1. Colligative Properties Chemistry GT 5/11/15

2. Drill List the four colligative properties. Give a real world example of each. What is the molality of a solution made with 25.0 g LiBr and 750. g of water? HW: 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 Finish Colligative Properties Notes Colligative Properties Calculations Exit Ticket

5. Boiling and Freezing Point Calculations

6. BEFORE WE CALCULATE… WE NEED TO TALK ABOUT THE VAN’T HOFF FACTOR

7. 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)

8. 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

9. 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

10. 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

11. 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

12. Example problems Take out the Freezing Point and Boiling Point practice WS Work the odd # problems on your own paper—check as you go

13. 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