Please answer with calculations and use accompanying table at bottom, thank you!
ID: 780381 • Letter: P
Question
Please answer with calculations and use accompanying table at bottom, thank you!
1) 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
2) Calculate the composition in weight percentage of the vapor that is in equilibrium with the solution
BENZENE TOLUENE
temp (celsius) mmHg temp (celsius) mmHg
30 120 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
Explanation / Answer
a) To determine the mole fraction you first need to determine the moles of benzene and toluene:
C6H6: (3.9 g) x [(1 mole C6H6)/(78.11 g)] = 0.050 moles C6H6
C7H8: (4.6 g) x [(1 mole C7H8)/(92.14 g)] = 0.050 moles C7H8
Thus the mole fraction of each is 0.050/0.100 = 0.5
b) Raoult's law states that the sum of the product of the mole fractions by the partial pressures equals the total pressure of the system:
P(tot) = 0.5(270 mmHg) + 0.5(95 mmHg) = 182.5 mmHg
c) If you plot the vapor pressure (calculated for the mixture) versus temperature and find an exponential regression you'll get the following relationship:
Pressure = [68.418e^(0.027*T)] - 80.617
Plugging the value for the pressure and solving for T gives:
760 = 68.418e^(0.027*T) - 80.617
840.617 = 68.418e^(0.027*T)
12.286 = e^(0.027*T)
Ln(12.286) = 0.027*T
2.508 = 0.027*T
92.9 oC = T