Two liquids A and B form an ideal solution. At 320 K, the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is 500 mm Hg. At the same temperature, if 1 mol of A is further added to this solution, vapour pressure of the solution increases by 20 mm Hg. Vapour pressure (in mm Hg) of B in pure state is $\_\_\_\_$. (Nearest integer)

For an ideal solution, Raoult's law states:
$P_{total} = P_A^o x_A + P_B^o x_B$.
Initial condition: 3 mol of A and 1 mol of B give a total vapor pressure of 500 mm Hg.
Mole fractions are $x_A = \frac{3}{4} = 0.75$ and $x_B = \frac{1}{4} = 0.25$.
This gives: $500 = 0.75 P_A^o + 0.25 P_B^o$ ... (1).
After adding 1 mol of A: 4 mol of A and 1 mol of B give a total vapor pressure of 520 mm Hg.
Mole fractions are $x_A = \frac{4}{5} = 0.8$ and $x_B = \frac{1}{5} = 0.2$.
This gives: $520 = 0.8 P_A^o + 0.2 P_B^o$ ... (2).
From equation (1) multiplied by 4: $2000 = 3 P_A^o + P_B^o$.
From equation (2) multiplied by 5: $2600 = 4 P_A^o + P_B^o$.
Subtracting: $600 = P_A^o$.
Substituting back into equation (1): $2000 = 3(600) + P_B^o$, so $P_B^o = 200$ mm Hg.