Freezing
Point Depression
Water freezes at 0°C by forming a highly ordered crystal.
If another substance added to water, it becomes more difficult to crystallize.
Below is an illustration representing water crystals (squares), sugar crystals
(triangles), and water/sugar crystals (squares/triangles).
Because of its disorder, the mixture is more difficult
to crystallize. Therefore, mixtures always freeze at lower temperature
than pure substances. One mol solute particles per kg water depresses F.P.
by 1.86°C. This value known as molal F.P. constant (Kf).
DTf
= imKf
DTf
= freezing point depression
i =
van't Hoff factor (number of ions from 1 molecule)
m = molality
Kb/Kf
Values
| Substance |
Kb |
Kf |
| Benzene (C6H6) |
2.53 |
5.07 |
| Camphor (C10H16O) |
5.95 |
37.7 |
| Chloroform (CHCl3) |
3.63 |
4.70 |
| Diethyl ether (C4H10O) |
2.02 |
1.79 |
| Ethanol (C2H6O) |
1.22 |
1.99 |
| Water |
0.51 |
1.86 |
Problem #1
Calculate freezing point for 2.50m Al2(SO4)3
DTf
= imKf = (5)(2.50)(1.86) = 23.3
F.P. = -23.3°C
Problem #2
If 160g molecular solute
and 500g chlorofom depresses F.P. of chlorofom by 20.0°C, calculate
molar mass of solute.
DTf
= mKf
20 = [160/mm/0.500](4.70)
mm = 75