1. Molarity and Colligative Properties Chemistry GT 5/8/15

2. Drill Use the table from pg. 9 to give the amount of substances that will dissolve in 100 g of H 2 O: –NH 3 at 10°C and 80°C –Why does solubility of NH 3 decrease at higher temperatures? –KCl at 10°C and 70°C –If 50 g of KCl were dissolved in 100 g of water, and it was cooled to 50°C, what kind of solution would you have? HW: –pg. 5 Molality Problems –Effect of a Solute on FP and BP

3. Objectives Today I will be able to: –Calculate the molality of a solution –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

4. Agenda Drill Molality Notes & Example Colligative Properties Research Chart Colligative Properties Notes Colligative Properties Calculations Exit Ticket

5. Molality Don’t you mean molarity?? Nope! Molality is another way to measure concentration. Equation: m = n m n = moles of solute m = mass of solvent in kilograms

6. Why?? Molarity changes with the density of the solvent, so at different temperatures, you have different molarities. Molality does NOT change with density or temperature! Let’s do #1 on pg. 5

7. COLLIGATIVE PROPERTIES! What’s next?

8. Colligative Properties Chart You have 15 minutes to fill in the chart of notes. Split the work up at your table—each person can take one quadrant, and then share. You may use a textbook or BYOD 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

13. Vapor Pressure Solute particles take up surface area and lower the 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. 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 (CaCl 2 ) is used on ice - The salt solution that forms has a lower freezing point than the original ice

17. 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

18. Osmotic Pressure The tendency for a solution to take in water due to osmosis –This is why cells have to maintain their internal pressures—if they stop pumping water out, they will fill up and pop!

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 particles that are present when a compound dissolves in a solution Covalent compounds do not dissociate –C 12 H 22 O 11 1 mole (Van’t Hoff Factor = 1, the same for all nonelectrolytes) Ionic Compounds can dissociate –NaCl  Na + + Cl - 2 moles of ions (Van’t Hoff Factor = 2) –CaCl 2  Ca +2 + 2 Cl - 3 moles of ions (Van’t Hoff Factor = 3)

22. Determine the Van’t Hoff Factor for the following Compounds C 6 H 12 O 6 KCl Al 2 O 3 P 2 O 5

23. Calculating Boiling and Freezing Points  T b = K b  m  i K b is the molal boiling point elevation constant, which is a property of the solvent –K b (H 2 O) = 0.52°C/m m is molality i is the Van’t Hoff Factor  T b is added to the normal boiling point

24. Calculating Boiling and Freezing Points  T f = K f  m  i K f is the molal freezing point depression constant, which is a property of the solvent –K f (H 2 O) = 1.86°C/m m is molality i is the Van’t Hoff Factor  T f is subtracted from the normal freezing point

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(NO 3 ) 2, in 1.00 kg of water? ∆T f = K f  m  i 62.5 g Ba(NO 3 ) 2 .239 moles.239 moles/1.00 kg =.239 m 1.86°C/m x.239 m =.444°C Ba(NO 3 ) 2  Ba +2 + 2 NO 3 -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. –AlCl 3 –Mg 3 (PO 4 ) 2 –C 6 H 12 O 6