1. Chapter 12 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

5. If solvent is water, the solution is called an aqueous solution.

6. Liquor BeerWine Ethanol Concentration

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

8. Four Concentrations Unit: none (3) Unit: none

9. Four Concentrations Unit: mol/kg (4)

10. 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 Example 12.4 on page 533 and compare your results with the answers.

11. Electrical Conductivity of Aqueous Solutions

12. solute strong electrolyte weak electrolyte nonelectrolyte strong acids strong bases salts weak acids weak bases many organic compounds

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

14. MgBr 2 MgSO 4 FeCl 3 Glucose Mg 3 (PO 4 ) 2 NaOH Hexane

15. 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

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

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

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. 180 g/mol

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

26. 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

27. 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. 776 g/mol

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

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 i ― van’t Hoff factor

33. Π = iMRT Units Π ― atm M ― mol/L R ― atm·L·K −1 ·mol −1 T ― K i ― none

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. 1.66 x 10 4 g/mol

36. Practice on Example 12.10 on page 546 and compare your results with the answers.

39. What concentration of NaCl in water is needed to produce an aqueous solution isotonic with blood ( Π = 7.70 atm at 25 °C)? 0.158 mol/L

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

42. Lowering Vapor Pressure Nonvolatile solute to volatile solvent

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

44. pure solvent Liquid Surface Surface Molecules

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

46. Four Concentrations Unit: none (3) Unit: none

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

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

49. Calculate the vapor pressure at 25 °C of a solution containing 99.5 g of sucrose (C 12 H 22 O 11 ) and 300 mL of water. The vapor pressure of pure water at 25 °C is 23.8 torr. Assume the density of water to be 1.00 g/mL. Example 12.6, page 537 23.4 torr

50. 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.

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

52. 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. 22.1 torr

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

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

55. 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?

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

57. At 25 °C of a solution is prepared by dissolving 99.5 g of sucrose (nonelectrolyte, nonvolatile) into 300 mL of water. The vapor pressure of the solution and pure water are 23.4 torr and 23.8 torr, respectively. Assume the density of water to be 1.00 g/mL. Calculate the molar mass of sucrose. Modified Example 12.6, page 537

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

60. 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.

61. What kind of solution is ideal?

62. 10% P0P0 # of molecules in vapor = 100 x 1 x 10% = 10 χ pure solvent

63. 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 P sln

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

66. A solution contains 3.95 g of carbon disulfide (CS 2 ) and 2.43 g of acetone (CH 3 COCH 3 ). The vapor pressures at 35  C of pure carbon disulfide and pure acetone are 515 torr and 332 torr, respectively. Assuming ideal behavior, calculate the vapor pressures of each of the components and the total vapor pressure above the solution. The experimentally measured total vapor pressure of the solution at 35  C is 645 torr. Is the solution ideal? If not, what can you say about the relative strength of carbon disulfide–acetone interactions compared to the acetone–acetone and carbon disulfide– carbon disulfide interactions? Example 12.7, page 540

67. (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.