Question

A solution of hydrofluoric acid was prepared by dissolving 0.5mol HF in 2*10^2 g water. What is the boiling points of this solution if only 50% of the acid dissociates? The K boiling for water is 0.512c/m

Answer 1

Solution :-

0.50 mol HF dissolved in 2*10^2 g water

50 % dissociation

Means 0.5*50 % /100 % = 0.25

So the moles of H+ and F- formed are 0.25 each

And HF moles remain = 0.5-0.25 = 0.25

So the total particles in the solution = 0.25 + (0.25*2) = 0.75

Now lets calculate the molaity

Molality = moles / kg solvent

                = 0.75 mol / 0.200 kg

                = 3.75 m

Now lets calculate the change in the boiling point.

Delta Tb = Kb * m

              = 0.512 C/m * 3.75 m

              = 1.92 C

Boiling point of solution = boiling point of pure water + delta Tb           

                                             = 100 C + 1.92 C

                                            = 101.92 C