Question

Assuming 100% dissociation, calculate the freezing point and boiling point of 3.37 m AgNO3(aq).

Answer 1

we have elevation in BP formula as dT = i x Kb x m

where i = vantoff factor = 2 for AgNO3 (since AgNo3 gives 2 ions per complete dissociation)

m = molality = 3.37 ,   Kb = boiling point constant = 0.512C/m

we get dT   = 2 x0.512 C/m x 3.37 m

dT = 3.45 = Tb ( solution) - Tb ( solvent water)

      3.45 = Tb ( solution) - 100

Tb solution = 103.45 C

2) now we have formula for depression of freezing point dT = i x Kf x m

where Kf = freezing point constant = 1.86 C/m ,

dT = 2 x 1.86 C/m x 3.37

dT = 12.54 C = T( freexing point of solvent) - T ( freezing point of solution)

12.54 = 0 - T( solution)

Freezing point of solution = - 12.54 C