For 100% dissociation:
Freezing point:
Tf sol'n = Tf solvent - (m x kf )
kf for water is -1.86 C/m and Tf for water is 0 C
Tf sol'n = 0 C - (3.28 m x (1.86 C/m))
Tf sol'n = -6.1008 C
Boiling point:
kb for water is 0.512, kb of water is 100 C
Tb sol'n = Tb solvent - (kb x m)
Tb sol'n = 100 + (0.512 x 3.28)
Tb sol'n = 101.68 C
