The molality of an aqueous solution of a nonelectrolyte is 0.940 m. For pure wat

ID: 982909 • Letter: T

Question

The molality of an aqueous solution of a nonelectrolyte is 0.940 m.

For pure water at 20 oC:

a) What is the vapor pressure of the solution at 20.0 oC?

P = torr

b) What is the boiling point of the solution to the nearest 0.1 oC?

t = oC

c) What is the freezing point of the solution to the nearest 0.1 oC?

t = oC

d) Assume that the molality and molarity of the solution are equal to estimate the osmotic pressure that is developed by the solution at 20.0 oC?

P = atm

Po = 17.5 torr kf = 1.86 oC/m kb = 0.563 oC/m

m = 0.94

mol = 0.95

kg solvent = 1

mol water = mass/MW = 1000/18 = 55.555

dP = x*P°

x = 0.94/(0.94+55.55) =0.016640

dP = (0.016640)(17.5) = 0.2912

P = 17.5-0.2912 = 17.2088 torr

Tb =Kb*m

Tb = 0.512*0.94 = 0.48128°C

Tb = 100+0.48128 = 100.48128

Tf = -KF*m

Tf = -1.86*0.94 = -1.7484

Tf = -1.7484°C

PI = MRT

PI = 0.94*0.082*(20+273) = 22.58444 atm

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