Question

A solution is made by dissolving 0.543 mol of nonelectrolyte solute in 817 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution. Constants may be found here.

Answer 1

Solution :-

converting gram to kg

817 g * 1 kg / 1000 g = 0.817 kg benzene

Lets first calculate the moality of the solution

molality = moles of solute / kg solvent

               = 0.543 mol / 0.817 kg

               = 0.6646 m

now lets find the freezing point of the solution

freezing point constat of benzene Kf = 5.12 C per m

freezing point of benzene = 5.5 C

Delta Tf = Kf*m

               = 5.12 C/m * 0.6646 m

               = 3.40 C

Freezing point of solution = freezing point of pure benzene - delta Tf

                                              = 5.5 C - 3.40 C

                                              = 2.10 C

Therefore freezing point of the solution = 2.10 ^oC

Now lets calculate the boiling point of the solution

Boling point constant of the benzene Kb = 2.65 C/m

boiling point of pure benzene = 80.1 C

Delta Tb= Kb* m

               = 2.65 C/m * 0.6646 m

               = 1.76 C

Therefore boiling point of the solution = boiling point of pure benzene + Delta Tb

                                                                     = 80.1 C + 1.76 C

                                                                     = 81.86 C

Therefore the boiling point of the solution = 81.86 ^oC