The vapor pressure of pure liquids A and B are 450 and 750 mm Hg respectively, at 350K. Find out the composition of the liquid mixture if total vapor pressure is 600 mm Hg. Also find the composition of the vapor phase.
Given that, P0A=450mmHgPA0=450mmHg P0B=750mmHgPB0=750mmHg PT=600mmHgPT=600mmHg Now, from Raoult’s law; PA=P0AxAPA=PA0xA PB=P0BxB=P0B(1−xA)PB=PB0xB=PB0(1−xA) So, PT=PA+PBPT=PA+PB Now, PT=P0AxA+P0B(1−xA) xA=PT−P0BP0A−P0B From this, xA=0.5 xB=0.5 This gives us; PA=450×0.5=225mmHg PB=750×0.5=375mmHg Now, Composition in vapor phase is given as; yA=PA/PT=225/600=0.375 yA=1−yA=1−0.375=0.625
