Question

Bromine is sometimes used as a solution in tetrachloromethane, CCl4. What is the vapor pressure (in mm Hg) of a solution of 6.40 g of Br₂ in 114.5 g of CCl₄ at 300K? The vapor pressure of pure bromine at 300K is 30.5 kPa, and the vapor pressure of CCl₄ is 16.5 kPa.

Answer 1

Solution :-

Lets calculate the moles of Br2 and CCl4

Moles of of Br2 = 6.40 g Br2 / 159.808 g per mol = 0.04005 mol Br2

Moles of CCl4 = 114.5 g CCl4 / 153.82 g per mol = 0.74437 mol CCl4

Now lets calculate the mole fraction of each

Mole fraction of Br2 = 0.04005 mol Br 2 / (0.04005 mol + 0.74437 mol )

                                      = 0.05106

Mole fraction of CCl4 = 1- 0.05106 = 0.94894

Now lets calculate the parital pressure of each

Partial pressure of Br2 = mole fraction * vapor pressure of pure Br2

                                          = 0.05106 * 30.5 kpa

                                         =1.56 kpa

Vapor pressure of CCl4 = 0.94894 * 16.5 kpa

                                         = 15.7 kpa

Vapor pressure of solution = parital pressure of Br2 + partial pressure of CCl4

                                               = 1.56 kpa + 15.7 kpa

                                             = 17.26 kpa