Question

The vapor pressure of pure carbon tetrachloride, CCl4, (molar mass 153.82 g/mol) is 0.354 atm, and the vapor pressure of pure chloroform, CHCl3 (molar mass 119.38 g/mol), is 0.526 atm at 316 K. A solution is prepared from equal masses of the two compounds at this temperature. If the vapor over the original solution is condensed and isolated into a separate flask, what with the vapor pressure of the chloroform be above this new solution?

Answer 1

Solution :-

Lets assume we have 100 g of each component in the solution

Now lets calculate the moles of each

Moles = mass / molar mass

100 g CCl4 / 153.8 g/mol = 0.650 moles CCl4
100 g CHCl3 / 119.4 g/mol CHCl3 = 0.838 moles CHCl3

Total moles = 0.650 mol + 0.838 mol = 1.488 mol

Now lets find the mole fraction of each component

Mole fraction = moles of x /total moles

Mole fraction of the CCl4 = 0.650 mol / 1.488 mol = 0.437

Mole fraction of the CHCl3 = 1 – 0.437 = 0.563

Now lets calculate the total pressure above the solution using the mole fractions and the vapor pressure of pure liquids

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

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

                               = 0.155 + 0.296

                               = 0.451 atm

Now we get the total pressure above the solution

When we condense the vapor in the separate in the flask then the mole fractions of the CCl4 and CHCl3 remain same

Now lets calculate the mole fraction of the CHCl3 in the vapor = 0.296 / 0.451 = 0.656

Now the vapor pressure of the CHCl3 in the new flask = mole fraction * vapor pressure of pure CHCl3

                                                                                                  = 0.656 * 0.526 atm

                                                                                                   = 0.345 atm

So the vapor pressure of the CHCl3 in the separate flask = 0.345 atm