Question

Assuming 100% dissociation, calculate the freezing point and boiling point of 2.27 m Na2SO4(aq). Constants may be found here.

Answer 1

Answer - We are given, molality of Na₂SO₄ = 2.27 m ,

We know, normal boiling point of water, H2O = 100.00 °C

Kb = 0.512 °C/m,  

The Van’t off factor i for Na₂SO₄ = 3

We know, ∆Tb = i*Kb*m

                         = 3*0.512 °C/m*2.27 m

                         = 3.49^oC

boiling point of the solution = pure solvent boiling point +∆Tb

                                             = 100 + 3.49

                                             = 103.5^oC

Now for freezing point

We know, Kf = 1.86 °C/m,  

Now we know,

ΔT_f = i*K_f*m

ΔT_f = 3*1.86^oC/m * 2.27 m

        = 12.7 ^oC

So, freezing point of solution = Pure solvent freezing point - ΔT_f

                                              = 0.00^oC – 12.7^oC

                                              = -12.7 ^oC