Question

part A The normal boiling point of water, H2O is 100.00 °C and its Kbp value is 0.512 °C/m. Assuming complete dissociation of the electrolyte, if 11.22 grams of sodium nitrate (NaNO3, 85.00 g/mol) are dissolved in 236.2 grams of water, what is the boiling point of the solution?

part b The normal freezing point of water, H₂O is 0.00 °C and its K_fp value is 1.86 °C/m.

Assuming complete dissociation of the electrolyte, if 10.25 grams of potassium iodide (KI, 166.0 g/mol) are dissolved in 178.3 grams of water what is the freezing point of the solution?

Answer 1

Answer – Part A) We are given, normal boiling point of water, H2O = 100.00 °C

Kb = 0.512 °C/m, mass of NaNO₃ = 11.22 g , mass of water = 236.2 g

Molar mass of NaNO₃ = 85.00 g/mol

First we need to calculate the moles of NaNO₃

Moles of NaNO₃ = 11.22 g / 85.0 g.mol^-1

                            = 0.132 moles

Molality of NaNO₃ = 0.132 moles / 0.2362 kg

                            = 0.559 m

The Van’t off factor i for NaNO₃ = 2

We know, ∆Tb = i*Kb*m

                         = 2*0.512 °C/m*0.559 m

                         = 0.572^oC

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

                                             = 100 + 0.572

                                             = 100.6^oC

Part B) Given, mass of KI = 10.25 g, mass of water = 178.3 g

Kf = 1.86 °C/m, Molar mass of KI = 166.00 g/mol

First we need to calculate the moles of KI

Moles of KI = 10.25 g / 166.0 g.mol^-1

                    = 0.0617 moles

Molality of KI = 0.0617 moles / 0.1783 kg

                      = 0.346 m

Now we know,

ΔT_f = i*K_f*m

ΔT_f = 2*1.86^oC/m * 0.346 m

        = 1.29 ^oC

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

                                               = 0.00^oC – 1.291^oC

                                               = -1.29 ^oC