Question

A solution at 30 ° C is prepared by dissolving 3 g of urea in 100 ml of water. Under these conditions, calculate for the solution:

a) the boiling point

b) the freezing point

c) the vapor pressure

d) the osmotic pressure

e) the ebulloscopic and cryoscopic constant (from the solvent data).

Answer 1

a)

mass of urea = 3 g

moles of urea = 3 / 60.06 = 0.050 mol

molality = moles / mass of solvent

            = 0.050 / 0.10

molality = 0.50 m

delta Tb = Kb x m

Tb - 100 = 0.512 x 0.50

Tb = 100.26

boiling point Tb = 100.26 oC

b)

delta Tf = Kf x m

0 - Tf = 1.86 x 0.50

Tf = - 0.93 oC

freezing point = - 0.93 oC

c)

moles of water = 100 / 18.02 = 5.55 mol

mole fraction of urea = 0.05 / 0.05 + 5.55 = 0.00893

Po - Ps / Po = Xsolute

31.8 - Ps/ 31.8 = 0.0089

Ps = 31.5 torr

vapor pressure of solution = 31.5 torr

d)

osmatic pressure = M x S x T

                           = 0.5 x 0.0821 x 303

osmatic pressure = 12.4 atm