A solution is prepared by mixing 235 mL of carbon tetrachlor

A solution is prepared by mixing 23.5 mL of carbon tetrachloride and 35.8 mL of 2,5- dichlorohexane. What is the expected vapor pressure of this solution? What is the mole fraction of carbon tetrachloride in the vapor above the solution? The density of carbon tetrachloride is 1.594 g/mL and its vapor pressure is 91 mmHg. The density of 2,5-dichlorohexane is 1.0 g/mL and its vapor pressure is 2.3 mmHg.

Solution

Mass of CCl4 = Density*Volume = 23.5*1.594 = 37.459 g

Moles of CCl4 = Mass/MMW = 37.459/153.82 = 0.243

Mass of dichlorohexane = Density*Volume = 35.8*1 = 35.8 g

Moles of dichlorohexane = Mass/MMW = 35.8/155.06 = 0.231

Using Raoult\'s law:

P = P_A⁰X_A + P_B⁰X_B

Putting values:

P = 91*(0.243/(0.243+0.231)) + 2.3*(0.231/(0.243+0.231)) = 47.77 mm Hg

Mole fraction of CCl4 in vapor phase = (91*(0.243/(0.243+0.231)))/47.77 = 0.976

Hope this helps !