1. Colligative Properties

2. Calculate the Van’t Hoff Factor for a Compound
Objective:
Today I will be able to:
Describe the 4 colligative properties of vapor pressure, boiling point, freezing point and osmotic pressure
Calculate the Van’t Hoff Factor for a Compound
Calculate the freezing point depression and boiling point elevation of a solute
Evaluation/Assessment:
Informal Assessment – monitoring student progress as they complete the summary and practice problems
Formal Assessment – analyzing student responses to the activities and exit ticket
Common Core Connection
Make sense of problem and persevere in solving them
Look for and make use of structure
Use appropriate tools strategically

3. Lesson Sequence Evaluate: Warm – Up
Explore: Colligative Properties Research
Explain: Colligative Properties Notes
Elaborate: Colligative Properties Calculations
Evaluate: Exit Ticket

4. Warm - Up
Take a worksheet from the front desk. Interpret the graph on the page.

5. Objective Today I will be able to:
Describe the 4 colligative properties of vapor pressure, boiling point, freezing point and osmotic pressure
Calculate the Van’t Hoff Factor for a Compound
Calculate the freezing point depression and boiling point elevation of a solute

6. Homework

7. Agenda Warm-Up Colligative Properties Research Chart
Colligative Properties Notes
Colligative Properties Calculations
Exit Ticket

8. Colligative Properties Chart
You have 15 minutes to fill in the chart of notes. We will review the chart together

9. Colligative Properties Notes

10. Colligative Properties
A property that depends on the number of molecules present, but not on their chemical nature

11. There are 4 colligative properties
Vapor Pressure
Boiling Point
Freezing Point
Osmotic Pressure

12. Vapor Pressure
Liquid molecules at the surface of a liquid can escape to the gas phase
- This process is reversible (g  l)
Vapor pressure of a solution containing a nonvolatile solute is less than the pressure of the solvent alone
Nonvolatile solute – typically has a boiling point less than 100 C, does not want to vaporize

13. Vapor Pressure

14. Vapor Pressure
Vapor pressure reduction is proportional to the concentration of the solution
When the concentration goes up, the vapor pressure is reduced

15. Vapor Pressure
This partially explains why The Great Salt Lake has a lower evaporation rate than expected. The salt concentration is so high that the vapor pressure (and evaporation) has been lowered

16. Boiling Point
The boiling point of a solution is always higher than that of the solvent alone
Boiling Point Elevation
You continue to use antifreeze in the summer, because you want the coolant to boil at a higher temperature so it will absorb the engine heat

17. Boiling Point - Example
Placing salt into cooking water causes the water to boil at a higher temperature
Allows more energy to be transferred to the food
The water will take longer to boil, but the food will cook faster

18. Freezing Point
The freezing point of a solution is always lower than that of the solvent alone
This is called Freezing Point Depression
Explains why salt (CaCl2) is used on ice
- The salt solution that forms has a lower freezing point than the original ice

19. Boiling and Freezing Point Calculations

20. Before we calculate… we need to talk about the Van’t Hoff Factor

21. Van’t Hoff Factor
Determines the moles of ions that are present when a compound dissolves in a solution
Covalent compounds do not dissociate
C12H22O11
1 mole (same for all nonelectrolytes)
Ionic Compounds can dissociate
NaCl  Na+ + Cl-
2 moles of ions (Van’t Hoff Factor = 2)
CaCl2  Ca Cl-
3 moles of ions (Van’t Hoff Factor = 3)

22. Determine the Van’t Hoff Factor for the following Compounds
C6H12O6
KCl
Al2O3
P2O5

23. Calculating Boiling and Freezing Points
Tb = Kb  m i
Kb is the molal boiling point elevation constant, which is a property of the solvent
Tb is added to the normal boiling point
i is the Van’t Hoff Factor

24. Calculating Boiling and Freezing Points
Tf = Kf  m  i
Kf is the molal freezing point depression constant, which is a property of the solvent
Tf is subtracted from the normal freezing point
i is the Van’t Hoff Factor

25. Boiling and Freezing Points and Electrolytes
What is the expected change in the freezing point of water in a solution of 62.5 grams of barium nitrate, Ba(NO3)2, in 1.00 kg of water?
∆Tf = Kf  m  i
62.5 g Ba(NO3)2  .239 moles
.239 moles/1.00 kg = .239 m
1.86°C/m x .239 m = .444°C
Ba(NO3)2  Ba NO3-1 = 3 moles of ions (i value)
.444°C x 3 = 1.33°C
0°C – 1.33°C = -1.33°C

26. Exit Ticket
Determine the Van’t Hoff Factor for the following compounds.
AlCl3
Mg3(PO4)2
C6H12O6