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)
A. A
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 = 3 4 = 0.75 and x_B = 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 = 4 5 = 0.8 and x_B = 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.
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