1. Chapter 13 Section 5: Colligative Properties
Colligative properties depend on quantity of solute molecules.
Boiling Point Elevation
When a solute is dissolved in a solvent, the solute reduce the ability of the surface solvent molecules to escape the liquid.
The vapor pressure of the solvent is lowered, therefore the temperature at which the solution will boil will be higher than before.
Example: Sugar water will boil at a temperature above 100 ºC.
Why?
The sugar lowers the vapor pressure of water. So the water molecules near the surface of the liquid have a more difficult time escaping into the vapor phase and the temperature has to be higher in order for them to break away.

2. Vapor Pressure of Solutions

3. Calculating Boiling Points of Solutions
To calculate how much the boiling point temperature will increase, we must know the concentration of the solution in units of molality.
Molality (m) = moles of solute/ kg of solvent
You must also know if the substance breaks apart into ions.
If it does, this will increase the # of particles in the solution, therefore increasing the molality of particles and increasing the boiling point even further.
Here’s the formula… ∆Tb = (Kb)(m)(i)
Molal boiling-point-elevation constant, Kb, expresses how much Tb changes with molality, m.
The variable “i” represents the # of particles present in the solution if ionic. (This variable is not given on the AP formula sheet!)
Examples: NaCl Na+ + Cl− …i= CaCl2 Ca Cl− …i=3
Sugar and other molecular nonelectrolytes don’t break into ions, so i=1.

4. Freezing Points of Solutions
Freezing Point Depression
When a solute is dissolved in a solvent, the solute gets in the way so the solvent can’t form intermolecular bonds as well.
The liquid can’t form solid particles and so the freezing point is lowered.
Example: Sugar water will freeze at a temperature below 0 ºC.
Why?
The sugar gets in the way of the hydrogen bonds that water tries to form. So the water molecules can’t form ice crystals until they are even colder than 0 ºC.

5. Calculating Freezing Points of Solutions
The same information about boiling point elevations must be known in order to calculate freezing point depressions.
Here’s the formula… ∆Tf = (Kf)(m)(i)
Only the name of the constant, Kf, changes! It’s called the molal freezing-point-depression constant.
Let’s Practice!!

6. molality = 0.75 moles of glucose/0.75 kg of water = 1 m
Practice Problems
If grams of glucose, (M.W. = 180 g/mole), is dissolved in grams of water, what is the new F.P and B.P. of the solution?
135.0 g of glucose x
molality = 0.75 moles of glucose/0.75 kg of water = 1 m
∆Tf = (Kf)(m)(i)
∆Tf = (1.86 ºC/m)(1 m )(1) = 1.86 ºC
…so New F.P. = ºC
∆Tb = (Kb)(m)(i)
∆Tb =( 0.51 ºC/m)(1 m )(1) = 0.51ºC
…so New B.P. = ºC
1 mole
= 0.75 moles of glucose
180 g