Given:

1 mole of volatile liquid A

3 moles of volatile liquid B

Vapor pressure of pure A, $ P_A^o = 200 $ mm Hg

Vapor pressure of the solution, $ P_{S} = 500 $ mm Hg

We apply Raoult's law, which states:

$ P_{S} = P_A^o \cdot X_A + P_B^o \cdot X_B $

Where:

$ X_A $ is the mole fraction of A

$ X_B $ is the mole fraction of B

$ P_B^o $ is the vapor pressure of pure liquid B

Calculate the mole fractions:

$ X_A = \frac{1}{1+3} = \frac{1}{4} $

$ X_B = \frac{3}{1+3} = \frac{3}{4} $

Plug these into the equation:

$ 500 = 200 \times \frac{1}{4} + P_B^o \times \frac{3}{4} $

Simplifying:

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

Subtract 50 from both sides:

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

Multiply both sides by $\frac{4}{3}$ to solve for $P_B^o$:

$ P_B^o = 600 \, \text{mm Hg} $

Since $ P_A^o < P_B^o $, liquid A is the least volatile component.

In conclusion:

The vapor pressure of pure B, $ P_B^o $, is 600 mm Hg.

The least volatile component is A.