Please help! Have no clue how to approach #9 and 10 using this provided graph Fo
ID: 1071259 • Letter: P
Question
Please help! Have no clue how to approach #9 and 10 using this provided graph
For a liquid mixture prepared as 0.54 mole fraction in toluene with 10 moles of benzene what are the mole fractions of the liquid and vapor phases? What fraction of the mixture is in the liquid phase and in the vapor phase? How many moles of toluene are there in the vapor phase, and in the liquid phase? When benzoic acid is added to a mixture of two immiscible liquids an equilibrium is reached where the mole fraction of benzoic acid in the first layer is 0.15 and in the second as 0.25. When 4.0 additional grams of benzoic acid is added to the mixture the first layer is found to have increased to 0.21 mole fraction what is the mole fraction of benzoic acid in the second layer then?Explanation / Answer
as shown in the figure,at 0.54, draw a line so as to intersect at 368K. this corresponds to a liquid fraction of 0.4 and 0.601 vapor composition for toluene
total moles of Benzene = 10moles, Mole fraction Benzene =1-0.54= 0.46
Total moles of mixture =10/0.46=21.74, moles Benzene= 21.74-10= 11.74
From material balance, V= L+V =21.74 ( L and V are liquid and vapor flows), V= 21.74-L (1)
Writing toluene balance, 11.74= L*0.4+0.601*V (2)
Eq.2 now becomes ( from Eq.1) , 11.74 = L*0.4+0.601*(21.74-L)
11.74= 0.4L +0.601*21.74-0.601L, L= 6.6 and V= 21.74-6= 15.74 moles
V/F= 15.74/21.74 = 0.72 and L/F= 0.28
mole fractions :Tolune : Liquid phase : 0.4 Vapor phase =0.601, Benzene =1-0.4=0.6 and vapor phase = 1-0.601=0.399
Moles : Liquide phase : toluene =6.6*0.4= 2.64, Benzene =6.6-2.64=3.96, Benzene = 0.399*15.74=6.28 and toluene= 15.74-6.28=9.46