1-propanol (p* 1= 20.9 torr at 25 degree C) and 2-propanol (p*2 = 45.2 torr at 2
ID: 898507 • Letter: 1
Question
1-propanol (p* 1= 20.9 torr at 25 degree C) and 2-propanol (p*2 = 45.2 torr at 25 degree C) form ideal solutions in all proportions. Let xi and x2 represent the mole fractions of 1-propanol and 2-propanol, respectively, in a liquid mixture and yi and y2 represent the mole fractions of each in the vapor phase For a solution of these liquids with xi = 0.480, calculate the composition of the vapor phase at 25 degree C if the vapor above the liquid phase is composed only of 1-propanol and 2-propanol.Explanation / Answer
Pº1 = 20.9 torr
Pº2 = 45.2 torr
x1 = 0.48
x2 = 0.52
Find vapor phases
Apply Raoult Equation
x1*P1 = y1*PT
x2*P2 = y2*PT
Subsitutte all in
x1*P1 = y1*PT
0.48*20.9 = y1*PT
0.52*45.2 = (1-y1)PT
Solve for PT or y1
10.03= y1*PT ......(1)
23.5 = (1-y1)PT......(2)
Pt = 10.03/Y1
subtitute in (2)
23.5 = (1-y1)(10.03)/y1
23.5 = 10.03/y1 - 10.03
(23.5+10.03) = 10.03/y1
y1 = (10.03)/(33.53 ) = 0.2999
therefore
y2 = 1-y1 = 1-0.299 = 0.7001
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