Question

You have a 20.0 g of a salt, which you spread 1.0 kg of ice to melt it. What would the new freezing point of the solution be if salt is: NaCl, KCl, CaCl2?

what happens to the concentration of the solution as it freezes?

Answer 1

First calculate the molality of NaCl, KCl and CaCl2

Number of moles = amount in g/ molar mass

NaCl = 20.0 g/ 58.44 g/mol

= 0.342 moles NaCl

KCl = 20.0 g/ 74.5513 g/mol

= 0.268 moles KCl

CaCl2 = 20.0 g/ 110.98 g/mol

= 0.180 moles CaCl2

Given that 1.0 kg of ice

Therefore

molality of NaCl= 0.342 m

molality of KCl= 0.268 m

molality of CaCl2= 0.180 m

and van't hoff factor for non-volatile solute added in the solvent which is fro NaCl and KCl = 2

and CaCl2 = 3

for NaCl:

T_f = iK_fm

i = van't hoff factor for non-volatile solute added in the solvent.

K_f = freezing point depression constant.

m = molality of non-volatile solute added.

T_f = depression in freezing point of solvent.

i = 2

m = 0.342

K_f = 1.86

T_f = 2*1.86*0.342

= 1.27

As T_f is depression in freezing point , so freezing point will be reduced by 1.27 , that is

= - 1.27 ⁰C

for KCl:

T_f = iK_fm

i = van't hoff factor for non-volatile solute added in the solvent.

K_f = freezing point depression constant.

m = molality of non-volatile solute added.

T_f = depression in freezing point of solvent.

i = 2

m 0.2638

K_f = 1.86

T_f = 2*1.86*0.268

= 0.997

As T_f is depression in freezing point , so freezing point will be reduced by 0.997, that is

= -0.997⁰C

for CaCl2:

T_f = iK_fm

i = van't hoff factor for non-volatile solute added in the solvent.

K_f = freezing point depression constant.

m = molality of non-volatile solute added.

T_f = depression in freezing point of solvent.

i = 3

m =0.180

K_f = 1.86

T_f = 3*1.86*0.180

= 1.0044

As T_f is depression in freezing point , so freezing point will be reduced by 1.0044, that is

= -1.0044⁰C