What are the boiling point and freezing point of a 0.625m aqueous solution of any nonvolatile, nonelectrolyte solute?

1. The boiling point elevation is directly proportional to the molality of the solution according to the equation: ΔTb = Kb · b.
ΔTb = 0,512°C·kg/mol · 0.625 mol/kg = 0.32°C.
Tb = 100°C + 0.32°C = 100.32°C.
2. ΔTf = Kf · b.
ΔTf = 1.86 °C/m · 0.625 m = 1.1625°C.
Tf = 0°C - 1.1625°C = -1.1625°C; freezing point of a 0.625m aqueous solution of any nonvolatile, nonelectrolyte solute.