In the accompanying chart are appropriate vapor pressures for benzene and toluene at various temperatures:
Temp (C) mmHg Temp (C) mmHg
Benzene 30 120 Toluene 30 37
40 180 40 60
50 270 50 95
60 390 60 140
70 550 70 200
80 760 80 290
90 1010 90 405
100 1340 100 560
110 760
d. Calculate the composition of the vapor (mole fraction of each component) that is in equilibrium in the solution at the boiling point of this solution.
Also this is part e) of same question
e. Calculate the composition in weight percentage of the vapor that is in equilibrium with the solution.
To calculate the composition of the vapor at the boiling point of the solution, we need to determine the mole fraction of each component (benzene and toluene) in the vapor phase. We can use Raoult's law, which states that the vapor pressure of each component in a mixture is proportional to its mole fraction.
First, let's find the vapor pressure of benzene and toluene at their respective boiling points. Using the given data, we can see that the boiling point of benzene is 80°C, and the boiling point of toluene is 110°C.
From the chart, we see that the vapor pressure of benzene at 80°C is 760 mmHg and the vapor pressure of toluene at 110°C is 760 mmHg.
Now, we can calculate the mole fraction of benzene and toluene in the vapor phase at their respective boiling points.
Mole fraction of benzene (xb):
xb = Pb / Ptot
where Pb is the vapor pressure of benzene and Ptot is the total vapor pressure at the boiling point.
xb = 760 mmHg / 760 mmHg = 1
Mole fraction of toluene (xt):
xt = Pt / Ptot
where Pt is the vapor pressure of toluene and Ptot is the total vapor pressure at the boiling point.
xt = 760 mmHg / 760 mmHg = 1
Therefore, at the boiling point of the solution, the mole fraction of both benzene and toluene in the vapor phase is 1 each.