Chapter 12 Solutions

1. Chapter 12 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

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

6. Liquor Beer Wine Ethanol Concentration

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

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

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

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

11. Electrical Conductivity of Aqueous Solutions

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

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

14. MgBr2 MgSO4 FeCl3 Mg3(PO4)2 NaOH Hexane Glucose

15. 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

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

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

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. 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 freezing point = 0 °C water = solvent freezing point < 0 °C water + salt = solution Solution compared to pure solvent

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

26. Freezing point depression can be used to find molar mass of solute. ∆Tf ― experiments i ― electrolyte or nonelectrolyte Kf ― 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 water 0 °C 100 °C < 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 i ― van’t Hoff factor M ― molarity R ― ideal gas constant T ― temperature

33. Π = iMRT Units Π ― atm i ― none 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. 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 104 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. Nonvolatile solute to volatile solvent Lowering Vapor Pressure

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

44. Surface Molecules Liquid Surface pure solvent

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

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

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

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

49. Example 12.6, page 537 Calculate the vapor pressure at 25 °C of a solution containing 99.5 g of sucrose (C12H22O11) 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. 23.4 torr

50. 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.