Colligative properties are the physical changes that result from adding solute to a solvent. Colligative Properties depend on how many solute particles are present as well as the solvent amount, but they do NOT depend on the type of solute particles, although do depend on the type of solvent.
Solvent molecules are attracted together to a certain degree. The tighter the attraction (called intermolecular force) the less molecules that will escape liquid into gas phase = vapor pressure
If the Solvent is attracted to the solute particles more than to the other solvent molecules, then the vapor pressure will lower because the solvent molecules are held more tightly in the liquid phase.
A lower vapor pressure = a higher temperature is required to reach boiling of the solution (ΔTb).
Molal boiling-point-elevation constant, Kb, expresses how much ΔTb changes with molality, m and ionization, i.
The equation for boiling point elevation is: ΔTb = Kbmi
for H2O Kb = 0.51°C/m
i is the number of ions created when the substance is dissolved. For example, NaCl splits into Na+ and Cl-, so i = 2.
Example: Automotive antifreeze consists of ethylene glycol (C2H6O2) a non volatile nonelectrolyte. Calculate the new boiling point of the water in a radiator if 250. g of ethylene glycol are added to 750. g H2O in the radiator.
Find the molality of solution
m = moles solute/ kg solvent = moles C2H6O2 / kg H2O
m = 250. g / 62.08 g/mol / 0.750 kg = 5.37 m
ΔTb = Kbmi = (0.51°C/m) (5.37 m)(1) = 2.74°C
new bp = 100 + 2.74 = 102.74°C
The solute prevents the solvent from forming bonds with itself required to form a solid. Therefore, the solution freezes at a lower temperature (ΔTf) than the pure solvent.
Decrease in freezing point (ΔTf) is directly proportional to molality. The equation for freezing point depression is similar to boiling point elevation:
ΔTf = Kfmi
Kf is the molal freezing-point-depression constant
Kf for water = 1.86°C/m
Example: The automotive antifreeze also alters the freezing point of the water in a car’s radiator. If 150. g of C2H6O2 are dissolved in 850. g H2O, what will the new freezing point of the solution be?
Find the molality of solution
m = moles solute/ kg solvent = moles C2H6O2 / kg H2O
m = 150. g / 62.08 g/mol / 0.850 kg = 2.84 m
ΔTf = Kfmi = (1.86°C/m) (2.84 m)(1) = 5.28°C
new fp = 0 – 5.28 = -5.28°C
You may recall osmosis as being the movement of water through a semi-permeable membrane from Biology class. This is actually a colligative property.
Osmotic pressure, π, is the pressure required to stop osmosis which can be calculated using the following formula:
π = MRT = molarity x ideal gas law constant x Temp in K