Question

In: Chemistry

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

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.

Solutions

Expert Solution

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


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