COLLIGATIVE PROPERTIES

1. Colligative Properties

2. Colligative Properties
These are the effects that a solute has on
a solvent.
When water has something dissolved in it,
its physical properties change.
It will no longer boil at 100oC and it will no
longer freeze at 0oC like pure water.

3. Three Main Effects
Lowers the vapor pressure of a solvent.
– Lower vapor pressure means that fewer
water molecules can escape from the liquid
phase into the gas phase at given
temperature. Remember, a lower vapor
pressure means a higher boiling point!
2) Raises the boiling point of a solvent.
3) Lowers the freezing point of a
solvent.

4. Freezing Point Depression

5. Boiling Point Elevation

6. Applications
–salting icy roads
–making ice cream
–antifreeze
cars (-64°C to 136°C)

7. Colligative Properties
-These depend only on the
number of dissolved particles
-Not on what kind of particle
General Rule: The more solute
particles that are present in a
solvent, the greater the effect.

8. # of Particles
–Nonelectrolytes (covalent)
remain intact when dissolved
1 particle
–Electrolytes (ionic)
dissociate into ions when dissolved
2 or more particles
– Electrolytes have a stronger affect in
lowering the freezing point and elevating the
boiling point because it puts more particles
into the solution.

9. Remember the rule, the more particles,
the greater the effect!
The Dissociation Factor (d.f.) for an
electrolyte is the number of ions a
compound dissociates into.
NaCl gives Na+ ions and Cl- ions, which is
2 particles, therefore d.f. = 2
What is “d.f. “ for Al(NO3)3 ?
Al(NO3)3  Al3+ + 3 NO3
- = 4 particles

10. Non-electrolytes
A compound that does not conduct electricity
when dissolved in water.
Examples are glucose or any other sugars and
alcohols, such as ethanol (CH3CH2OH)
The d.f. = 1 for any non-electrolyte.
Why?
Covalent molecules do not break apart when
they become solvated by water molecules.

11. Calculations
t: change in temperature (°C)
K: constant based on the solvent (°C·kg/mol)
Use Kb for boiling point elevation and Kf for
freezing point depression. Each solvent has
it’s own unique factors!
m: molality (m) = moles solute/Kg solvent
d.f.: # of particles
t = m · d.f. · K

12. Example 1:
If 48 moles of ethylene glycol, C2H4(OH)2, is
dissolved in 5.0kg of water. What is the boiling
point of the solution?
∆Tbp = m  d.f.  Kb
d.f. = 1 ( it is a non-eletrolyte)
Kb = 0.52oC/m (a constant for water)
Molality = m = moles solute/kg solvent
∆Tbp = (48 moles) (1) (0.52oC/molal)
(5.0 Kg)
∆Tbp = 5.0 oC New BP = 100 + 5 = 105 oC

13. Example 2:
55.5 grams of CaCl2 are dissolved in 2.0kg of
water. What is the freezing point of this
solution?
∆Tfp = m  d.f.  Kf
d.f. = 3 (CaCl2 produces 3 particles)
CaCl2 Ca2+ + 2 Cl-
Kf = 1.86oC/m (a constant for water)
∆Tfp = (0.25molal) (3) (1.86oC/molal)
Must calculate molality
Molality = mass/molar mass
Kg Solvent
= (55.5 gram/111g/mole)
2.0 kg
= 0.25m
= 1.4 oC
New FP = -1.4 oC