JEE Chemistry 2026 Question with Solution

Given: The solution is ideal and follows Raoult's law. Initially, $n_A = 3$, $n_B = 1$, and total vapour pressure is $500 \, \text{mm Hg}$. After adding $1$ mol of A, $n_A = 4$, $n_B = 1$, and total vapour pressure is $520 \, \text{mm Hg}$.

Find: Vapour pressure of pure B, that is $P_B^\circ$.

According to Raoult's law,

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

Case I:

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

So,

$$500 = \frac{3}{4}P_A^\circ + \frac{1}{4}P_B^\circ$$

Multiplying by $4$,

$$2000 = 3P_A^\circ + P_B^\circ$$

This is equation $(1)$.

Case II: After adding $1$ mol of A,

$$x_A = \frac{4}{5}, \qquad x_B = \frac{1}{5}$$

So,

$$520 = \frac{4}{5}P_A^\circ + \frac{1}{5}P_B^\circ$$

Multiplying by $5$,

$$2600 = 4P_A^\circ + P_B^\circ$$

This is equation $(2)$.

Subtract equation $(1)$ from equation $(2)$:

$$600 = P_A^\circ$$

Substitute $P_A^\circ = 600$ into equation $(1)$:

$$2000 = 3(600) + P_B^\circ$$ $$2000 = 1800 + P_B^\circ$$ $$P_B^\circ = 200 \, \text{mm Hg}$$

Therefore, the vapour pressure of B in pure state is $200 \, \text{mm Hg}$. Hence, the numerical answer is 200.

Given: Two observations of total vapour pressure are provided for the same ideal binary solution.

Find: The pure vapour pressure $P_B^\circ$.

For an ideal binary solution, total vapour pressure depends linearly on mole fractions of the two components. Using the two compositions gives two linear equations in $P_A^\circ$ and $P_B^\circ$.

From the first composition:

$$500 = \frac{3}{4}P_A^\circ + \frac{1}{4}P_B^\circ$$ $$2000 = 3P_A^\circ + P_B^\circ$$

From the second composition:

$$520 = \frac{4}{5}P_A^\circ + \frac{1}{5}P_B^\circ$$ $$2600 = 4P_A^\circ + P_B^\circ$$

Now eliminate $P_B^\circ$ by subtraction:

$$(4P_A^\circ + P_B^\circ) - (3P_A^\circ + P_B^\circ) = 2600 - 2000$$ $$P_A^\circ = 600$$

Put this in the first equation:

$$3(600) + P_B^\circ = 2000$$ $$1800 + P_B^\circ = 2000$$ $$P_B^\circ = 200$$

Thus, the vapour pressure of pure B is $200 \, \text{mm Hg}$.