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Chapter 9 Solutions

Chapter 9 Solutions. 9.1 Properties of Water 9.2 Solutions 9.3 Electrolytes and Nonelectrolytes 9.6 Percent Concentration 9.7 Molarity. 9.1 Water. Water is the most common solvent. The water molecule is polar.

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Chapter 9 Solutions

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  1. Chapter 9 Solutions 9.1 Properties of Water 9.2 Solutions 9.3 Electrolytes and Nonelectrolytes 9.6 Percent Concentration 9.7 Molarity

  2. 9.1 Water • Water is the most common solvent. • The water molecule is polar. • Hydrogen bonds form between the hydrogen atom in one molecule and the oxygen atom in a different water molecule.

  3. Water

  4. Water in Foods • Water for the body is obtained from fluids as well as foods. • Some foods have a high percentage of water.

  5. Surface Tension Water molecules • At the surface form hydrogen bonds with molecules on or below the surface, which pulls them closer. • At the surface behave like a thin, elastic membrane, or “skin.” • Cannot hydrogen bond when compounds called surfactants are added.

  6. 9.2 Solutions: Solute and Solvent Solutions • Are homogeneous mixtures of two or more substances. • Consist of a solvent and one or more solutes.

  7. Nature of Solutes in Solutions Solutes • Spread evenly throughout the solution. • Cannot be separated by filtration. • Can be separated by evaporation. • Are not visible, but can give a color to the solution.

  8. Examples of Solutions • The solute and solvent in a solution can be a solid, liquid, and/or a gas.

  9. Learning Check Identify the solute and the solvent in each. A. brass: 20 g zinc + 50 g copper solute = 1) zinc 2) copper solvent = 1) zinc 2) copper B. 100 g H2O+5 g KCl solute = 1) KCl 2) H2O solvent = 1) KCl 2) H2O

  10. Solution Identify the solute and the solvent in each. A. brass: 20 g zinc + 50 g copper solute = 1) zinc solvent = 2) copper B. 100 g H2O+5 g KCl solute = 1) KCl solvent = 2) H2O

  11. Learning Check Identify the solute in each of the following solutions: A. 2 g sugar (1) and 100 mL water (2) B. 60.0 mL of ethyl alcohol (1) and 30.0 mL of methyl alcohol (2) C. 55.0 mL water (1) and 1.50 g NaCl (2) D. Air: 200 mL O2 (1) and 800 mL N2 (2)

  12. Solution Identify the solute in each of the following solutions: A. 2 g sugar (1) B. 30.0 mL of methyl alcohol (2) C. 1.5 g NaCl (2) D. 200 mL O2 (1)

  13. Like Dissolves Like • A solution forms when there is an attraction between the particles of the solute and solvent. • A polar solvent such as water dissolves polar solutes such as sugar and ionic solutes such as NaCl. • A nonpolar solvent such as hexane (C6H14)dissolves nonpolar solutes such as oil or grease.

  14. Examples of Like Dissolves Like Solvents Solutes Water (polar) Ni(NO3)2 (ionic) CH2Cl2 (nonpolar) I2 (nonpolar)

  15. Learning Check Which of the following solutes will dissolve in water? Why? 1) Na2SO4 2) gasoline (nonpolar) 3) I2 4) HCl

  16. Solution Which of the following solutes will dissolve in water? Why? 1) Na2SO4Yes, ionic 2) gasoline No, nonnpolar 3) I2 No, nonpolar 4) HCl Yes, polar Most polar and ionic solutes dissolve in water because water is a polar solvent.

  17. Formation of a Solution • Na+ and Cl- ions on the surface of a NaCl crystal are attracted to polar water molecules. • In solution, the ions are hydrated as several H2O molecules surround each.

  18. Equations for Solution Formation • When NaCl(s) dissolves in water, the reaction can be written as H2O NaCl(s) Na+(aq) + Cl– (aq) solid separation of ions

  19. Learning Check Solid LiCl is added to water. It dissolves because A. The Li+ ions are attracted to the 1) oxygen atom ( –) of water. 2) hydrogen atom (+) of water. B. The Cl- ions are attracted to the 1) oxygen atom ( –) of water. 2) hydrogen atom (+) of water.

  20. Solution Solid LiCl is added to water. It dissolves because A. The Li+ ions are attracted to the 1) oxygen atom ( –) of water. B. The Cl- ions are attracted to the 2) hydrogen atom ( +) of water.

  21. Hydrates • Hydrates are solid compounds that contain water molecules as part of the crystal structure. • Heating a hydrate releases the water of hydration to give the anhydrate salt.

  22. Examples of Hydrates

  23. Learning Check Write the equation for the dehydration of AlCl3 •6H2O.

  24. Solution Write the equation for the dehydration of AlCl3 •6H2O. AlCl3 •6H2O AlCl3 + 6H2O

  25. 9.3 Electrolytes Electrolytes • Produce positive (+) and negative (-) ions when they dissolve in water. • In water conduct an electric current.

  26. Strong Electrolytes • Strong electrolytes ionize 100% in solution. • Equations for the dissociation of strong electrolytes show the formation of ions in aqueous (aq) solutions. H2O 100% ions NaCl(s) Na+(aq) + Cl-(aq) H2O CaBr2(s) Ca2+(aq) + 2Br- (aq)

  27. Learning Check Complete each of the following dissociation equations for strong electrolytes dissolved in water: H2O A. CaCl2 (s) 1) CaCl2 2) Ca2+ + Cl2- 3) Ca2+ + 2Cl- H2O B. K3PO4 (s) 1) 3K+ + PO43- 2) K3PO4 3) K3+ + P3- + O4-

  28. Solution Complete each of the following dissociation equations for strong electrolytes dissolved in water: H2O A. 3) CaCl2 (s) Ca2+ + 2Cl- H2O B. 1) K3PO4 (s) 3K+ + PO43-

  29. Weak Electrolytes • A weak electrolyte • Dissolves mostly as molecules in solution. • Produces only a few ions in aqueous solutions. • Has an equilibrium that favors the reactants.HF + H2O H3O+(aq) + F- (aq)NH3 + H2O NH4+(aq) + OH- (aq)

  30. Nonelectrolytes • Nonelectrolytes • Form only molecules in water. • Do not produce ions in water. • Do not conduct an electric current.

  31. 9.6 Percent Concentration • The concentration of a solution is the amount of solute dissolved in a specific amount of solution. amount of solute amount of solution • The percent concentration describes the amount of solute that is dissolved in 100 parts of solution. amount of solute 100 parts solution

  32. Mass Percent The mass percent (%m/m) • Concentration is the percent by mass of solute in a solution.mass percent = g of solute x 100% g of solution • Is the g of solute in 100 g of solution.mass percent = g of solute 100 g of solution

  33. Mass of Solution grams of solute + grams of solvent 50.0 g KCl solution

  34. Calculating Mass Percent • Mass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution). g of KCl = 8.0 g g of solvent (water) = 42.0 g g of KCl solution = 50.0 g 8.0 g KCl (solute) x 100 = 16% (m/m) KCl 50.0 g KCl solution

  35. Learning Check A solution is prepared by mixing 15 g Na2CO3 and 235 g of H2O. Calculate the mass percent (%m/m) of the solution. 1) 15% (m/m) Na2CO3 2) 6.4% (m/m) Na2CO3 3) 6.0% (m/m) Na2CO3

  36. Solution 3) 6.0% (m/m) Na2CO3 mass solute = 15 g Na2CO3 mass solution = 15 g + 235 g = 250 g mass %(m/m) = 15 g Na2CO3 x 100 250 g solution = 6.0% Na2CO3 solution

  37. Mass/Volume Percent The mass/volume percent (%m/v) • Concentration is the ratio of the mass in grams (g) of solute in a volume (mL) of solution. mass/volume % = g of solute x 100% mL of solution • Is the g of solute in 100 mL of solution. mass/volume % = g of solute 100 mL of solution

  38. Preparing a Solution with a Mass/Volume % Concentration • A percent mass/volume solution is prepared by weighing out the grams of solute (g) and adding water to give the final volume of the solution.

  39. Calculation of Mass/Volume Percent • Mass/volume percent (%m/v) is calculated from the grams of solute (g KCl) and the volume of solution (mL KCl solution). g of KI = 5.0 g KI mL of KI solution = 250.0 mL 5.0 g KI (solute) x 100 = 2.0%(m/v) KI 250.0 mL KI solution

  40. Learning Check A 500. mL samples of an IV glucose solution contains 25 g glucose (C6H12O6) in water. What is the mass/volume % (%m/v) of glucose of the IV solution?1) 5.0% 2) 20.% 3) 50.%

  41. Solution 1) 5.0% Mass/volume %(m/v) = 25 g glucose x 100 500. mL solution = 5.0 % (m/v) glucose solution

  42. Volume Percent The volume percent (%v/v) • Concentration is the percent volume (mL) of solute (liquid) to volume (mL) of solution. volume % (v/v) = mL of solute x 100% mL of solution • Is the mL of solute in 100 mL of solution. volume % (v/v) = mL of solute 100 mL of solution

  43. Percent Conversion Factors • Two conversion factors can be written for any type of % value.

  44. Learning Check Write two conversion factors for each solutions: A. 8%(m/v) NaOH B. 12%(v/v) ethyl alcohol

  45. Solution A. 8%(m/v) NaOH 8 g NaOH and 100 mL solution 100 mL solution 8 g NaOH B. 12%(v/v) ethyl alcohol 12 mL alcohol and 100 mL solution 100 mL solution 12 mL alcohol

  46. Using Percent Factors How many grams of NaCl are needed to prepare 250 g of a 10.0% (m/m) NaCl solution? 1. Write the 10.0 % (m/m) as conversion factors. 10.0 g NaCl and 100 g solution 100 g solution 10.0 g NaCl 2. Use the factor that cancels given (g solution). 250 g solution x 10.0 g NaCl = 25 g NaCl 100 g solution

  47. Learning Check How many grams of NaOH are needed to prepare 2.0 L of a 12%(m/v) NaOH solution? 1) 24 g NaOH 2) 240 g NaOH 3) 2400 g NaOH

  48. Solution 2) 240 g NaOH 2.0 L x 1000 mL = 2000 mL 1 L 2000 mL x 12 g NaOH = 240 g NaOH 100 mL 12 % (m/v) factor

  49. Learning Check How many milliliters of 5% (m/v) glucose solution are given if a patient receives 150 g of glucose? 1) 30 mL 2) 3000 mL 3) 7500 mL

  50. Solution 2) 3000 mL 150 g glucose x 100 mL = 3000 mL 5 g glucose 5% m/v factor (inverted)

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