JEE Chemistry 2025 Question with Solution

Given: One mole of volatile liquid $A$ is mixed with $3$ moles of volatile liquid $B$. Pure vapor pressure of $A$ is $200 \, \text{mm Hg}$ and total vapor pressure of the solution is $500 \, \text{mm Hg}$.

Find: The vapor pressure of pure $B$ and the least volatile component.

For an ideal solution, Raoult's law gives

$$P_{\text{total}} = P_A^{\circ} x_A + P_B^{\circ} x_B$$

The total number of moles is

$$n_{\text{total}} = 1 + 3 = 4$$

So the mole fractions are

$$x_A = \frac{1}{4}, \qquad x_B = \frac{3}{4}$$

Substitute the known values:

$$500 = 200 \left(\frac{1}{4}\right) + P_B^{\circ}\left(\frac{3}{4}\right)$$ $$500 = 50 + \frac{3}{4}P_B^{\circ}$$ $$450 = \frac{3}{4}P_B^{\circ}$$ $$P_B^{\circ} = 450 \times \frac{4}{3} = 600 \, \text{mm Hg}$$

Now compare the pure vapor pressures:

$$P_A^{\circ} = 200 \, \text{mm Hg}, \qquad P_B^{\circ} = 600 \, \text{mm Hg}$$

The component with lower pure vapor pressure is less volatile, so $A$ is the least volatile component.

Therefore, the vapor pressure of pure $B$ is $600 \, \text{mm Hg}$ and the least volatile component is $A$. The correct option is C.

The solution also contains a contradictory line marking option D, but the worked calculation clearly gives $600 \, \text{mm Hg}$ and least volatile component $A$.

Given: $n_A = 1$, $n_B = 3$, $P_A^{\circ} = 200 \, \text{mm Hg}$, and $P_{\text{total}} = 500 \, \text{mm Hg}$.

Find: $P_B^{\circ}$ and the least volatile component.

First find the partial pressure of $A$ in the solution:

$$P_A = P_A^{\circ} x_A = 200 \times \frac{1}{4} = 50 \, \text{mm Hg}$$

Then the partial pressure of $B$ is

$$P_B = P_{\text{total}} - P_A = 500 - 50 = 450 \, \text{mm Hg}$$

Using Raoult's law for $B$,

$$P_B = P_B^{\circ} x_B$$ $$450 = P_B^{\circ} \times \frac{3}{4}$$ $$P_B^{\circ} = 450 \times \frac{4}{3} = 600 \, \text{mm Hg}$$

Since $200 \, \text{mm Hg} < 600 \, \text{mm Hg}$, liquid $A$ has lower vapor pressure in pure state and is therefore less volatile.

Hence, the required pair is $600 \, \text{mm Hg}$, $A$.