Question

A Microscale Approach to Organic Laboratory Techniques (5th Edition), CH 15, #1. d. e. ONLY. PLEASE...

A Microscale Approach to Organic Laboratory Techniques (5th Edition), CH 15, #1. d. e. ONLY. PLEASE

given: 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

1.a. What is the mole fraction of each component if 3.9g of benzene C6H6is dissolved in 4.6g of tolueneC7H8?

The mole fraction of each is 0.050/0.100 = 0.5

       1.b. Assuming that this mixture is ideal – that is, it follow Rault’s Law, what is the partial vapor pressure of benzene in this mixture at 50oC?

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:

1.c. Estimate to the nearest degree the temperature at which the vapor pressure of the solution equals 1 atm (bp of the solution).

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


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.

e. Calculate the composition in weight percentage of the vapor that is in equilibrium with the solution.

Homework Answers

Answer #1

1b) partial vapour pressure of benzene is

270 x 0.5 = 135 mm Hg

1d) from the given data and excel function rend find the value of the valopur pressure at 92.5 oC

SO for benzene it is 1046 and for toluene it is 482

At bp the total pressure is 760

760 = 1046 * x + 482 * (1-x)

760 = 1046x + 482 - 482x

278 = 564x

x = 0.5 mol fraction of benzene

0.5 mol fraction is 0.5 x 78 = 39

0.5 x 92 = 46

39/(46+39) = 0.46 wt fraction benzene and 0.54 wt fraction toluene

So wt percentage of vapour is 46% benzene and 54% toluene

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
92.5 1046.31 110 760
92.5 482.7083
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