Question

A solution is prepared by dissolving 0.59 mol of MgCl2 in 0.25 kg of water. How many moles of ions are present in solution? What is the change in the boiling point of the aqueous solution?

Answer 1

MgCl2(aq) --------------> Mg^2+ (aq) + 2Cl^- (aq)

1 mole of Mg^2+ and 2 moles of Cl^- . So total 3 moles of ions.

3 moles of ions are present in MgCl2

i = 3

molality of MgCl2   = no of moles/weight of Kg

                              = 0.59/0.25   = 2.36m

Kb   = 0.512⁰C/m

Tb   = i*kb*m

            = 3*0.512*2.36

             = 3.625⁰C

The chang in boiling point of the aqueous solution = 3.625⁰C >>>>>>answer