1. Use Raoult’s law for an ideal solution

For a binary solution of two volatile liquids:

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

where:

• $x_A, x_B$ are mole fractions in liquid phase
• $P_A^0, P_B^0$ are vapour pressures of pure $A$ and $B$

2. Find mole fractions

Given:

• moles of $A = 1$
• moles of $B = 3$
• total moles $= 4$

So,

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

3. Substitute known values

Given:

• $P_A^0 = 200\ \text{mm Hg}$
• $P_{\text{solution}} = 500\ \text{mm Hg}$

Thus,

$500 = \frac{1}{4}(200) + \frac{3}{4}P_B^0$

$500 = 50 + \frac{3}{4}P_B^0$

$450 = \frac{3}{4}P_B^0$

$P_B^0 = 450 \times \frac{4}{3} = 600\ \text{mm Hg}$

So the vapour pressure of pure $B$ is:

$P_B^0 = 600\ \text{mm Hg}$

4. Find the least volatile component

Less volatile substance has lower vapour pressure.

Compare:

• $P_A^0 = 200\ \text{mm Hg}$
• $P_B^0 = 600\ \text{mm Hg}$

Since $200 < 600$, liquid $A$ is less volatile.

5. Check options

• A: $1400\ \text{mm Hg}, A$ — incorrect vapour pressure
• B: $1400\ \text{mm Hg}, B$ — incorrect vapour pressure and wrong component
• C: $600\ \text{mm Hg}, A$ — correct
• D: $600\ \text{mm Hg}, B$ — wrong least volatile component

Therefore, the correct option is C.