Question

Order the following solutions in ascending order of their boiling point: 0.0825 m NaCl, 0.050 m KCl, 0.050 M Na2SO4, 0.055m Ca (NO3) 2, 0.050 m Na3PO4. Determine the osmotic pressure of the solutions.

Answer 1

0.0825 m NaCl, 0.050 m KCl, 0.050 M Na2SO4, 0.055m Ca (NO3) 2, 0.050 m Na3PO4.

1)

delta Tb = i x Kb x m

here Kb value is same for all . boiling point depend on i x m value.

a) NaCl = 2 x 0.0825 = 0.165

b) KCl = 2 x 0.050 = 0.100

c) Na2SO4 = 3 x 0.050 = 0.150

d) Ca(NO3)2 = 3 x 0.055 = 0.165

e) Na3PO4 = 4 x 0.050 = 0.200

order of boiling point :

KCl < Na2SO4 < NaCl < Ca(NO3)2 < Na3PO4 .

2)

osmatic pressure = i x M x S x T

a) NaCl :

osmatic pressure = 2 x 0.0825 x 0.0821 x 298

                            = 4.04 atm

b)

KCl :

osmatic pressure = 2 x 0.05 x 0.0821 x 298

                            = 2.45 atm

c)

Na2SO4 :

osmatic pressure = 3 x 0.05 x 0.0821 x 298

                            = 3.67 atm

d)

Ca(NO3)2 :

osmatic pressure = 3 x 0.055 x 0.0821 x 298

                            = 4.04 atm

e) Na3PO4 :

osmatic pressure = 4 x 0.05 x 0.0821 x 298

                            = 4.89 atm