Question

3. A mixture of benzene and water distills at 83°C at 760 mmHg. At 83 C the vapor pressure of water is 401mmHg. What is the mole % of the vapor during distillation? What is the weight of benzene that is distilled with each gram of water? b. a.

Answer 1

Solution :-

Vapor pressure of water at 83 C= 401 mmHg

Vapor pressure of benzene = atmospheric pressure – vapor pressure of water

                                              = 760 mmHg – 401 mmHg

                                              = 359 mmHg

Part a ) mole fraction of vapor during the distillation

Mole fraction of vapor = vapor pressure of water /total pressure

                                         = 401 mmHg / 760 mmHg

                                        = 0.5276

% of mole vapor = 0.5276 * 100 % = 52.76 % mole vapor

Part b) using the mole fraction of the vapor and benzene we can find the mass of each

Mole fraction of vapor = 0.5276

Mole fraction of benzene = 1 – 0.5276 = 0.4724

Now lets consider the mole fraction as nothing but the moles of each

Mass= moles x molar mass

Mass of water vapor = 0.5276 mol * 18.016 g per mol = 9.51 g

Mass of benzene = 0.4724 mol * 78.11 g per mol = 36.9 g

Now lets find the ratio

Mass of benzene / mass of water vapor

36.9 g / 9.51 g = 3.88 g

Therefore for every gram of water vapor it distills out 3.88 g benzene.