Vapour Composition from Total Pressure

1. Setting Up the Total-Pressure Equation Incorrectly

The Mistake: Writing $P_{total}=P_A^\circ+P_B^\circ$ instead of $P_{total}=x_AP_A^\circ+x_BP_B^\circ$.

How to avoid: Always start from Raoult's law for each component and sum the partial pressures, using $x_B=1-x_A$ to keep one unknown.

2. Forgetting Which Mole Fraction Is Which

The Mistake: After solving $x_A$, forgetting that $x_B=1-x_A$, or mixing up which component is more volatile.

How to avoid: Since $P_B^\circ$(700) > $P_A^\circ$(450), B is the more volatile component and will be enriched in the vapour relative to the liquid.

3. Using Liquid Mole Fraction for Vapour Composition

The Mistake: Reporting $x_A=0.4$ as also being the vapour-phase composition. …