Question

The van ’t Hoff factors for ionic solutes are not necessarily integers. We’ve assumed that all soluble salts dissociate completely but that’s not necessarily the case at all concentrations for all salts. The freezing points of 0.0935 m ammonium chloride and 0.0378 m ammonium sulfate in water were found to be -0.322 ◦C and -0.173 ◦C, respectively. What are the values of the van ’t Hoff factors for these salts.

Answer 1

1) dTf = i*Kf*m

DTf = Tf-TS c

Tf = freezingpoint of solvent = 0 c

Ts = freezingpoint of solution = -0.322 c

i= vanthoff factor of solute(NH4Cl) = x

Kf of water = 1.86 c/m

m = molality = 0.0935 m

(0-(-0.322)) = x*1.86*0.0935

i = vanthoff factor of solute(NH(so4)2) = 1.85

2) dTf = i*Kf*m

DTf = Tf-TS c

Tf = freezingpoint of solvent = 0 c

Ts = freezingpoint of solution = -0.173 c

i= vanthoff factor of solute((NH4)2SO4) = x

Kf of water = 1.86 c/m

m = molality = 0.0378 m

(0-(-0.173)) = x*1.86*0.0378

i = vanthoff factor of solute((NH4)2SO4) = 2.46