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Chapter 12 Solutions. Classification of matter Matter Homogeneous (visibly indistinguishable) Heterogeneous (visibly distinguishable) Elements Compounds.

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Presentation on theme: "Chapter 12 Solutions. Classification of matter Matter Homogeneous (visibly indistinguishable) Heterogeneous (visibly distinguishable) Elements Compounds."— Presentation transcript:

1 Chapter 12 Solutions

2 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 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 and compare your results with the answers. 2.2 %, 0.018, 0.21 mol/L, 0.24 mol/kg

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 NaOH Hexane NaCl

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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 of solute m: molality K b : boiling-point elevation constant K b is characteristic of the solvent. Does not depend on solute. Units

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21 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

22 A solution was prepared by dissolving g glucose in g water. The resulting solution was found to have a boiling point of °C. Calculate the molar mass of glucose. Glucose is molecular solid that is present as individual molecules in solution. 180 g/mol

23 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

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

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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 g was dissolved in 15.0 g benzene, and the freezing-point depression was determined to be °C. Calculate the molar mass of the hormone. 776 g/mol

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30 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

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32 Experiment 23 this Thursday Read the lab manual before coming to the lab Bring the manual, a copy of the data form, a pair of goggles, and an apron to the lab Wash your apron before Thursday Dress properly according to the syllabus

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

34 Osmotic Pressure

35 Π = iMRT Π ― osmotic pressuue M ― molarity R ― ideal gas constant T ― absolute temperature i ― van’t Hoff factor of solute

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

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

38 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 x 10 4 g/mol

39 Practice on Example and compare your results with the answers.

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41 isotonic = isosmotic: same osmotic pressure

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

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

44 Lowering Vapor Pressure Nonvolatile solute to volatile solvent

45 Liquid Surface

46 pure solvent Liquid Surface Surface Molecules number of solvent molecules above the liquid 100 x 100% x 5% = 5

47 solvent + solute Liquid Surface pure solvent number of solvent molecules above the liquid pure solvent: 100 x 100% x 5% = 5 solution: 100 x 80% x 5% = 4

48 Four Concentrations Unit: none (3) Unit: none

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

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

51 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 torr molar mass of sucrose = g/mol

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53 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 23.8 torr.

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

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56 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 23.8 torr torr Try Example 12.12

57 ΔTbΔTb aqueous solution

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

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

60 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?

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

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64 What kind of solution is ideal if viewed at a molecular level?

65 10% P0P0 # of molecules in vapor = 100 x 1 x 10% = 10 χ pure solvent

66 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

67 What kind of solution is ideal? Solute-solute, solvent-solvent, and solute-solvent interactions are very similar.

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69 Try Example 12.7

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

71 (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.

72 Quiz 3 Use osmotic pressure to find molar mass of a solute Raoult’s law, ideal solution Find van’t Hoff factors from formulas


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