1. Unit 6: Solutions and Kinetics
Colligative Properties

2. Colligative Properties
Colligative property: a property that depends only on the number of solute particles, and not the type of particle
Examples:
1. Freezing Point Depression (lowering)
2. Boiling Point Elevation (raising)
3. Vapor Pressure Lowering
We will focus on Freezing Point and Boiling Point.

3. Molality
In order to work with colligative properties, we need to use a different type of concentration…
Molality (m): the number of moles of solute per kilogram of solvent
Units: m ex: 3.0 m is “3 molal”
Difference between molarity and molality:

4. Molality Example
What is the molality of a solution of 16.8 grams of lithium sulfate dissolved in kg of water?

5. Molality Practice
What is the molality of a solution of 47.3 grams of potassium iodide dissolved in kg of water?

6. Covalent compounds: will not dissociate → df = 1
Dissociation Factor
Dissociation factor: the number of particles a solute will break in to when dissolved in a solution
Dissociation Factor
Covalent compounds: will not dissociate → df = 1
Ionic compounds: will dissociate into ions → df = # of ions per molecule

7. Dissociation Factor Examples
What is the dissociation factor for each compound?
AlPO4
N2O4
LiCl
CaI2
PCl5
Pb(OH)4
XeF4
Cu2CO3 2 1 3 5

8. Freezing Point Depression
What happens when something freezes (for example, water)?
Decrease in energy slows molecules/atoms down so intermolecular forces have more effect (atoms have less energy to fight them)
Frozen water (ice) molecules are in an orderly pattern
What happens when you add a solute?
Adding Solute
The addition of a solute disrupts and prevents water molecules from forming an orderly pattern.

9. Freezing Point Depression
Freezing Point Depression: adding a substance to a pure solvent lowers the freezing point
To calculate the change in freezing point:
ΔTf = (m)(kf)(df)
Dissociation factor:
the number of particles the solute will break into in solution
molality
constant
Dissociation Factor
Covalent compounds: will not dissociate → d.f. = 1
Ionic compounds: will dissociate into ions → d.f. = # of ions per molecule

10. Freezing Point Depression Example
What is the freezing point of 10.2 grams of NaCl in 5.1 kg of water?
kf = 1.86 oC/m for water
ΔTf = (m)(kf)(df)
To find the new freezing point, SUBTRACT your answer from the normal freezing point

11. Boiling Point Elevation
Solution containing nonvolatile solute
Pure solvent
Solute particles also get in the way of a solvent’s ability to boil, thereby increasing the boiling temperature.

12. Boiling Point Elevation
Boiling Point Elevation: adding a substance to a pure solvent increases the boiling point
To calculate the change in boiling point:
ΔTb = (m)(kb)(df)
Dissociation factor:
how many particles the solute will break into in solution
molality
constant
Dissociation Factor
Covalent compounds: will not dissociate → d.f. = 1
Ionic compounds: will dissociate into ions → d.f. = # of ions per molecule

13. Boiling Point Elevation Example
What is the boiling point of a solution containing g MgCl2 dissolved in g of water?
kb = oC/m for water
ΔTb = (m)(kb)(df)
To find the new boiling point, ADD your answer to the normal boiling point

14. Summary Freezing point depression:ΔTf = (m)(kf)(df)
kf = 1.86 oC/m for water
To find the new freezing point, SUBTRACT your answer from the normal freezing point
Boiling point elevation:ΔTb = (m)(kb)(df)
kb = oC/m for water
To find the new boiling point, ADD your answer to the normal boiling point