1. Chapter 11 Properties of Solutions

2. From Chapter 1: Classification of matter Matter Homogeneous (visibly indistinguishable) Heterogeneous (visibly distinguishable) Elements Compounds Mixtures (multiple components) Pure Substances (one component) (Solutions)

3. Solution = Solute + Solvent

4. Vodka = ethanol + waterBrass = copper + zinc

6. Liquor BeerWine Ethanol Concentration

7. Four Concentrations Unit: none Unit: mol/L (1) (2)

8. Four Concentrations Unit: none Unit: mol/kg (3) (4)

9. A solution contains 5.0 g of toluene (C 7 H 8 ) and 225 g of benzene (C 6 H 6 ) and has a density of 0.876 g/mL. Calculate the mass percent and mole fraction of C 7 H 8, and the molarity and molality of the solution. Practice on Sample Exercise 11.1 on page 486 and compare your results with the answers.

10. Electrical Conductivity of Aqueous Solutions

11. solute strong electrolyte weak electrolyte nonelectrolyte strong acids strong bases most salts weak acids weak bases many organic compounds Chapter 4

12. van’t Hoff factor nonelectrolyte:i = 1 strong electrolyte: depends on chemical formula weak electrolyte: depends on degree of dissociation Unit: none

13. NaCl MgCl 2 MgSO 4 FeCl 3 HCl Glucose

15. Figure 11.22 In an Aqueous Solution a Few Ions Aggregate, Forming Ion Pairs that Behave as a Unit

16. Four properties of solutions (1) Boiling point elevation water = solvent water + sugar = solution Boiling point = 100 °C Boiling point > 100 °C Solution compared to pure solvent

17. Sugar Dissolved in Water to Make Candy Causes the Boiling Point to be Elevated

18. ∆T b = T b,solution − T b,solvent = i K b m i: van’t Hoff factor m: molality K b : molal boiling-point elevation constant K b is characteristic of the solvent. Does not depend on solute. Units

19. Table 11.5 Molal Boiling-Point Elevation Constants (K b ) and Freezing-Point Depression Constants (K f ) for Several Solvents

20. Boiling point elevation can be used to find molar mass of solute. ∆T b ― experiments i ― electrolyte or nonelectrolyte K b ― table or reference book

21. A solution was prepared by dissolving 18.00 g glucose in 150.0 g water. The resulting solution was found to have a boiling point of 100.34 °C. Calculate the molar mass of glucose. Glucose is molecular solid that is present as individual molecules in solution. Sample exercise 11.8, page 505

22. Four properties of solutions (1) Boiling point elevation (2) Freezing point depression water = solvent water + salt = solution freezing point = 0 °C freezing point < 0 °C Solution compared to pure solvent

23. Spreading Salt on a Highway

24. ∆T f = T f,solvent − T f,solution = i K f m i: van’t Hoff factor m: molality K f : molal freezing-point depression constant K f is characteristic of the solvent. Does not depend on solute. Units

25. Table 11.5 Molal Boiling-Point Elevation Constants (K b ) and Freezing-Point Depression Constants (K f ) for Several Solvents

26. The Addition of Antifreeze Lowers the Freezing Point of Water in a Car's Radiator 0 °C 100 °C water < 0 °C> 100 °C water + antifreeze

27. Freezing point depression can be used to find molar mass of solute. ∆T f ― experiments i ― electrolyte or nonelectrolyte K f ― table or reference book

28. A chemist is trying to identify a human hormone that controls metabolism by determining its molar mass. A sample weighing 0.546 g was dissolved in 15.0 g benzene, and the freezing-point depression was determined to be 0.240 °C. Calculate the molar mass of the hormone. Sample exercise 11.10, page 507

29. Table 11.5 Molal Boiling-Point Elevation Constants (K b ) and Freezing- Point Depression Constants (K f ) for Several Solvents

30. Four properties of solutions (1) Boiling point elevation (2) Freezing point depression (3) Osmotic pressure

31. Osmotic Pressure

32. Π = iMRT Π ― osmotic pressuue M ― molarity R ― ideal gas constant T ― temperature

33. Π = iMRT Π ― atm M ― mol/L R ― atm·L·K −1 ·mol −1 T ― K Units

34. Osmotic pressure can be used to find molar mass of solute. Π ― experiments i ― electrolyte or nonelectrolyte R ― constant T ― experiments

35. To determine the molar mass of a certain protein, 1.00 x 10 −3 g of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at 25.0 °C. Calculate the molar mass of the protein. Sample exercise 11.11, page 509

37. What concentration of NaCl in water is needed to produce an aqueous solution isotonic with blood ( Π = 7.70 atm at 25 °C)? Sample exercise 11.12, page 510

38. Four properties of solutions (1) Boiling point elevation (2) Freezing point depression (3) Osmotic pressure (4) Lowering the vapor pressure

40. Lowering Vapor Pressure Nonvolatile solute to volatile solvent

41. The Presence of a Nonvolatile Solute Lowers the Vapor Pressure of the Solvent

42. pure solvent Liquid Surface

43. When you count the number of solute particles, use van’t Hoff factor i. solvent + solute Liquid Surface

44. Show that the packing efficiency for body centered cubic unit cell is 68 %. Pop Quiz: 0.5 extra point

45. Raoult’s Law: Case 1 ― vapor pressure of solution ― vapor pressure of pure solvent ― mole fraction of solvent Nonvolatile solute in a Volatile solvent

46. Figure 11.11 For a Solution that Obeys Raoult's Law, a Plot of P soln Versus X solvent, Give a Straight Line

47. Calculate the expected vapor pressure at 25 °C for a solution prepared by dissolving 158.0 g of common table sugar (sucrose, molar mass = 342.3 g/mol) in 643.5 cm 3 of water. At 25 °C, the density of water is 0.9971 g/cm 3 and the vapor pressure is 23.76 torr. Sample exercise 11.5, page 499

48. Predict the vapor pressure of a solution prepared by mixing 35.0 g solid Na 2 SO 4 (molar mass = 142 g/mol) with 175 g water at 25 °C. The vapor pressure of pure water at 25 °C is 23.76 torr. Sample exercise 11.6, page 500

49. Raoult’s Law: Case 2 Volatile solute in a Volatile solvent Recall Dalton’s law of partial pressures

50. Vapor Pressure for a Solution of Two Volatile Liquids X A + X B = 1 0 1

51. A mixture of benzene (C 6 H 6 ) and toluene (C 7 H 8 ) containing 1.0 mol of benzene and 2.0 mol of toluene. At 20 °C the vapor pressures of pure benzene and toluene are 75 torr and 22 torr, respectively. What is the vapor pressure of the mixture? What is the mole fraction of benzene in the vapor?

52. Lowering vapor pressure can be used to find molar mass of solute. and ― experiments

53. At 25 °C a solution is prepared by dissolving 158.0 g of common table sugar (sucrose, nonelectrolyte, nonvolatile) in 643.5 cm 3 of water. The vapor pressure of this solution is 23.46 torr. At 25 °C, the density of water is 0.9971 g/cm 3 and the vapor pressure is 23.76 torr. Calculate the molar mass of sucrose. Modified sample exercise 11.5, page 499

54. A solution that obeys Raoult’s Law is called an ideal solution.

55. A solution is prepared by mixing 5.81 g acetone (molar mass = 58.1 g/mol) and 11.0 g chloroform (molar mass = 119.4 g/mol). At 35 °C, this solution has a total vapor pressure of 260. torr. Is this an ideal solution? The vapor pressure of pure acetone and pure chloroform at 35 °C are 345 torr and 293 torr, respectively. Sample exercise 11.7, page 503

56. What kind of solution is ideal?

57. 10% P0P0 # of molecules in vapor = 100 x 1 x 10% = 10 χ pure solvent

58. 10% 5% 15% # of molecules in vapor = 100 x 0.8 x 5% = 4 # of molecules in vapor = 100 x 0.8 x 15% = 12 # of molecules in vapor = 100 x 0.8 x 10% = 8 χ Raoult’s law: Deviate from Raoult’s law P0P0 solvent + solute

59. What kind of solution is ideal? Solute-solute, solvent-solvent, and solute-solvent interactions are very similar. Comparison to ideal gas.

60. Figure 11.13 a-c Vapor Pressure for a Solution of Two Volatile Liquids IdealNonideal

61. (1) Boiling point elevation:∆T b = i K b m (2) Freezing point depression: ∆T f = i K f m (3) Osmotic pressure:Π = iMRT (4) Lowering the vapor pressure: Four Colligative properties of solutions Colligative: depend on the quantity (number of particles, concentration) but not the kind or identity of the solute particles.

62. Rule of solubility: Like dissolves like Polarity

64. Henry’s Law: the amount of gas that dissolved in a solution is directly proportional to the pressure of the gas above the solution. C = kP

65. Figure 11.5 a-c Henry's Law

66. C = kP

67. What you must master in this chapter Four concentrations. Four colligative properties. A Quiz this week? Calculations associated with