Colligative Properties
Colligative Properties
Colligative Properties
� Colligative Properties – physical properties of
solutions that are affected by the number of
particles but not by the identity of the dissolved
solute
� Vapor pressure lowering
� Boiling point elevation
� Freezing point depression
� Osmotic pressure
Ionic vs. Covalent
� Ionic compounds – dissolve and dissociate into their
ions
� 1M NaCl – 1M Na+ and 1M Cl-
� Creates 2M ions total
� Covalent Compounds – dissolve but do not dissociate
� 1M C12H22O12 – 1M C12H22O12
� Creates 1M dissolved compound
� For now, we will only worry about COVALENT compounds
Boiling Point Elevation
� For a solvent to boil, the vapor pressure = atmospheric pressure
� For a solution containing the solvent, the vapor pressure temperature must be
raised to a higher degree to equal the atmospheric pressure
� Water alone = 100°C
� Water with sugar = 100.77°C
� The boiling point of a solution will always be HIGHER than that of the pure
solvent!
� 5Tb = Kbm
� 5Tb = Boiling point elevation
� Kb = Molal boiling point elevation constant (must be given to you!)
� m = molality
Boiling Point Elevation
� What is the boiling point of a 0.926 molal solution in which sucrose is
dissolved in water? (Kb water = 0.512°C/m)
� Given Find
0.926 m – molality Boiling Point
0.512°C/m - Kb
5Tb = Kbm
5Tb = (0.512°C/m) (0.926m)
5Tb = 0.474°C
Boiling point = normal B.P. + 5Tb = 100.0°C + 0.474°C = 100.474°C
Freezing Point Depression
� For a solution to freeze, KE < interparticle attractive forces
� In a solution, the solute particles interfere with the attractive forces of the solvent
which prevents the solvent from becoming a solid
� For a solution containing the solvent, the freezing point temperature must be
raised to a lower degree to overcome the solute particles interference
� Water alone = 0°C
� Water with sugar = -1.9°C
� The freezing point of a solution will always be LOWER than that of the pure
solvent!
� 5Tf = Kfm
� 5Tf = Freezing point depression
� Kf = Molal freezing point depression constant (must be given to you!)
� m = molality
Freezing Point Depression
� What is the freezing point of a 1.57 molal solution in which sucrose is
dissolved in water? (Kf water = 1.858°C/m)
� Given Find
1.57 m – molality Freezing Point
1.858°C/m - Kf
5Tf = Kfm
5Tf = (1.858°C/m) (1.57 m)
5Tf = 2.92°C
Freezing point = normal F.P. - 5Tf = 0.0°C – 2.92°C = -2.92°C
Solution Concentration
� As the concentration (molality) of a solution
increases:
� The boiling point INCREASES
� The freezing point DECREASES