Question

Distillation is a method of purification based on successive separations and recondensations of vapor above a solution.

Part A

The vapor pressure of carbon tetrachloride, CCl4, is 0.354 atmand the vapor pressure of chloroform, CHCl3, is 0.526 atm at 316 K. A solution is prepared from equal masses of these two compounds at this temperature. The mole fraction of chloroform in the vapor above this solution is χCHCl3 = 0.657. If the vapor above the original solution is condensed and isolated into a separate flask, the vapor pressure of chloroform above this new solution is 0.346 atm.

Calculate the mole fraction of chloroform in the vapor above a solution obtained by three successive separations and condensations of the vapors above the original solution of carbon tetrachloride and chloroform.

χCHCl3=

Part B

Show how this result explains the use of distillation as a separation method.

Answer 1

First find the mole fraction of each component in the solution.

let's you have a solution with 100 g of each compound.

The molar mass of CCl4 = 153.8 g/mole

the molar mass of CHCl3 = 119.4 g/mole.

100 g CCl4 x (1 mole CCl4 / 153.8 g CCl4) = 0.650 moles CCl4

100 g CHCl3 x (1 mole CHCl3 / 119.4 g CHCl3) = 0.838 moles CHCl3

Mole fraction CHCl3 in solution = (moles CHCl3 / total moles) = 0.838 / (0.838 + 0.650) = 0.563

Mole fraction CCl4 = 1 - 0.563 = 0.437

P above solution = (mole fraction CHCl3)(VP CHCl3) + (mole fraction CCl4)(VP CCl4)

                            = (0.563)(0.526 atm) + (0.437)(0.354 atm)

                             = 0.296 atm + 0.155 atm

                              = 0.451 atm

The mole fraction of CHCl3 in the vapor = 0.296 atm / 0.451 atm = 0.656

If you condensed this vapor, then the liquid would have the same

mole fractions of CHCl3 = (0.656)

CCl4 = (1 - 0.656 = 0.344).

When this liquid vaporizes,

the vapor pressure due to CHCl3 = (mole fraction CHCl3)(VP CHCl3)

                                                     = (0.656)(0.526 atm)

                                                     = 0.345 atm