Ch11 z5e solutions

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  • Section 11.1 Solution Compositions Z5e 514
  • Z5e 514 Section 11.1 Solution Composition
  • Z5e 514.
  • SE 11.1 Z5e 514 Molarity = 0.215 M Mass % = 0.990% Mole fraction = 0.00389 Molality - 0.217 m
  • SE 11.1 Z5e 514 Molarity = 0.101 L Mass % = 0.990% Mole fraction = 0.00389 Molality - 0.217 m
  • Section 11.2 The Energies of Solution Formation 5e 518
  • Fig. 11.1 Z5e 519
  • Rf. Fig. 11-2 Z5e 519
  • Rf. Z553 #35.
  • Rf. Z553 #35.
  • Rf. Z553 #35.
  • Section 11.3 Factors Affecting Solubility Z5e 522
  • Bottom diagram shows skeletal structure of soap with nonpolar and polar ends
  • Z5e 1133 Fig. 23.29.
  • Z5e 523.
  • Rf. Fig. 11.5 Z5e 523
  • Z5e 524 SE 11.4
  • Z5e 524 SE 11.4
  • Z5e 524 SE 11.4
  • Z5e 525.
  • Rf. Fig. 11.6 Z5e 525 Note: most become more soluble with higher temperature, but note cerium sulfate
  • Rf. Z5e 526 Warmer water on lakes blocks oxygen getting to older water below and the warmer water contains less oxygen.
  • Fig. 11.7 Z5e 527
  • Section 11.4 The Vapor Pressures of Solutions Z5e 538
  • Z5e 529
  • This is because the dissolved nonvolatile solute decreases the number of solvent molecules at the surface, and so proportionately lowers the escaping tendency of the solvent molecules. Rf. Z5e 528 and Fig. 11.9
  • Note: all substances are molecules (so no colligative property problem) Use mole fraction        77 b. 
  • Note: all substances are molecules (so no colligative property problem) Use mole fraction        77 b. 
  • Note: all substances are molecules (so no colligative property problem) Use mole fraction        77 b. 
  • Z5e 532
  • Z5e 554 # 51. Use density to get number of moles. Then find  of liquid pentane & hexane. Then use Raoult’s law to get P in torr of both partial and total pressure. b. Since partial pressure of gas is proportional to number of moles of gas present use mol pentane in vapor/total mol vapor to get  See solutions guide p. 267
  • Z5e 554 # 51. Use density to get number of moles. Then find  of liquid pentane & hexane. Then use Raoult’s law to get P in torr of both partial and total pressure. b. Since partial pressure of gas is proportional to number of moles of gas present use mol pentane in vapor/total mol vapor to get  See solutions guide p. 267
  • Use density to get number of moles. Then find  of liquid pentane & hexane. Then use Raoult’s law to get P in torr of both partial and total pressure. b. Since partial pressure of gas is proportional to number of moles of gas present use mol pentane in vapor/total mol vapor to get  See solutions guide p. 267
  • Z5e 554 # 51. Use density to get number of moles. Then find  of liquid pentane & hexane. Then use Raoult’s law to get P in torr of both partial and total pressure. b. Since partial pressure of gas is proportional to number of moles of gas present use mol pentane in vapor/total mol vapor to get  See solutions guide p. 267
  • Use density to get number of moles. Then find  of liquid pentane & hexane. Then use Raoult’s law to get P in torr of both partial and total pressure. b. Since partial pressure of gas is proportional to number of moles of gas present use mol pentane in vapor/total mol vapor to get  See solutions guide p. 267
  • Z5e 532
  • Z5e 532
  • Z5e 532 SE 11.6
  • Z5e 532
  • Section 11.5 Bp elevation & Fp depression Stays liquid longer because intermolecular forces between solute/solvent help hold the liquid together.
  • Stays liquid longer (before freezing) because IMF of solute/solution prevents forming orderly crystal
  • Fig. 11.14 Z5e 536 AP question will require you to diagram the change of addition of a solute to a solvent. The effect of adding a solute is to extend the range of a solvent.
  • Section 11.6 Osmotic Pressure Z5e 540
  • Z5e 540 fig. 11.16
  • Z5e 540 Fig. 11.17
  • Z5e 546 Section 11.7 Colligative Properties of Electrolyte Solutions
  • Ch11 z5e solutions

    1. 1. Ch11 Solutions pp
    2. 2. 11.1 Solution Composition <ul><li>Solutions occur in all phases </li></ul><ul><li>The solvent does the dissolving. </li></ul><ul><li>The solute is dissolved. </li></ul><ul><li>There are examples of all types of solvents dissolving all types of solutes. </li></ul><ul><li>We will focus on aqueous solutions. </li></ul>
    3. 3. Ways of Measuring <ul><li>Molarity = moles of solute Liters of solut ion </li></ul><ul><li>% mass = Mass of solute x 100% Mass of solut ion </li></ul><ul><li>Mole fraction of component A =  A = n A n A + n B </li></ul>
    4. 4. <ul><li>Mol al ity = moles of solute Kilograms of solv ent </li></ul><ul><li>Molality is abbreviated m . </li></ul><ul><li>Normality - read for general info. </li></ul>Ways of Measuring
    5. 5. <ul><li>Molarity </li></ul><ul><li>Mass % </li></ul><ul><li>Mole Fraction </li></ul><ul><li>Molality </li></ul><ul><li>Answers are . . . </li></ul>Given 1.00 g C 2 H 5 OH & 100. g H 2 O and final volume of 101 mL <ul><li>0.215 M </li></ul><ul><li>0.990% </li></ul><ul><li>0.00389 </li></ul><ul><li>0.217 m </li></ul>
    6. 6. 11.2 The Energies of Solution Formation pp <ul><li>Heat of solution (  H soln ) is the energy change for making a solution. </li></ul><ul><li>Most easily understood if broken into 3 energy steps (that must be added). </li></ul><ul><li>1.Break apart sol vent </li></ul><ul><li>2.Break apart sol ute </li></ul><ul><li>3. Mixing solvent and solute </li></ul>
    7. 7. 1. Break apart Solvent pp <ul><li>Have to overcome attractive forces. So,  H 1 > 0 (must put in energy). </li></ul><ul><li>2. Break apart Solute </li></ul><ul><li>Have to overcome attractive forces. So,  H 2 > 0 </li></ul>
    8. 8. 3. Mixing solvent and solute pp <ul><li> H 3 depends on what you are mixing. </li></ul><ul><li>If molecules can attract each other then  H 3 is large and negative . </li></ul><ul><li>If molecules can’t attract then  H 3 is small and negative (negligible interactions). </li></ul><ul><li>This explains the rule “ Like dissolves Like ” </li></ul>
    9. 9. Figure 11.1 The Steps in the Dissolving Process pp
    10. 10. <ul><li>Size of  H 3 determines whether a solution will form pp </li></ul>Energy Solute Solution  H 1  H 2  H 3 Solvent Solute and Solvent  H 3 No solution
    11. 11. Types of Solvent and solutes pp <ul><li>If  H soln is small and positive , a solution will still form because of entropy . </li></ul><ul><li>This is because there are many more ways for them to become mixed than there is for them to stay separate. </li></ul><ul><li>This why NaCl is so soluble in water even though its  H soln is + 3 kJ/mol </li></ul><ul><li>Memorize Table 11.3 p. 492 </li></ul>
    12. 12. Table 11.3 p.521 pp
    13. 13. Energetics of Solutions & Solubility pp <ul><li>The lattice energy of KCl is -715 kJ/mol, & the enthalpy of hydration is -684 kJ/mol. Calculate the enthalpy of solution per mole of solid KCl. Steps . . . </li></ul><ul><li>Lattice energy was defined in Chapter 8 as M + (g) + X - (g)  MX (s) (memorize this). </li></ul><ul><li>Use Hess’ law in two steps: 1st, take the solid and convert it into gaseous ions (opposite of lattice energy) to get . . . </li></ul>
    14. 14. Energetics of Solutions & Solubility pp <ul><li>KCl (s)  K 1+ (g) + Cl 1- (g) whose ∆H is the (-) of ∆H lattice energy = -(-) 715 kJ/mol </li></ul><ul><li>Next, hydrate the ions to get a solution: </li></ul><ul><li>K 1+ (g) + Cl 1- (g)  K 1+ ( aq ) + Cl 1- ( aq ) , where ∆H hydration = -684 kJ/mol </li></ul>
    15. 15. Energetics of Solutions & Solubility pp <ul><li>Add the energies to get . . . </li></ul><ul><li>KCl (s)  K 1+ (g) + Cl 1- (g) = -(-) 715 kJ/mol </li></ul><ul><li>K 1+ (g) + Cl 1- (g)  K 1+ ( aq ) + Cl 1- ( aq ) = -684 kJ/mol </li></ul><ul><li>KCl (s)  K 1+ ( aq ) + Cl 1- ( aq ) = 31 kJ/mol </li></ul><ul><li>∆H solution = 31 kJ/mol (endothermic). </li></ul><ul><li>You have a HW problem on this (#33). </li></ul><ul><li>You will have a test question on this!!! </li></ul>
    16. 16. 11.3 Factors Affecting Solubility Structure and Solubility <ul><li>Water soluble molecules must have dipole moments; i.e. , polar bonds. </li></ul><ul><li>To be soluble in non polar solvents the molecules must be non polar. </li></ul><ul><li>Read Vitamins A and C discussion on figure 11.4 pg. 493 on your own. </li></ul>
    17. 17. Soap P O - CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 O - O -
    18. 18. Soap <ul><li>Hydro phobic non polar end </li></ul>P O - CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 O - O -
    19. 19. Soap <ul><li>Hydro philic polar end </li></ul>P O - CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 O - O -
    20. 20. Skeletal structure of Soap _ P O - CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 O - O -
    21. 21. <ul><li>A drop of grease in water </li></ul><ul><li>Grease is non polar </li></ul><ul><li>Water is polar </li></ul><ul><li>Soap lets you dissolve the nonpolar in the polar. </li></ul>
    22. 22. <ul><li>Hydro phobic ends dissolve in grease </li></ul>
    23. 23. <ul><li>Hydro philic ends dissolve in water </li></ul>
    24. 24. <ul><li>Water molecules can surround and dissolve grease. </li></ul><ul><li>Helps get grease out of your way. </li></ul>
    25. 25. Figure 23.29 Soap Micelles
    26. 26. Pressure effects <ul><li>Changing the pressure doesn’t affect much the amount of solid or liquid that dissolves </li></ul><ul><li>They are basically incompressible. </li></ul><ul><li>But, changing pressure does affect gases (including those dissolved in liquids). </li></ul>
    27. 27. Dissolving Gases <ul><li>Pressure affects the amount of gas that can dissolve in a liquid. </li></ul><ul><li>The dissolved gas is at equilibrium with the gas above the liquid . </li></ul><ul><li>(Mentos video and http://eepybird.com/ ) </li></ul>
    28. 28. <ul><li>The gas is at equilibrium with the dissolved gas in this solution. </li></ul><ul><li>The equilibrium is dynamic . </li></ul>
    29. 29. <ul><li>If you increase the pressure the gas molecules dissolve faster than they escape. </li></ul><ul><li>The equilibrium is disturbed . </li></ul>
    30. 30. <ul><li>The system reaches a new equilibrium with more gas dissolved. </li></ul><ul><li>This is Henry’s Law. </li></ul><ul><li>P= kC </li></ul><ul><li>Pressure = constant x Concentration of gas </li></ul>
    31. 31. Calculations using Henry’s Law pp <ul><li>A bottle of soft drink at 25 o Celsius contains CO 2 at a pressure of 5.0 atm over the liquid. </li></ul><ul><li>Assuming that the partial pressure of CO 2 in the atmosphere (that is, in an open system) is 4.0 x 10 -4 atm, </li></ul><ul><li>Calculate the equilibrium concentrations of CO 2 both before and after the bottle is opened. Henry’s constant here is 32 L atm/mol. . . </li></ul>
    32. 32. Calculations using Henry’s Law pp <ul><li>Henry’s Law is P = kC </li></ul><ul><li>For this problem it is P CO2 = k CO2 C CO2 </li></ul><ul><li>In the unopened bottle, solve for C CO2 to get . . . </li></ul><ul><li>C CO2 = P CO2 /k CO2 = 5.0 atm/32 L atm/mol = 0.16 mol/L. </li></ul>
    33. 33. Calculations using Henry’s Law pp <ul><li>In the opened bottle the gas will reach equilibrium with the atmosphere </li></ul><ul><li>So P CO2 = the partial pressure = 4.0 x 10 -4 atm. </li></ul><ul><li>C CO2 = P CO2 /k CO2 = 4.0 x 10 -4 atm /32 L atm/mol = 1.2 x 10 -5 mol/L . </li></ul><ul><li>This large change in concentration (from 0.16 mol/L to 1.2 x 10 -5 mol/L) explains why soda goes flat. </li></ul>
    34. 34. Temperature Effects <ul><li>Increased temperature increases the rate at which a solid dissolves. </li></ul><ul><li>We can not predict whether it will increase the amount of solid that dissolves (only the rate). </li></ul><ul><li>We must read it from a graph of experimental data. </li></ul>
    35. 35. 20 40 60 80 100
    36. 36. Gases are predictable <ul><li>As temperature increases, solubility decreases. </li></ul><ul><li>So, gas molecules can move fast enough to escape. </li></ul><ul><li>Causes thermal pollution. </li></ul>
    37. 37. Figure 11.7 The Solubilities of Several Gases in Water
    38. 38. 11.4 The Vapor Pressures of Solutions <ul><li>A non volatile sol ute lowers the vapor pressure of the sol vent . </li></ul><ul><li>The molecules of the solvent must overcome the force of both the other solvent molecules and the solute molecules. </li></ul>
    39. 39. Raoult’s Law: <ul><li>P solu tion = P o sol vent x  sol vent </li></ul><ul><li>Form of y = mx + b, plot of P soln and  sol vent is straight line with slope P o </li></ul><ul><li>Vapor pressure of the solution = mole fraction of solvent x vapor pressure of the pure solvent </li></ul><ul><li>Applies only to an ideal solution where the solute itself doesn’t contribute to the vapor pressure ( i.e ., the solute is non volatile. </li></ul>
    40. 40. <ul><li>Pure water has a higher vapor pressure than a solution </li></ul>Aqueous Solution Pure water
    41. 41. <ul><li>Water evaporates faster from pure water than from a solution </li></ul>Aqueous Solution Pure water
    42. 42. <ul><li>The water condenses faster in the solution so it should all end up there. </li></ul>Aqueous Solution Pure water
    43. 43. <ul><li>What is the percent composition of a pentane-hexane solution (not the gas) that has a vapor pressure of 350 torr at 25ºC ? The vapor pressures at 25ºC are </li></ul><ul><ul><li>pentane 511 torr </li></ul></ul><ul><ul><li>hexane 150 torr. </li></ul></ul><ul><li>Try first, then see hints next slide: </li></ul>Review Question pp
    44. 44. <ul><li>% Comp. Of pentane-hexane solution with V p soln = 350 and V p pen = 511, V p hex 150? </li></ul><ul><li>All are molecules, so no colligative property problem. </li></ul><ul><li>Use  can do w/ P since proportional to moles) </li></ul><ul><li>Answers are . . . </li></ul><ul><li>Hexane - 22.7% (150/(511+150) x 100%) </li></ul><ul><li>Pentane - 77.3% (100 - 22.7%) </li></ul>Review Question pp
    45. 45. <ul><li>What is the composition (hexane & pentane pressures) in torr of the vapor ? </li></ul><ul><li>Remember, % Comp. of pentane-hexane is hexane 22.7%, pentane 77.3% and V p solution is 350 torr. Use mole fraction. </li></ul><ul><li>Answers are . . . </li></ul><ul><li>Hexane 79 torr and pentane 271 torr Vapor composition hexane = (.227)(350) = 79 torr, so  that of pentane = 271 </li></ul>Review Question pp
    46. 46. <ul><li>Liquid-liquid solutions where both are volatile. </li></ul><ul><li>Modify Raoult’s Law to . . . </li></ul><ul><li>P total = P A + P B =  A P A 0 +  B P B 0 </li></ul><ul><li>P total = vapor pressure of mixture </li></ul><ul><li>P A 0 = vapor pressure of pure A (etc.) </li></ul><ul><li>If this equation works then the solution is ideal. </li></ul><ul><li>Solvent and solute are alike (as are the interactions between all species).. </li></ul>Ideal/Non-ideal solutions
    47. 47. Test-type Question (will be on our test!) pp <ul><li>Pentane (C 5 H 12 ) & Hexane (C 6 H 14 ) form an ideal solution with V p 25° of 511 & 150. Torr respectively. A solution is made from 25 ml pentane (d = 0.63 g/mL) & 45 mL hexane (d = 0.66 g/mL) </li></ul><ul><li>(a) What is V p of the resulting solution ? </li></ul><ul><li>(b) What is composition by mole fraction of pentane in the vapor that is in equilibrium with this solution? </li></ul><ul><li>Reference: Z7e p. 521, # 49 </li></ul>
    48. 48. Test-type Question pp <ul><li>Pentane (C 5 H 12 ) & Hexane (C 6 H 14 ) form an ideal solution with V p 25° of 511 & 150. Torr respectively. A solution is made from 25 ml pen (d = 0.63 g/mL) & 45 mL hex (d = 0.66 g/ML </li></ul><ul><li>(a) What is V p of the resulting solution ? </li></ul><ul><li>Use density to get number of moles. </li></ul><ul><li>Then find  of liquid pentane & hexane. </li></ul><ul><li>Then use Raoult’s law to get P in torr of both partial and total pressure </li></ul>
    49. 49. Test-type Question rf. Z7e #49 pp <ul><li>25 ml Pen x 0.63g/ml x 1 mol/72.15 g = 0.22 mol Pen (and 0.34 mol Hex) </li></ul><ul><li> pen liq = 0.22/0.56 = 0.39 and  hex liq = 1.00 - 0.39 = 0.61 </li></ul><ul><li>P pen = (  pen liq )(P pen o ) = (0.39)(511 torr) = 2.0 x 10 2 torr ; </li></ul><ul><li>ditto for P hex = 92 torr </li></ul><ul><li>P total = 292 torr = 290 torr (sig figs) </li></ul>
    50. 50. Test-type Question pp <ul><li>Pentane (C 5 H 12 ) & Hexane (C 6 H 14 ) form an ideal solution with V p 25° of 511 & 150. Torr respectively. A solution is made from 25 ml pen (d = 0.63 g/mL) & 45 mL hex (d = 0.66 g/ML </li></ul><ul><li>(b) What is composition by mole fraction of pentane in the vapor that is in equilibrium with this solution? </li></ul><ul><li>Hint: Since partial pressure of gas is proportional to number of moles of gas present, use mol pentane in vapor ÷ total mol vapor to get  </li></ul>
    51. 51. Test-type Question pp <ul><li>Pentane (C 5 H 12 ) & Hexane (C 6 H 14 ) form an ideal solution with V p 25° of 511 & 150. Torr respectively. A solution is made from 25 ml pen (d = 0.63 g/mL) & 45 mL hex (d = 0.66 g/ML </li></ul><ul><li>(b) What is composition by mole fraction of pentane in the vapor that is in equilibrium with this solution? </li></ul><ul><li>The answer is . . . </li></ul><ul><li> pen v = P pen /P total = 2.0 x 10 2 torr/290 torr = 0.69. </li></ul>
    52. 52. Deviations to non-ideality <ul><li>If Solvent has a strong affinity for solute (H bonding). </li></ul><ul><li>Lowers solvent’s ability to escape. </li></ul><ul><li>Lower vapor pressure than expected. </li></ul><ul><li>If (-) deviation from Raoult’s law then . . . </li></ul><ul><li> H soln is large and (-), so exothermic . </li></ul><ul><li>An Endothermic  H soln indicates positive deviation. </li></ul>
    53. 53. Colligative Properties <ul><li>Dissolved particles affect vapor pressure, so they affect phase changes. </li></ul><ul><li>Colligative properties depend only on the number - not the kind of solute particles present </li></ul><ul><li>The 3 colligative properties are Bp , Fp and osmotic pressure </li></ul><ul><li>Useful to determine molar mass (Lab 11) </li></ul><ul><li>M = (K fp x g solute )/(kg solvent x ∆T fp ) </li></ul>
    54. 54. Colligative Properties pp <ul><li>See SE 11.6 p. 500 ( this type will be on test) </li></ul><ul><li>Predict the V p of a solution of 35.0 g Na 2 SO 4 (M = 142 g/mol) & 175 g H 2 0, where V p H 2 O = 23.76 torr at 25 o C. . . </li></ul><ul><li>Strategy: use Raoult’s Law P soln =  H 2 O P o H 2 O </li></ul><ul><li> H 2 O = n H2O /(n H2O + n solute ) </li></ul><ul><li>But . . . Since the solute falls apart into 3 particles, the n solute has to be the moles of Na 2 SO 4 x 3 . </li></ul>
    55. 55. Colligative Properties pp <ul><li>See SE 11.6 p. 532 ( this type will be on test) </li></ul><ul><li>n H2O = 175 g/18.0 g/mol = 9.72 mol H 2 O </li></ul><ul><li>n Na2SO4 = 35.0 g/142 g/mol = 0.246 </li></ul><ul><li>n solute = 3 x n Na2SO4 = 0.738 </li></ul><ul><li> H2O = 9.72/(0.738 + 9.72) = 0.929 </li></ul><ul><li>P soln =  H2O P o H2O = (0.929)(23.76 torr) = 22.1 torr </li></ul>
    56. 56. 11.5 Boiling point Elevation & Freezing Point Depression <ul><li>Because a nonvolatile solute lowers the vapor pressure it raises the boiling point. </li></ul><ul><li>Stays liquid longer because intermolecular forces between solute/solvent help hold the liquid together. </li></ul><ul><li>The equation is:  T = K b m solute (for a molecular solute; otherwise use van’t Hoff factor, i , for electrolyte solutions). </li></ul><ul><li> T is the change in the boiling point </li></ul><ul><li>K b is a constant determined by the solvent (look up in a table). </li></ul><ul><li>m solute is the molality of the solute </li></ul>
    57. 57. Freezing point Depression <ul><li>Because a nonvolatile solute lowers the vapor pressure of the solution it lowers the freezing point. </li></ul><ul><li>Stays liquid longer (before freezing) because IMF of solute/solution prevents forming orderly crystal </li></ul><ul><li>The equation is:  T = K f m solute </li></ul><ul><li> T is the change in the freezing point </li></ul><ul><li>K f is a constant determined by the solvent </li></ul><ul><li>m solute is the molality of the solute </li></ul>
    58. 58. 1 atm Vapor Pressure of solution Vapor Pressure of pure water
    59. 59. 1 atm Freezing and boiling points of water
    60. 60. 1 atm Freezing and boiling points of solution
    61. 61. 1 atm  T f  T b
    62. 62. Figure 11.14 p. 536 The Development of Osmotic Pressure pp AP requires you to diagram the change when adding solute to a solvent The effect of adding a solute is to extend the range of a solvent.
    63. 63. 11.6 Osmotic Pressure <ul><li>Osmosis : The flow of solvent into the solution through a semi permeable membrane. </li></ul><ul><li>Osmotic Pressure : The excess hydrostatic pressure on the solution compared to the pure solvent. </li></ul><ul><li>Osmotic Pressure π = M RT </li></ul><ul><li>Watch units for R! </li></ul>
    64. 64. Figure 11.16 Osmotic Pressure. Net transfer of solvent molecules until hydrostatic pressure equalizes the solvent flow in both directions
    65. 65. Figure 11.18 Osmosis at Equilibrium
    66. 66. <ul><li>If the external pressure is larger than the osmotic pressure, reverse osmosis occurs. </li></ul><ul><li>One application is desalination of seawater . </li></ul>
    67. 67. Figure 11.17 Osmosis When pressure exceeds this then reverse osmosis occurs.
    68. 68. Review Problems Bp, Fp, π <ul><li>Try SE 11.8 p. 537 (& Text # 60) </li></ul><ul><li>Try SE 11.10 p. 539 (& Text # 65) </li></ul><ul><li>Try SE 11.11 p. 541 (& Text # 68) </li></ul><ul><li>Try SE 11.12 (& Text # 70) </li></ul><ul><li>#60 = 498 g/mol </li></ul><ul><li>#65 = 456 g/mol </li></ul><ul><li>#68 = 27 000 g/mol </li></ul><ul><li>#70 = Dissolve 18 g in 1 L of solution </li></ul>
    69. 69. 11.7 Colligative Properties of Electrolyte Solutions <ul><li>Since colligative properties only depend on the number of particles: </li></ul><ul><li>Ionic compounds should have a bigger effect because . . . </li></ul><ul><li>When they dissolve they dissociate. </li></ul><ul><li>Individual Na and Cl ions fall apart. </li></ul><ul><li>1 mole of NaCl makes 2 moles of ions. </li></ul><ul><li>1 mole Al(NO 3 ) 3 makes 4 moles ions. </li></ul>
    70. 70. <ul><li>Electrolytes have a bigger impact on on melting and freezing points per mole because they make more pieces. </li></ul><ul><li>Relationship is expressed using the van’t Hoff factor i </li></ul><ul><li>i = Moles of particles in solution </li></ul><ul><li>Moles of solute dissolved </li></ul><ul><li>The expected value can be determined from the above formula. </li></ul>
    71. 71. <ul><li>The actual value is usually less because: </li></ul><ul><li>At any given instant some of the ions in solution will be paired. </li></ul><ul><li>Ion pairing increases with concentration. </li></ul><ul><li>i decreases with increased concentration. </li></ul><ul><li>We can change our generic formulae to </li></ul><ul><li> H = i K m </li></ul>
    72. 72. Colligative Properties of Electrolytes <ul><li> T = i K b m solute = Bp elevation </li></ul><ul><li> T = i K f m solute = Fp depression </li></ul><ul><li>π = iM RT = Osmotic Pressure </li></ul><ul><li>11.8 - Tyndall effect = the scattering of light by particles (read on own). </li></ul><ul><li>End, Chapter 11. </li></ul>
    73. 73. <ul><li>Label your solutions, in the flasks and the ice cube trays. </li></ul><ul><li>Final conclusion will be to compare the actual freezing point depression to the theoretical. </li></ul><ul><li>Give reasons for any differences. </li></ul>

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