Mixture of liquid toluene (T) and benzene (B) form ideal solutions at 60 C. The
ID: 505267 • Letter: M
Question
Mixture of liquid toluene (T) and benzene (B) form ideal solutions at 60 C. The vapour pressures of the pure liquids at this temperature are 0.265 bar for T and 0.535 bar for B. a) Determine the mole fraction of B in a liquid mixture which has vapour pressure of 0.400 bar at 60 C. b) What is the mole fraction of B in the vapour of the same mixture? c) Why T and B constitute an ideal solution (i.e. what is the physical reason)? [20] Mixture of liquid toluene CT) and benzene (B) form ideal solutions at 60 C. The vapour pressures of the pure liquids at this temperature are 0.265 bar for T and 0.535 bar for B. a) Determine the mole fraction of Bin a liquid mixture which has vapour pressure of 0.400 bar at 60 C. b) What is the mole fraction of Bin the vapour of the same mixture? c) Why T and B constitute an ideal solution (i.e. what is the physical reason)?Explanation / Answer
P0 T = 0.265bar
P0 B = 0.535 bar
From Raoults law vapor pressue of solution of two volatile liwuids is
P = P0 T x molfe fraction of T +P0 B x mole fraction of B
We know that M.F T + M.F B = 1
a) Thus 0.400 bar = 0.265 (1-mfB ) + 0.535 (mfB)
solving mole fraction of B in liquid = 0.5
b) vapor pressure of B in the mixture = mole fraction x P0 B
= 0.5 x 0.535
= 0.2675 bar
The vapour follows Dalton's law.
According to Daltons law Partial pressure of agas = mole fraction (in vapour) x total pressure
Thus 0.2675 = mole fraction of B in vapour x 0.400bar
mole fraction of B = 0.2675 /0.4
= 0.66875
C) Benzene and toluene are homologous aromatic compounds with similar nature . Thus they both constitute ideal solutions.