Chapter 11 Properties of Solutions

1. Chapter 11 Properties of Solutions

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

3. Solution = Solute + Solvent

4. Vodka = ethanol + water Brass = copper + zinc

6. Liquor Beer Wine Ethanol Concentration

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

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

9. A solution contains 5.0 g of toluene (C7H8) and 225 g of benzene (C6H6) and has a density of 0.876 g/mL. Calculate the mass percent and mole fraction of C7H8, 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. Chapter 4 most salts strong acids strong electrolyte strong bases weak acids weak electrolyte solute weak bases many organic compounds nonelectrolyte

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

13. NaCl MgCl2 MgSO4 FeCl3 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 Boiling point = 100 °C water + sugar = solution 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. ∆Tb = Tb,solution − Tb,solvent = i Kb m i: van’t Hoff factor m: molality Kb: molal boiling-point elevation constant Units Kb is characteristic of the solvent. Does not depend on solute.

19. Table 11.5 Molal Boiling-Point Elevation Constants (Kb) and Freezing-Point Depression Constants (Kf) for Several Solvents

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

21. Sample exercise 11.8, page 505 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.

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

23. Spreading Salt on a Highway

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

25. Table 11.5 Molal Boiling-Point Elevation Constants (Kb) and Freezing-Point Depression Constants (Kf) for Several Solvents

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

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

28. Sample exercise 11.10, page 507 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.

29. Table 11.5 Molal Boiling-Point Elevation Constants (Kb) and Freezing-Point Depression Constants (Kf) 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 Units Π ― atm M ― mol/L R ― atm·L·K−1·mol−1 T ― K

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

35. Sample exercise 11.11, page 509 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.

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

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

40. Nonvolatile solute to volatile solvent Lowering Vapor Pressure

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

42. Liquid Surface pure solvent

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

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

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

46. Figure 11.11 For a Solution that Obeys Raoult's Law, a Plot of Psoln Versus Xsolvent, Give a Straight Line

47. Sample exercise 11.5, page 499 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 cm3 of water. At 25 °C, the density of water is 0.9971 g/cm3 and the vapor pressure is 23.76 torr.

48. Sample exercise 11.6, page 500 Predict the vapor pressure of a solution prepared by mixing 35.0 g solid Na2SO4 (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.

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 1 0 XA + XB = 1