question_answer 8)

The vapour pressure of pure liquids A and B are 450 and 700 mrn Hg at 350 K respectively. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.  

Answer:

Here, Ptotal = 600 mm Applying Raoults' law, pA = xA x pTotal = pA + pB   Substituting the given values, we get 600 = 700 + (450 - 700)xA or 250xA = 100 or xA =  = 0.40 Thus, composition of the liquid mixture will be xA (mole fraction of A) = 0.40 xB (mole fraction of B) = 1 - 0.40 = 0.60 PA = xA x = 0.40 x 450 mm = 180 mm   = 420 mm Mole fraction of A in the vapour phase   Mole fraction of B in the vapour phase yB = 1 - 0.30 = 0.70