1. Physical Properties of Solutions Chapter 13

2. Colligative Properties of Solutions Colligative properties - properties that depend only on the number of solute particles in solution and not on the nature of the solute particles Vapor-Pressure Lowering P 1 = X 1 P 1 o Boiling-Point Elevation  T b = K b m Freezing-Point Depression  T f = K f m Osmotic Pressure (  )  = MRT

3. Vapor-Pressure Lowering Raoult’s law If the solution contains only one solute: X 1 = 1 – X 2 P 1 o - P 1 = ΔP = X 2 P 1 o o = vapor pressure of pure solvent X 1 = mole fraction of the solvent X 2 = mole fraction of the solute P 1 = X 1 P 1 o Fig 13.20

4. Boiling-Point Elevation ΔT b = T b – T b o T b ≡ boiling point of pure solvent o T b ≡ boiling point of solution ΔT b = K b m K b ≡ molal boiling-point elevation constant (°C/m) Fig 13.22

5. Boiling-Point Elevation ΔT b = K b m  T b is added to the normal boiling point of the solvent

6. Freezing-Point Depression ΔT f = K f m  T f is subtracted from the normal freezing point of the solvent

7. Osmotic Pressure (  ) Osmosis - selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. Semipermeable membrane - allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure (  ) - pressure required to stop osmosis. dilute more concentrated

8. Osmotic Pressure (  )  = MRT M is the molarity of the solution R is the gas constant = 0.08206 (L atm)/(mol K) T is the temperature (in K) Chemistry In Action: RO Water

9. A cell in an: isotonic solution Fig 13.24 hypotonic solution hypertonic solution Red blood cell

10. Colligative Properties of Electrolyte Solutions 0.1 m NaCl solution 0.1 m Na + ions & 0.1 m Cl - ions Colligative properties ≡ properties that depend only on the number of solute particles and not on their nature 0.1 m NaCl solution0.2 m ions in solution van’t Hoff factor (i) = actual number of particles in soln after dissociation number of formula units initially dissolved in soln nonelectrolytes NaCl CaCl 2 i should be 1 2 3

11. Boiling-Point ElevationΔT b = i K b m Freezing-Point DepressionΔT f = i K f m Osmotic Pressure π = iMRT Colligative Properties of Electrolyte Solutions

12. Sample Exercise 13.13 Molar Mass from Osmotic Pressure The osmotic pressure of an aqueous solution of a certain protein was measured to determine the protein’s molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution at 25 °C was found to be 1.54 torr. Treating the protein as a nonelectrolyte, calculate its molar mass.

13. Solutions Colloids Table 13.6 Types of Colloids Colloid versus solution Collodial particles are much larger than solute molecules Collodial suspension is not as homogeneous as a solution

14. Solutions Tyndall Effect  The scatter of rays of light by particles in a colloidal suspension  Light is NOT scattered by particles in a solution Fig 13.26

15. Solutions Colloids in Biological Systems E.g.: Molecules in enzymes and antibodies Have a polar, hydrophilic (water-loving) end and A non-polar, hydrophobic (water-hating) end Fig 13.28

16. Fig 13.29 Hydrophobic colloids

17. Solutions Colloids in Biological Systems Sodium stearate

18. Solutions Colloids in Biological Systems Fig 13.31 Stabilization of an emulsion of oil in water by stearate ions

19. Exam #2 Chapters 11 and 13 Thirty multiple choice (60%) Thirty multiple choice (60%) Terminology, concepts, properties, etc. Terminology, concepts, properties, etc. Calculations (40%) Calculations (40%) Heating/cooling and phase transitions Heating/cooling and phase transitions Calc. concentration (M, m, or %(w/w) from given data Calc. concentration (M, m, or %(w/w) from given data Det. molar mass from colligative property data Det. molar mass from colligative property data

20. How much heat to convert 180. g H 2 O from -10 °C to 150. °C? sp. ht. (ice) = 2.03 J/(g °C)ΔH fus = 6.01 kJ/mol sp. ht. (water) = 4.18 J/(g °C) sp. ht. (steam) = 1.99 J/(g °C) ΔH vap = 40.8 kJ/mol Ans: 561 kJ q = mcΔT ΔH fus ΔH vap

21. Sample Exercise 13.7 Calculation of Molality and Molarity Using the Density of a Solution A solution with a density of 0.876 g/mL contains 5.0 g of toluene (C 7 H 8 ) and 225 g of benzene. Calculate the molarity of the solution.

22. A 202-mL benzene solution containing 2.47 g of an organic polymer has an osmotic pressure of 8.83 mm Hg at 21.0 °C. Calculate the molar mass of the polymer. π = MRT M = π /RT = (4.81 x 10 -4 mol/L) (0.202 L) = 9.72 x 10 -5 mol 9.72 x 10 -5 mol 2.47 g = 2.54 x 10 4 g/mol = 4.81 x 10 -4 M