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