The vapor pressure of carbon tetrachloride, CCl₄, is 0.354 atm and 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. a) Calculate the mole fraction of the chloroform in the vapor above the solution. b) If the vapor above the original solution is condensed and isolated into a separate flask, what would the vapor pressure of chloroform be above this new solution?

a) 0.65

b) 0.342 atm

Explanation:

a) First, we need to know the molar mass of the compounds. By periodic table, the molar mass of the elements are:

C = 12 g/mol; Cl = 35.5 g/mol; H = 1 g/mol. So:

CCl4 = 12 + 4x35.5 = 154 g/mol

CHCl3 = 12 + 1 + 3x35.5 = 119.5 g/mol

They both have the same mass, so we can choose the basis of calculus as 100 g (you can choose any other basis, the result will be the same because the fraction will be the same!)

The number of moles is:

n = mass/molar mass

nCCl4 = 100/154 = 0.649 mol

nCHCl3 = 100/119.5 = 0.837 mol

So, the total number of moles is nt = nCCl4 + nCHCl3 = 1.486

Then, the molar fractions in the solution will be:

xCHCl3 = nCHCl3/nt = 0.837/1.486 = 0.56

xCCl4 = 1 - 0.56 = 0.44

By Dalton's Law

Pt = PCCl4*xCCl4 + PCHCl3*xCHCl3

Where Pt is the total pressure of the vapor, and PCCl4 and PCHCl3 are the vapor pressure of the compounds. So:

Pt = 0.44*0.354 + 0.56*0.526 = 0.451 atm

The molar fraction of the vapor will be:

yCHCl3 = (xCHCl3*PCHCl3) / Pt

yCHCl3 = (0.56*0.526) / 0.451 = 0.65

b) When the vapor is condensed, the molar fraction of the vapor phase will be the molar fraction of the solution, so xCHCl3 = 0.65

P = molar fraction x vapor pressure

P = 0.65 x 0.526

P = 0.342 atm