Question

Calculate the freezing point and boiling point in each solution, assuming complete dissociation of the solute. a.Calculate the freezing point of a solution containing 13.0 g FeCl3 in 164 g water. b.Calculate the boiling point of a solution above. c.Calculate the freezing point of a solution containing 3.9 % KCl by mass (in water). Express your answer using two significant figures. d.Calculate the boiling point of a solution above. e.Calculate the freezing point of a solution containing 0.162 m MgF2. f.Calculate the boiling point of a solution above.

Answer 1

This is a typical example of colligative properties.

Recall that a solute ( non volatyle ) 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 concentation.

At the end:

Tf mix = Tf solvent - dTf

Tb mix = Tb solvent - dTb

a)

m = 13 g of FeCl3; m = 164 g water

mol of FeCl3 = mass/MW = 13/162.2 = 0.0801479 mol

total ions = 4; Fe+3 and 3 Cl-

kg of solvnet = 164 g = 0.164 kg

Kf = -1.86

dTf = -Kf*molality * i

dTf = -1.86*0.0801479 / 0.164 * 4

dTf = -3.63 °C

Tf = 0 -3.63 = -3.63 °C

b)

BP for this solution:

dTb = Kb*molality * i

dTb = 0.512*0.0801479 / 0.164 * 4

dTb = -1.0008 °C

Tf = 100 + 1.0008 = 101.0008 °C

c)

for a solution of 3.9 % w/w,

assume a basis of 100 g of solution, so

3.9 g KCl; (100-3.9 ) = 96.1 g of water; = 0.0961 kg

mol of KCl = mass/MW = 3.9/74.5513 = 0.0523 mol

total ions 2; K+ and Cl

dTf = -Kf*molality * i

dTf = -1.86*0.0523 / 0.0961 *2

dTf -2.024

Tf = -2.024 °C

d)

now, get Tb

dTb = Kb*molality * i

dTb = 0.512*0.0523 / 0.0961 *2

dTb = 0.5572

Tb = 100 + 0.5572 = 100.5572 °C

e)

0.162 m f MgF2

total ions 3; Mg+2 and 2F-

dTf = -Kf*molality * i

dTf = -1.86*0.162 *3

dTf -0.90396

Tf = -0.90396 °C

f)

get BP

0.162 m f MgF2

total ions 3; Mg+2 and 2F-

dTbf = Kb*molality * i

dTb = 0.512*0.162 *3

dTbf 0.248832

Tb = 100 + 0.248832 °C

Tb = 100.248832 °C