Raoult Law Vapour Composition in Ideal Solution

Learn how to relate mole fractions in liquid and vapour phases using Raoult’s law for ideal solutions. This method helps solve JEE Chemistry solution.

❓ Question

Liquid A and B form an ideal solution. The vapour pressures of pure liquids A and B are 350 and 750 mm Hg respectively at the same temperature. If xA and xB are the mole fraction of A and B in solution while yA and yB are the mole fraction of A and B in vapour phase then

🖼️ Question Image

JEE Chemistry Trick: Vapour Phase Composition from Raoult’s Law Explained 💡

✍️ Short Explanation

This problem is based on:

👉 Raoult’s law
👉 Ideal solutions
👉 Vapour phase composition.

Main idea:

For ideal solution:

$\boxed{ P_A=x_AP_A^\circ }$ $\boxed{ P_B=x_BP_B^\circ }$

and vapour mole fraction:

$\boxed{ y_A=\frac{P_A}{P_A+P_B} }$

Raoult Law Vapour Composition in Ideal Solution

🔷 Step 1 — Ratio in Vapour Phase 💯

Using Dalton’s law:

$\frac{y_A}{y_B} = \frac{P_A}{P_B}$

Now apply Raoult’s law:

$P_A=x_AP_A^\circ$ $P_B=x_BP_B^\circ$

Thus:

$\frac{y_A}{y_B} = \frac{x_AP_A^\circ}{x_BP_B^\circ}$

Substitute values:

$= \frac{x_A(350)}{x_B(750)}$ $= \frac{7}{15}\cdot\frac{x_A}{x_B}$

🔷 Step 2 — Rearrangement

$\frac{x_A}{x_B} = \frac{15}{7}\cdot\frac{y_A}{y_B}$

Thus:

$\boxed{ \frac{x_A}{x_B} > \frac{y_A}{y_B} }$

because:

$\frac{15}{7}>1$

🔷 Step 3 — Physical Meaning

Since:

$P_B^\circ > P_A^\circ$

liquid B is more volatile.

Hence vapour phase contains relatively more B.

Therefore:

$\frac{y_A}{y_B} < \frac{x_A}{x_B}$

🔷 Step 4 — JEE Trap Alert 🚨

❌ Vapour mole fraction ko liquid mole fraction ke equal maan lena

❌ More volatile component identify na kar pana

Remember:

More volatile component

$\Rightarrow \text{greater vapour-phase concentration}$

✅ Final Answer