1. Colligative Properties Properties of solutions that depend on the concentration of solute particles. (Molarity or Molality!)

2. Colligative Properties Boiling Point Elevation & Freezing Point Depression Adding a solute to a liquid solvent can change the temperatures at which it boils or freezes.

3. ΔT = The change in temperature i = van’t Hoff factor (# of solute particles) K = Constant (Pg. 522 boiling, 526 freezing) m = Molality

4. [http://wps.prenhall.com/wps/media/objects/602/616516/Media_Assets/Chapter11/Text_Images/FG11_TB04.JPG]

5. Colligative Properties Which has the higher van’t Hoff factor: NaCl or CaCl 2 ? NaCl(s)  Na + (aq) + Cl - (aq) 2 particles CaCl 2 (s)  Ca +2 (aq) + 2 Cl - (aq) 3 particles

6. Colligative Properties Example: What is the new freezing point of water when 117 g of NaCl is added to 500.0 g of water?

7. ΔT = (i)(K)(m) 1 st Step – Find the molality = 2.00 mol 117 g NaCl 1 mol NaCl 58.5 g NaCl = 4.00 mol/kg 2.00 mol.5000 kg

8. 2 nd Step – Find the van’t Hoff factor NaCl(s)  Na + (aq) + Cl - (aq) So….i = 2 ΔT = (2)(K)(4.00) ΔT = (i)(K)(4.00 mol/kg)

9. [http://wps.prenhall.com/wps/media/objects/602/616516/Media_Assets/Chapter11/Text_Images/FG11_TB04.JPG] ΔT = (2)(K)(4.00)

10. Colligative Properties ΔT = (2)(1.86)(4.00) ΔT = 14.9 ºC But since H 2 O normally freezes at 0ºC, the freezing point is –14.9 ºC Click Here for an Interactive Demo