JEE Main 2025 — Solutions Question with Solution

Given:

1 mole of volatile liquid A

3 moles of volatile liquid B

Vapor pressure of pure A, $P_A^o=200$ mm Hg

Vapor pressure of the solution, $P_S=500$ mm Hg

We apply Raoult's law, which states:

$P_S=P_A^o\cdot X_A+P_B^o\cdot X_B$

Where:

$X_A$ is the mole fraction of A

$X_B$ is the mole fraction of B

$P_B^o$ is the vapor pressure of pure liquid B

Calculate the mole fractions:

$X_A=\frac{1}{1+3}=\frac{1}{4}$

$X_B=\frac{3}{1+3}=\frac{3}{4}$

Plug these into the equation:

$500=200\times \frac{1}{4}+P_B^o\times \frac{3}{4}$

Simplifying:

$500=50+\frac{3}{4}{P}_B^o$

Subtract 50 from both sides:

$450=\frac{3}{4}{P}_B^o$

Multiply both sides by $\frac{4}{3}$ to solve for $P_B^o$:

$P_B^o=600\text{mm Hg}$

Since $P_A^o<P_B^o$, liquid A is the least volatile component.

In conclusion:

The vapor pressure of pure B, $P_B^o$, is 600 mm Hg.

The least volatile component is A.