Question

I have to estimate, to the nearest degree, the temp. at which the vapor pressure of a solution (3.9g benzene, 4.6g toluene) reaches 1 atm. I believe the mole fractions are .73 for benzene, and .26 for toluene.

Answer 1

Molar fraction of Benzene and toluene, which you have guessed is correct. This is the following way to calculate the mole fraction.

Molar mass of Benzene = 78.1 g and Molar mass of Toluene = 92.1 g

Moles of Benzene is 3.9/78.1 = 0.049 and Moles of toluene = 4.6/92.1 = 0.049; Total moles = 0.049+0.049 = 0.0798

Mole fraction of Benzene = 0.049/0.0798 = 0.5; Mole fraction of Toluene = 1-0.5 = 0.5

Vapor Pressure of Benzene and Toluene at 50 degrees C (according to the example provided in the book from this question has been took - Pavia) is 270 and 95 mmHg respectively.

Thus, Partial vapor pressure of benzene is 0.5 X 270 = 135 mmHg and for Tolune is 0.5 X 95 = 47.5 mmHg. The total vapor presure of the mixture is 182.5 mmHg.

Mole fraction of Benzene in mixture is 135/182.5 = 0.73 and toluene is 47.5/182.5 = 0.26.

After this, you can calculate the temperature by plotting temperature vs vapor pressure of the mixtute and by performing exponential regression analysis or you can simply use Clausisus Clayperon equation which is

T1 = unknown; T2 = 110 degrees (Boiling point of the solution); R= 8.314 KJ/mol; P2 = 1 atm (760 mmHg); P1 = 182.5 mmHg; vapH = 0.025

Substitung all the values, gives the temperature of 92.9 degrees Celsius.