1. Osmotic Pressure Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 1 of 46 πV = nRT π = RT n V = MRT For dilute solutions of nonelectrolytes:

2. Osmotic Pressure 2. A 0.426 g sample of an organic compound is dissolved in 225 mL of solvent at 24.5 0 C to produce a solution with an osmotic pressure of 3.36 mm Hg. What is the molar mass of the organic compound? 3. What additional information would be required to determine the molecular formula as well as the molar mass?

3. Henry’s Law 4. At 0 0 C a 1.00L aqueous solution contains 48.98 mL of dissolved oxygen when the O 2 (g) pressure above the solution is 1.00 atm. What would be the molarity of oxygen in the solution if the oxygen gas pressure above the solution were instead 4.15 atm?

4. Freezing-Point Depression and Boiling Point Elevation of Nonelectrolyte Solutions A Colligative property. – Depends on the number of particles present. Vapor pressure is lowered when a solute is present. – This results in boiling point elevation. – Freezing point is also effected and is lowered. Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 4 of 46

5. Vapor-pressure lowering by a nonvolatile solute FIGURE 13-19 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 5 of 46 ΔT f = -K f  m ΔT b = +K b  m

6. Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 6 of 46 ΔT f = -K f  mΔT b = +K b  m

7. Practical Applications Copyright © 2011 Pearson Canada Inc. Slide 7 of 46General Chemistry: Chapter 13

8. Solutions of Electrolytes Svante Arrhenius – Nobel Prize 1903. – Ions form when electrolytes dissolve in solution. – Explained anomalous colligative properties. Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13Slide 8 of 46 ΔT f = -K f  m = -1.86°C m -1  0.0100 m = -0.0186°C Compare 0.0100 m aqueous urea to 0.0100 m NaCl (aq) Freezing point depression for NaCl is -0.0361°C.

9. Van’t Hoff Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13Slide 9 of 46 ΔT f = -i  K f  m i = == 1.98 measured ΔT f ΔT b = i  K b  m expected ΔT f 0.0361°C 0.0186°C π = i  M  RT

10. Interionic Interactions Copyright © 2011 Pearson Canada Inc. Slide 10 of 46General Chemistry: Chapter 13 Arrhenius theory does not correctly predict the conductivity of concentrated electrolytes.

11. Interionic Interactions The data on the previous slide show that solutions of electrolytes behave less ideally as (1) ionic charges increase and (2) electrolyte concentraion increases. Are there correspondences to nonideal gas behaviour? The latter manifests itself at high P and low T (high gas “concentration”) and as molecules become more polar or polarizable.

12. Ideal van’t Hoff Factors (Aqueous Solutions): We need to (a) identify covalent and ionic substances by inspecting their chemical formulas and (b) for ionic compounds (and strong acids) determine how many ions are formed in solution per formula unit of solute dissolved. Examples are given on the next slide.

13. Van’t Hoff “i Factors”: Chemical NameChemical FormulaIonic/Covalent ?Ideal “i” value SucroseC 12 H 22 O 11 Covalent1 Potassium nitrateKNO 3 Ionic2 Urea(NH 2 ) 2 C=OCovalent1 Magnesium chloride MgCl 2 Ionic3 Aluminum nitrateAl(NO 3 ) 3 Ionic4 Lead (II) iodidePbI 2 Ionic3 Copper (II) nitrate hexahydrate Cu(NO 3 ) 2 ∙6H 2 OIonic Hydrochloric acidHCl(aq) Potassium hydroxide

14. Boiling Pt. and Freezing Pt. Changes Ex. 2. If 3.30 g of ammonium nitrate were dissolved in 475 g of water what would you expect the boiling point and freezing point of the solution to be? (K b and K f values for water are 0.512 0 C m -1 and 1.86 0 C m -1.) (Mention different form of “freezing pt. equation” in some texts?)

15. Freezing and Boiling Points Ex. 3. Which aqueous solution would have a higher boiling point? (a) 1.00 m sucrose (C 12 H 22 O 11 ) or (b) 0.400 m calcium nitrate? Ex. 4. Which aqueous solution would have the lowest freezing point? (a) 1.20 m sucrose, (b) 0.600 m sodium chloride (c) 0.400 m magnesium nitrate or (d) 0.700 m copper (II) sulfate? Why?