JEE Chemistry 2026 Question with Solution

Given: At temperature $T$ K, $2$ moles of liquid $A$ and $3$ moles of liquid $B$ form an ideal solution with total vapour pressure $320$ mm Hg. Then $1$ mole each of $A$ and $B$ is added, and the new vapour pressure becomes $328.6$ mm Hg.

Find: The vapour pressures of pure $A$ and pure $B$, namely $P_A^0$ and $P_B^0$.

For an ideal solution, Raoult’s law applies:

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

Step 1: Initial mixture

$$n_A = 2, \quad n_B = 3, \quad n_{\text{total}} = 5$$

Hence, the mole fractions are

$$x_A = \frac{2}{5}, \quad x_B = \frac{3}{5}$$

Using the given vapour pressure,

$$\frac{2}{5}P_A^0 + \frac{3}{5}P_B^0 = 320$$

So,

$$2P_A^0 + 3P_B^0 = 1600 \qquad (1)$$

Step 2: After adding $1$ mole each of $A$ and $B$

$$n_A = 3, \quad n_B = 4, \quad n_{\text{total}} = 7$$

Thus,

$$x_A = \frac{3}{7}, \quad x_B = \frac{4}{7}$$

Using the new vapour pressure,

$$\frac{3}{7}P_A^0 + \frac{4}{7}P_B^0 = 328.6$$

So,

$$3P_A^0 + 4P_B^0 = 2300 \qquad (2)$$

Step 3: Solve equations $(1)$ and $(2)$ Multiply equation $(1)$ by $3$:

$$6P_A^0 + 9P_B^0 = 4800$$

Multiply equation $(2)$ by $2$:

$$6P_A^0 + 8P_B^0 = 4600$$

Subtracting gives

$$P_B^0 = 200 \; \text{mm Hg}$$

Substitute into equation $(1)$:

$$2P_A^0 + 600 = 1600$$ $$2P_A^0 = 1000$$ $$P_A^0 = 500 \; \text{mm Hg}$$

Therefore, the vapour pressures of pure $A$ and pure $B$ are $500 \; \text{mm Hg}$ and $200 \; \text{mm Hg}$ respectively. The correct option is B.

Given: The solution is ideal, so total vapour pressure depends on mole fractions and pure-component vapour pressures.

Find: $P_A^0$ and $P_B^0$.

Write the two Raoult’s law equations directly from the two mixture compositions:

$$\frac{2}{5}P_A^0 + \frac{3}{5}P_B^0 = 320$$ $$\frac{3}{7}P_A^0 + \frac{4}{7}P_B^0 = 328.6$$

Clear fractions:

$$2P_A^0 + 3P_B^0 = 1600$$ $$3P_A^0 + 4P_B^0 = 2300$$

Now eliminate one variable. Using subtraction after matching coefficients of $P_A^0$:

$$6P_A^0 + 9P_B^0 = 4800$$ $$6P_A^0 + 8P_B^0 = 4600$$

Therefore,

$$P_B^0 = 200$$

Substitute back:

$$2P_A^0 + 3(200) = 1600$$ $$2P_A^0 = 1000$$ $$P_A^0 = 500$$

Hence, the required vapour pressures are $500$ and $200$ mm Hg, so the correct option is B.