DTb = Kb
msolute
i
where
Kb = molal boiling point elevation constant (for water = 0.51˚C/m)
Now try a problem that deals with freezing point depression and boiling point elevation.
Example
Calculate the freezing point and boiling point of a solution of 100 g of ethylene glycol (C2H6O2) in 900 g of water.
Explanation
Calculate molality:
Freezing point depression = (m)(Kf)(i)
Tf = (1.79)(1.86)(1) = 3.33ºC
Freezing point = 0ºC - 3.33ºC = -3.33ºC
Boiling point elevation = (m)(Km)(i)
Tb = (1.79)(0.51)(1) = 0.91ºC
Boiling point = 100ºC + 0.91ºC = 100.91ºC
