Question

Calculate the freezing point of a solution containing 104 g FeCl3 in 153 g water.

Calculate the boiling point of a solution above.

Calculate the freezing point of a solution containing 55.2 % KCl by mass (in water).

Calculate the freezing point of a solution containing 3.34 m MgF2.

Calculate the boiling point of a solution above

Answer 1

Part a

freezing point of a solution containing 104 g FeCl3 in 153 g water

FeCl3 = Fe3+ + 3Cl-

Van't Hoff factor i = 4

Moles of FeCl3 = mass/molecular weight

= (104g) /(162.2 g/mol) = 0.6412 mol

Mass of water in kg = 153 g x 1kg/1000g = 0.153 kg

Molality = moles of solute / mass of water

= 0.6412/0.153

= 4.191

Freezing point of water Tf = 0°C

Freezing point depression

Tf - Ts = i x Kf x m

0 - Ts = 4 x 1.86 x 4.191

Ts = - 31.181 °C

Part b

Boiling point of a solution containing 104 g FeCl3 in 153 g water

FeCl3 = Fe3+ + 3Cl-

Van't Hoff factor i = 4

Moles of FeCl3 = mass/molecular weight

= (104g) /(162.2 g/mol) = 0.6412 mol

Mass of water in kg = 153 g x 1kg/1000g = 0.153 kg

Molality = moles of solute / mass of water

= 0.6412/0.153

= 4.191

Boiling point of water Tb = 0°C

Boiling point elevation

Ts - Tb = i x Kb x m

Ts - 100 = 4 x 0.512 x 4.191

Ts = 108.583 °C

Part c

freezing point of a solution containing 55.2 % KCl by mass (in water)

KCl = K+ + Cl-

Van't Hoff factor i = 2

Moles of KCl = mass/molecular weight

= (55.2g) /(74.55 g/mol) = 0.7404 mol

Mass of water in kg = 100 g x 1kg/1000g = 0.100 kg

Molality = moles of solute / mass of water

= 0.7404/0.100

= 7.404

Freezing point of water Tf = 0°C

Freezing point depression

Tf - Ts = i x Kf x m

0 - Ts = 2 x 1.86 x 7.404

Ts = - 27.543 °C

Part d

Ts - Tb = i x Kb x m

Ts = 100 + 2*0.512*7.404

Ts = 107.582 °C

Part e

freezing point of a solution containing 3.34 m MgF2

MgF2 = Mg2+ + 2F-

Tf - Ts = i x Kf x m

0 - Ts = 3 x 1.86 x 3.34

Ts = - 18.637 °C

Part F

Ts = Tb + i x Kb x m

= 100 + 3*0.512*3.34

= 105.130 °C