Two liquids $$X$$ and $$Y$$ form an ideal solution. At $$300$$ K, vapour pressure of the solution containing $$1$$ mol of $$X$$ and $$3$$ mol of $$Y$$ is $$550$$ mmHg. At the same temperature, if $$1$$ mol of $$Y$$ is further added to this solution, vapour pressure of the solution increases by $$10$$ mmHg. Vapour pressure (in mmHg) of $$X$$ and $$Y$$ in their pure states will be, respectively :

For an ideal liquid solution, Raoult’s law gives the total vapour pressure as the mole-fraction-weighted sum of the pure-component vapour pressures:

$$P_{\text{total}} = x_X P_X^{\,0} + x_Y P_Y^{\,0}$$

where $$P_X^{\,0}$$ and $$P_Y^{\,0}$$ are the vapour pressures of pure $$X$$ and pure $$Y$$, and $$x_X,\,x_Y$$ are their mole fractions in the liquid phase.

Case 1: Initial mixture (1 mol $$X$$ + 3 mol $$Y$$)

Total moles $$n_1 = 1 + 3 = 4$$, so

$$x_X = \frac{1}{4} = 0.25, \qquad x_Y = \frac{3}{4} = 0.75$$

Given $$P_{\text{total}} = 550 \,\text{mmHg}$$, hence

$$0.25\,P_X^{\,0} + 0.75\,P_Y^{\,0} = 550 \quad -(1)$$

Case 2: After adding 1 mol of $$Y$$ (now 1 mol $$X$$ + 4 mol $$Y$$)

Total moles $$n_2 = 1 + 4 = 5$$, so

$$x_X' = \frac{1}{5} = 0.20, \qquad x_Y' = \frac{4}{5} = 0.80$$

The vapour pressure increases by $$10$$ mmHg, so $$P_{\text{total}} = 550 + 10 = 560 \,\text{mmHg}$$. Therefore

$$0.20\,P_X^{\,0} + 0.80\,P_Y^{\,0} = 560 \quad -(2)$$

Solving equations (1) and (2)

Multiply (1) by 4: $$P_X^{\,0} + 3P_Y^{\,0} = 2200$$

Multiply (2) by 5: $$P_X^{\,0} + 4P_Y^{\,0} = 2800$$

Subtract the first from the second:

$$(P_X^{\,0} + 4P_Y^{\,0}) - (P_X^{\,0} + 3P_Y^{\,0}) = 2800 - 2200$$ $$\Rightarrow \; P_Y^{\,0} = 600 \,\text{mmHg}$$

Substitute $$P_Y^{\,0}=600$$ into $$P_X^{\,0} + 3P_Y^{\,0} = 2200$$:

$$P_X^{\,0} + 3(600) = 2200 \;\Longrightarrow\; P_X^{\,0} = 2200 - 1800 = 400 \,\text{mmHg}$$

Thus the vapour pressures of the pure liquids are

$$P_X^{\,0} = 400 \,\text{mmHg}, \qquad P_Y^{\,0} = 600 \,\text{mmHg}$$

Option C which is: 400 and 600