1. Chapter 19: Molality and Colligative Properties Chapter 14 —Big Book p. 487 & 14.1 (p. 498-504) HW Ch. 19 Blue Book: #1-17, 19 Chapter 14 —Big Book p. 487 & 14.1 (p. 498-504) HW Ch. 19 Blue Book: #1-17, 19

2. Molality… a little review to start http://www.youtube.com/watch?v=WNr SexmBDXU&feature=related

3. Colligative Properties What are they? What are they? – The word Colligative means “depending on the collection” – Change the physical properties of the solvent. – Depends on the number of particles of the solute NOT which solute is used!

4. Colligative Properties Lowers the vapor pressure! Raises the boiling point!boiling point Lowers or depresses the freezing point!depresses the freezing point! Why?

5. Colligative Properties When a solute is dissolved in a solvent, the vapor pressure of the solvent is reduced. The reduction depends on the number of solute particles in a given amount of solvent. The French chemist, Raoult, first discovered the vapor pressure lowering relationship experimentally in 1882 which lead to…

6. Colligative Properties Raoult’s Law: Any nonvolatile solute at a specific concentration lowers the vapor pressure of the solvent by an amount that is characteristic of that solvent.

7. Antifreeze, Electrolytes…: – http://www.youtube.com/watch?v=n0W7Y2G wi2E&feature=related http://www.youtube.com/watch?v=n0W7Y2G wi2E&feature=related

8. Vapor Pressure Lowering The vapor pressure above a liquid is lowered due to the attractive forces of the solvent on the dissolved solute particles. Because of this, less solvent particles have the energy to transition to the gaseous state (evaporate), and therefore the vapor pressure is lower. So… The greater the number of solute particles in a solvent, the lower the VP

9. #1 – solvent has a large surface area to evaporate from #2 – mixed with solute = fewer solvent particles at surface #1 – solvent has a large surface area to evaporate from #2 – mixed with solute = fewer solvent particles at surface Beaker #1Beaker #2 Pure SolventSolution Which one has lower VP?

10. Boiling Point Elevation Similar factors (as with the vapor pressure lowering), contribute to the increase of the boiling point of a solvent. – The more solute particles the higher the BP (the lower the VP) Practical application – adding salt to water to increase the BP of water to cook foods.

11. Boiling Point Elevation

12. Freezing Point Depression Freezing occurs when the particles no longer have the energy to overcome their interparticle attractive forces – they organize and solidify (molecules slow way down, loss of kinetic energy). Adding solute to a pure solvent lowers the FP! – WHY? Because the solute interferes with the solvents interparticle attractions, therefore the solid forms at cooler or lower temperature. So… the FP of a solution is always lower than the FP of a pure solvent.

13. Freezing Point Depression

14. 0oC0oC 100 o C ___ = Pure Solvent ---- = Solution

15. BP elevation & FP depression – http://www.youtube.com/watch?v=tjHaIDSzH so&feature=related http://www.youtube.com/watch?v=tjHaIDSzH so&feature=related

16. Colligative Properties (now the math) The change in the freezing and boiling pts varies directly with the concentration of particles. Molal freezing pt constant: 1.86C˚ for water. Each mole of solute causes the freezing pt of water to drop by this much. Molal boiling pt constant: 0.512C˚ for water. Each mole of solute causes the boiling point to rise by this much. MEMORIZE THESE 2 CONSTANTS!! MEMORIZE THESE 2 CONSTANTS!!

17. Colligative Properties These can be used to determine: The freezing point of the water The boiling point of the water The molecular mass of the solute from the freezing point or the boiling point (see table 19-1 p. 166 in Blue Book for other constants)

18. Colligative Properties Ex. 6 Calculate the freezing point of a solution containing 5.70 g of sugar, C 12 H 22 O 11, in 50.0 g of water. (Molal freezing pt constant: 1.86C˚ for water. ) Convert grams of solute per gram of water to moles of solute per kg of water (molality). Then multiply by the conversion ratio to obtain the change in FP 5.70 g C 12 H 22 O 11 10 3 g H 2 O 1 mol C 12 H 22 O 11 1.86C˚ 50.0 g H 2 O 1kg H 2 O 342 g C 12 H 22 O 11 1 m = 0.620C˚, To determine the FP, subtract this from the FP of water 0 o C – 0.620 = - 0.620 o C

19. Calculating Molecular Mass Ex. 7 When 72.0 g of dextrose were dissolved in 100.0 g of water, the boiling point of the solution was observed to be 102.05˚ C. What is the molecular mass of dextrose? Step 1: determine the molality of the solution 100 o C - 102.05˚C = 2.05˚C determine the  T b 2.05 o C m = 4.00 m 0.512 ˚C molal boiling pt. constant for H 2 O Step 2: determine the grams per mole 72.0 g dextrose 1 kg H 2 O = 180 g 0.100 kg H 2 O 4.00 mol mol