Calculate the freezing point and boiling point of each aqueous solution, assumin

Question

Calculate the freezing point and boiling point of each aqueous solution, assuming complete dissociation of the solute.

a)Calculate the freezing point of the solution containing 0.111

m K2S.

b)Calculate the boiling point of the solution above.

c)Calculate the freezing point of the solution containing 23.9

g of CuCl2 in 451 g water.

d)Calculate the boiling point of the solution above.

E)Calculate the freezing point of the solution containing 7.4

% NaNO3 by mass (in water).

f)Calculate the boiling point of the solution above.

Explanation / Answer

a)Calculate the freezing point of the solution containing 0.111 m K2S.

Solution :- Kf of water = 1.86 C/m

K2S is ionic compound it can form 3 ions on dissociation

Therefore the i = 3

Formula to calculate the change in freezing point is as follows.

Delta Tf = Kf*m *I

Delta Tf = 1.86 C per m * 0.111 m * 3

              = 0.620 oC

Freezing point of pure water = 0 oC

Freezing point of solution = 0 oC - 0.620 oC = -0.620 oC

b)Calculate the boiling point of the solution above.

Solution :- Kb of water is 0.512 oC/m

Delta Tb = Kb*m*i

               = 0.512 oC*0.111 m* 3

               = 0.170 oC

Boiling point of the solution = 100 oC +0.170 oC =100.17 oC

c)Calculate the freezing point of the solution containing 23.9 g of CuCl2 in 451 g water.

Solution :- First need to calculate the molality of the CuCl2 solution

Moles = mass/ molar mass

Moles of CuCl2 = 23.9 g / 134.452 g per mol = 0.178 mol

Molality = moles / kg solvent

               = 0.178 mol / 0.451 kg

               = 0.395 m

CuCl2 can form 3 ions so i=3

Lets calculate the freezing point

Delta Tf = Kf*m*I

             = 1.86 oC/m * 0.395 m * 3

             = 2.20 oC

Freezing point of solution = 0 oC- 2.20 oC = - 2.20 oC

d)Calculate the boiling point of the solution above.

Solution :-

Delta Tb = Kb*m*i

               =0.512 oC/m * 0.395 m*3

              = 0.610 oC

Boiling point of solution = 100 oC+0.610 oC = 100.610 oC

E)Calculate the freezing point of the solution containing 7.4 % NaNO3 by mass (in water).

Solution :- 7.4 % by mass solution means

7.4 g NaNO3 and 92.6 g water

Lets first calculate the molality

Moles of NaNO3 = 7.4 g / 85 g per mol = 0.0871 mol

Molality = 0.0871 mol / 0.0926 kg

               = 0.941 m

Lets calculate the freezing point

Delta Tf= Kf*m*i

            = 1.86 oC /m * 0.941 m*2                           (NaNO3 forms 2 ions therefore i=2)

          = 3.50 oC

Freezing point of the solution = 0 oC – 3.50 oC = -3.50 oC

f)Calculate the boiling point of the solution above

Solution :-

Delta Tb= Kb*m*i

             = 0.512 oC /m * 0.941 m* 2

             = 0.963 oC

Boiling point of solution =100 oC +0.963 oC = 100.963 oC

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