A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure B and the least volatile component of the solution, respectively, are:

According to Raoult's law, the partial vapour pressure of each volatile component in an ideal solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution.

Given:

• Moles of liquid $$A = 1$$ mole
• Moles of liquid $$B = 3$$ moles
• Vapour pressure of pure $$A = 200\ \text{mm Hg}$$
• Total vapour pressure of the solution $$= 500\ \text{mm Hg}$$

The total number of moles in the solution is

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

The mole fractions are therefore

$$x_A = \frac{1}{4} = 0.25$$

$$x_B = \frac{3}{4} = 0.75$$

Using Raoult's law,

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

Substituting the given values,

$$500 = (200 \times 0.25) + (P_B^\circ \times 0.75)$$

$$500 = 50 + 0.75P_B^\circ$$

$$450 = 0.75P_B^\circ$$

$$P_B^\circ = \frac{450}{0.75} = 600\ \text{mm Hg}$$

To determine the least volatile component, compare the vapour pressures of the pure liquids.

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

A substance with a lower vapour pressure is less volatile because it vaporizes less readily.

Since $$200 < 600$$, liquid A is the least volatile component.

Therefore, the vapour pressure of pure $$B$$ is $$600\ \text{mm Hg}$$, and the least volatile component is A.

Hence, the correct answer is Option (D).