Question
The molality of an aqueous solution of a nonelectrolyte is 1.270 m.For pure water at 20 oC: Po = 17.5 torr kf = 1.86 oC/m kb = 0.563 oC/m a) What is the vapor pressure of the solution at 20.0oC?
P = torr
b) What is the boiling point of the solution to the nearest 0.1oC?
t = oC
c) What is the freezing point of the solution to the nearest 0.1oC?
t = oC
d) Assume that the molality and molarity of the solution are equalto estimate the osmotic pressure that is developed by the solutionat 20.0 oC?
P = atm
Po = 17.5 torr kf = 1.86 oC/m kb = 0.563 oC/m
Explanation / Answer
a) Vapour pressure of solution = vapor pressure of puresolvent * mole fraction of solvent. Given molality = 1.18 m i.e 1.18 moles in 1000 gof water Mole fraction of solute = Moles of solute / Total number ofmoles = 1.18 / {1.18 + (1000 g/ 18 g/mol) } =1.18 / 56.73 =0.02 Mole fraction of solvent = 1 - mole fraction of solute = 1 - 0.02 =0.98 Vapour pressure of solution = 17.5 torr * 0.98 = 17.15 torr b) Boiling point elevation, = Kb . molality =0.563 oC/m * 1.18 m =0.6643 C Boiling point = 100 C + 0.6643 C = 100.7 oC c) Freezing point depression = Kf . molality = 1.86 oC/m * 1.18 m = 2.1948 C Freezing point = 0 C - 2.1948 C =- 2.2 oC d) Molarity, C = molality = 1.18 M Temperature = 20 C = 293 K Osmotic pressure, = CRT = 1.18 M * 0.0821 L.atm/K.mol * 293 K = 28.38 atm