SolutionHard
Question
Mole fraction of the toluene in the vapour phase which is in equilibrium with a solution of benzene (po = 120 Torr) and toluene (po = 80 Torr) having 2.0 mol of each is
Options
A.0.50
B.0.25
C.0.60
D.0.40
Solution
PT = XA poA + XB poB
=
× 80 + 
× 120 = 100 Torr
Now mole fraction in vapour phase =
= 0.4.
=
× 80 + × 120 = 100 TorrNow mole fraction in vapour phase =
= 0.4.Create a free account to view solution
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