1. Raoult’s Law – Vapor Pressure Lowering: For a mixture of two substances Raoult’s Law tells us that P total = χ A P A 0 + χ B P B 0 where P A 0 and P B 0 are the vapor pressure of the two pure substances (at a particular T). If A is totally involatile, P A 0 = 0, then a mixture of A and B will have a total vapor pressure less than that of A. We can roughly estimate molar masses using this effect.

2. Raoult’s Law Example: 44.2 g of a simple carbohydrate (known to be either C 12 H 22 O 11 or C 6 H 12 O 6 ) are dissolved in 99.5 g of water at 24.0 o C. The vapor pressures of pure water and the solution at 24.0 o C are 2.986 kPa and 2.859 kPa respectively. Find the molar mass of the carbohydrate and hence its identity.

3. Osmotic Pressure Interesting changes occur when two solutions of different concentrations are placed in close contact. The next slide shows what happens when two aqueous solutions of different concentrations, each containing a nonvolatile solute are placed under a transparent enclosure. Water moves over time from the dilute solution (A) to the concentrated solution. Why? (Vapor pressure H 2 O higher above the left hand container. Why?)

4. Osmotic Pressure Copyright © 2011 Pearson Canada Inc. Slide 4 of 46 FIGURE 13-16 Observing the direction of flow of water vapor General Chemistry: Chapter 13

5. Osmotic Pressure – cont’d: On the previous slide there is a net transfer of water (solvent) molecules between the two containers until both containers have the same concentration or the vapor pressure of H 2 O is the same over both containers. Transfer of solvent need not occur through the gas phase! When two solutions of different concentration are separated by a solvent permeable membrane there is a surprise!

6. Osmosis FIGURE 13-17 Copyright © 2011 Pearson Canada Inc. Slide 6 of 46 General Chemistry: Chapter 13 Does pure water or sugar water have the higher vapor pressure. Mechanical analogy?

7. Osmotic Pressure – cont’d: On the previous slide the movement of solvent molecules generates enough pressure to lift solution through the tube. The size of this osmotic pressure can be calculated with a “familiar looking” equation. Here n is the number of moles of solute (initially a nonelectrolyte!!!!) and V is the solution volume (in Liters).

8. Osmotic Pressure Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 8 of 46 πV = nRT π = RT n V = MRT For dilute solutions of nonelectrolytes:

9. Osmotic Pressure – Applications : Osmotic pressure effects normally move solvent molecules from a dilute to a concentrated solution. In water desalination a large “external” pressure is used to push solvent molecules (water) from sea water through a membrane to produce drinking water. The process is shown on the next slide.

10. Practical Applications Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 10 of 46 Desalination of saltwater by reverse osmosis

11. Isotonic Solutions Two solutions with equal osmotic pressure are termed isotonic. In medical applications some care has to be taken when adding solutions to the body or osmotic pressure effects can distort cells (or worse!).

12. Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 13 Slide 12 of 46 Isotonic Saline 0.92% m/V Hypotonic < 0.92% m/V water flows into the cells causing rupture Hypertonic > 0.92% m/V water flows out of the cells causing crenation

13. The proper concentration of lens cleaning solution is important. Why?

14. Osmotic Pressure – Example: 1. An aqueous solution of urea, (NH 2 ) 2 C=O (lawn fertilizer), has an osmotic pressure of 3.00 atm at 19.1 0 C. What mass of urea is contained in 325 mL of this solution?

15. 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?

16. 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?