Question

A solution is made by dissolving 0.0500 mol of HF in 1.00 kg of water. The solution was found to freeze at -0.104C. Calculate the percent dissociation of HF in this solution.

Answer 1

Answer –

We are given, moles of HF = 0.0500 mol , mass of water = 1.00 kg

Freezing point = -0.104 ^oC,

First we need to calculate freezing point depression from the freezing point of solution

We know the pure solvent freezing point

ΔTf = freezing point of pure solvent – freezing point of solution

        = 0.00 ^oC –(-0.104 ^oC)

      = 0.104 ^oC

Now we need to calculate molality of solution

ΔTf = Kf * m

We know for water , Kf = 1.86 ^oC/m

So, m = ΔTf / Kf

         = 0.104 ^oC / 1.86 ^oC.m^-1

         = 0.0559 m

Now we need to calculate molality from the given moles and mass of solvent

Molality = moles / kg of solvent

               = 0.0500 mole / 1.00 kg

              = 0.0500 m

moles of orginal HF = 0.0500 m * 1.00 kg

                             = 0.0500

mole of HF in the solution after dissolving in the water = 0.0559 *1.00 kg

                                                                              = 0.0559 moles

so mole are dissociate = 0.0559-0.0500 = 0.0059 moles

,so percent dissociation = 0.0059 / 0.05 * 100 %

                                      = 12 %