In the accompanying chart are appropriate vapor pressures for benzene and toluen

Question

In the accompanying chart are appropriate vapor pressures for benzene and toluene at various temperatures: Benzene: Temp (C)....mmHg 30..........120 40..........180 50..........270 60..........390 70..........550 80..........760 90..........1010 100.........1340 110.........760 Toluene: Temp (C)....mmHg 30..........37 40..........60 50..........95 60..........40 70..........200 80..........290 90..........405 100.........560 a. What is the mole fraction of each component if 3.9 g of benzene (C6H6) is dissolved in 4.6 g of toluene (C7H8)? b. Assuming that this mixture is ideal, that is, it follows Raoult's Law, what is the partial vapor pressure of benzene in this mixture at 50 C? c. Estimate to the nearest degree the temperature at which the vapor pressure of the solution equals 1 atm (bp of the solution). 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.

Explanation / Answer

Mole fraction = moles of A / [moles of A + moles of B]

moles of benzene = 3.9 / 78 = 0.05

moles of toulene = 4.6 / 92 = 0.05

mole fraction of benzene = 0.5 and toulene = 0.

partial pressure of benzene = mole fraction * presurre at 50

= 0.5 * 270 = 136 mm hg

Vapor pressure of solution = 1atm = 760mm Hg

d] from the chart we know boiling point of benzene is 80C since it has vapor pressure 760mm hg

delta T = i kb m

m -molality

i = van hoff factor =1

kb = constant = 6.26

m = moles of solute / weight of solvent in kg = 12.82

boiling point of the solution is = 80+12.82 = 92.82 C

4] taking 92.82 as 90C

given 760mm hg of solution

mole fraction of vapor content = Total pressure of solution / the partial vapor pressure of component

at 90C

benzene :

1010*0.5 / 760 = 0.664

toulene = 405*0.5 / 760 = 0.266

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