Calculate vapour pressure of pure volatile liquid B at given temperature if mole faction of liquid B and vapour pressure of pure volatile liquid A are 0.4 and 400 mm Hg respectively. [ P _ toal =600 ~mm Hg ]

Step-by-step solution

Vapor pressure of pure volatile liquid B Using Raoult's law for a mixture of two volatile liquids, the total vapor pressure is P_ total = X_A P_A^0 + X_B P_B^0 , where X_A and X_B are the mole fractions, and P_A^0 and P_B^0 are the pure vapor pressures. Given X_B = 0.4 , P_A^0 = 400 , mm Hg , and P_ total = 600 , mm Hg , determine X_A = 1 - X_B = 0.6 . Substitute the known values: 600 = (0.6)(400) + (0.4) P_B^0 Simplify the first term: 600 = 240 + 0.4 P_B^0 Isolate P_B^0 : 0.4 P_B^0 = 360 Solve: P_B^0 = 360 0.4 = 9