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 0.114 m K2S.

B.) Calculate the boiling point of the solution above.

C.) Calculate the freezing point of the solution containing 22.1 g of CuCl2 in 459 g water.

D.) Calculate the boiling point of the solution above.

E.) Calculate the freezing point of the solution containing 5.6 % NaNO3 by mass (in water). Express your answer using two significant figures.

F.) Calculate the boiling point of the solution above.

Answer 1

Apply Colligative properties

This is a typical example of colligative properties.

Recall that a solute ( non volatile ) can make a depression/increase in the freezing/boiling point via:

dTf = -Kf*molality * i

dTb = Kb*molality * i

where:

Kf = freezing point constant for the SOLVENT; Kb = boiling point constant for the SOLVENT;

molality = moles of SOLUTE / kg of SOLVENT

i = vant hoff coefficient, typically the total ion/molecular concentration.

At the end:

Tf mix = Tf solvent - dTf

Tb mix = Tb solvent - dTb

for water, Kb = 0.512 and Kf = 1.86

a)

Tf mix = Tsolvent - Kf*m*i

K2S = 3 ions, so i = 3

Tf mix = 0 - 1.86*0.114*3 = -0.63612 °C

b)

Tb mix = Tbsolvent + Kb*m*i

K2S = 3 ions, so i = 3

Tb mix = 100 + 0.512*0.114*3 = 100.1751 °C

c)

molality = mol of solute / kg of solvent = (mass of solute / MW ) /(kg) = (22.1/134.45 ) / (0.459) = 0.3581119

ions = 3, Cu+2 and 2Cl-

Tf mix = Tsolvent - Kf*m*i

Tf mix = 0 - 1.86*0.3581119 *3 = -1.99 °C

d)

Tb mix = Tbsolvent + Kb*m*i

CuCl2 = 3 ions, so i = 3

Tb mix = 100 + 0.512*0.3581119 *3 = 100.550 °C

e)

5.% by mas..

100 g --> 5.6 g is NaNO3, (100-5.6) = 94.4 g is water

mol of NaNO3 = mass/MW = (5.6/84.9947 ) = 0.06589; kg = 0.0944

molal = 0.06589/0.0944 = 0.6979

Tf mix = Tsolvent - Kf*m*i

Tf mix = 0 - 1.86*0.6979*2 = -2.59 °C

f)

Tb mix = Tbsolvent + Kb*m*i

i = 2

Tb mix = 100 + 0.512*0.6979*2 = 100.71 °C