Question

Data for benzene and toluene are given below. Calculate the vapor pressure over a solution made by mixing 20.0 grams of benzene with 80.0 grams of toluene when the temperature is 40.0ºC.

Compound      Molar Mass                 Vapor Pressure at 40.0ºC

Benzene          78.11 g/mol                 183 mm Hg

Toluene           92.14 g/mol                 59 mm Hg

Answer 1

Data given: mass of benzene in liquid = 20 grams

                mass of toluene = 80 grams

molar mass of benzene = 78.11 grams/mole

molar mass of toluene = 92.14 grams/mole

number of moles of benzene in liquid solution = 20/78.11 =0.256 moles

number of moles of toluene in liquid solution = 80/92.14 = 0.8682 moles

Total moles in liquid = moles of benzene + moles of toluene = 0.256 +0.8682 = 1.1242 moles

Let benzene be component 1 and toluene be component 2

moles fraction of benzene in liquid (X1) = 0.256/1.1242 =0.2277

mole fraction of toluene in liquid (X2) = 0.8682/1.1242 = 0.7723

Saturation pressure of benzene (P1^sat) = 183 mmHg

Saturation pressure of toluene (P2^sat) = 59 mmHg

We will apply Raoult's law , to calculate the total pressure (P) , that is exerted over a solution.

Raoult's law is given by

Yi P=Xi Pi^sat

Where, Yi = mole fraction of component 'i' in vapour phase

            P = total pressure of the system

          Xi = mole fraction of component 'i' in liquid phase

        Pi^sat = vapor pressure of component 'i'

Raoult's law for component 1

Y1 P=X1 P1^sat                    ................(1)

Y2 P=X2 P2^sat                    ................(2)

Adding equation (1) and equation (2)

(Y1+Y2) P = X1 P1^sat +X2 P2^sat           ...........(3)

For a binary mixture,

Y1+Y2 =1       ................(4)

Substituting equation (4) in equation (3)

P = X1 P1^sat +X2 P2^sat   

Substituting all the known values to obtain the value of P

P = 0.2277 *183 + 0.7723*59 = 87.2348 mmHg

Hence vapor pressure exerted over a solution is 87.2348 mmHg