1. Chapter 13 Properties of Solutions
Chemistry, The Central Science, 11th edition
Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten
Chapter 13 Properties of Solutions
John D. Bookstaver
St. Charles Community College
Cottleville, MO
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2. Overview of Chapter 13 - Solutions
Review/On your own:
describe the three factors involved in solubility (solute-solute, solvent-solvent and solute-solvent interactions)
explain what factors affect solubility and predict solubilities of solutes in solvents
define and apply the terms miscible and immiscible, saturated, unsaturated and supersaturated solutions
use Henry’s Law to calculate the solubility or pressure of a gas in liquid solution
calculate solubilities (g/100 g H2O)
interpret solubility graphs
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3. Overview of Chapter 13 - Solutions
New Material
calculate concentration using mass percentage, ppm, ppb, mole fraction, molarity and molality and convert amongst them
define colligative properties
use Raoult’s Law to calculate vapor pressures and mole fractions of solutions
calculate freezing point depression, boiling point elevation and osmotic pressure
calculate the molar mass from any of the four colligative properties
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4. Solutions
Solutions are homogeneous mixtures of two or more pure substances.
In a solution, the solute is dispersed uniformly throughout the solvent.
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5. Solutions
The intermolecular forces between solute and solvent particles must be strong enough to compete with those between solute particles and those between solvent particles.
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6. How Does a Solution Form?
As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them.
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7. How Does a Solution Form
If an ionic salt is soluble in water, it is because the ion-dipole interactions are strong enough to overcome the lattice energy of the salt crystal.
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8. Energy Changes in Solution
Simply put, three processes affect the energetics of solution:
separation of solute particles,
separation of solvent particles,
new interactions between solute and solvent.
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9. Energy Changes in Solution
The enthalpy change of the overall process depends on H for each of these steps.
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10. Why Do Endothermic Processes Occur?
Things do not tend to occur spontaneously (i.e., without outside intervention) unless the energy of the system is lowered.
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11. Why Do Endothermic Processes Occur?
Yet we know the in some processes, like the dissolution of NH4NO3 in water, heat is absorbed, not released.
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12. Enthalpy Is Only Part of the Picture
The reason is that increasing the disorder or randomness (known as entropy) of a system tends to lower the energy of the system.
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13. Enthalpy Is Only Part of the Picture
So even though enthalpy may increase, the overall energy of the system can still decrease if the system becomes more disordered.
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14. Student, Beware!
Just because a substance disappears when it comes in contact with a solvent, it doesn’t mean the substance dissolved.
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15. Student, Beware!
Dissolution is a physical change — you can get back the original solute by evaporating the solvent.
If you can’t, the substance didn’t dissolve, it reacted.
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16. Sample Exercise 13.1 (p. 534)
By the process illustrated below, water vapor reacts with excess solid sodium sulfate to form the hydrated form of the salt. The chemical reaction is
Na2SO4(s) + 10 H2O(g)  Na2SO4• 10 H2O(s)
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17. Sample Exercise 13.1 (p. 534)
Essentially all of the water vapor in the closed container is consumed in this reaction. If we consider our system to consist initially of Na2SO4(s) and 10 H2O(g)
does the system become more or less ordered in this process, and
b) does the entropy of the system increase or decrease?
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18. Practice Exercise 13.1
Does the entropy of the system increase or decrease when the stopcock is opened to allow mixing of the two gases in the apparatus?
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19. Types of Solutions Saturated
In a saturated solution, the solvent holds as much solute as is possible at that temperature.
Dissolved solute is in dynamic equilibrium with solid solute particles.
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20. Types of Solutions Unsaturated
If a solution is unsaturated, less solute than can dissolve in the solvent at that temperature is dissolved in the solvent.
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21. Types of Solutions Supersaturated
In supersaturated solutions, the solvent holds more solute than is normally possible at that temperature.
These solutions are unstable; crystallization can usually be stimulated by adding a “seed crystal” or scratching the side of the flask.
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22. Factors Affecting Solubility
Chemists use the axiom “like dissolves like."
Polar substances tend to dissolve in polar solvents.
Nonpolar substances tend to dissolve in nonpolar solvents.
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23. Factors Affecting Solubility
The more similar the intermolecular attractions, the more likely one substance is to be soluble in another.
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24. Factors Affecting Solubility
Glucose (which has hydrogen bonding) is very soluble in water, while cyclohexane (which only has dispersion forces) is not.
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25. Factors Affecting Solubility
Vitamin A is soluble in nonpolar compounds (like fats).
Vitamin C is soluble in water.
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26. Sample Exercise 13.2 (p. 539)
Predict whether each of the following substances is more likely to dissolve in carbon tetrachloride (CCl4) or in water:
C7H16, Na2SO4, HCl, and I2.
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27. Practice Exercise 13.2
Arrange the following substances in order of increasing solubility in water:
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28. Gases in Solution
In general, the solubility of gases in water increases with increasing mass.
Larger molecules have stronger dispersion forces.
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29. Gases in Solution
The solubility of liquids and solids does not change appreciably with pressure.
The solubility of a gas in a liquid is directly proportional to its pressure.
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30. Henry’s Law Sg = kPg where Sg is the solubility of the gas,
k is the Henry’s Law constant for that gas in that solvent, and
Pg is the partial pressure of the gas above the liquid.
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31. Sample Exercise (p. 540)
Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25oC. The Henry’s Law constant for CO2 in water at this temperature is 3.1 x 10-2 mol/L-atm.
(0.12 M)
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32. Practice Exercise 13.3
Calculate the concentration of CO2 in a soft drink after the bottle is opened and equilibrates at 25oC under a CO2 partial pressure of 3.0 x 10-4 atm.
(9.3 x 10-6 M)
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33. Temperature
Generally, the solubility of solid solutes in liquid solvents increases with increasing temperature.
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34. Temperature The opposite is true of gases.
Carbonated soft drinks are more “bubbly” if stored in the refrigerator.
Warm lakes have less O2 dissolved in them than cool lakes.
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35. Ways of Expressing Concentrations of Solutions
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36. Ways of Expressing Concentration
All methods involve quantifying the amount of solute per amount of solvent, or solution.
Dilute and concentrated are qualitative ways to describe concentration.
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37. Quantitative Expressions of Concentration
Require specific information regarding such quantities as
Masses
Moles
Liters of the solute, solvent, or solution
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38. Quantitative Expressions of Concentration
Mass percentage, which includes ppm, ppb
Mole fraction
Molarity
Molality
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39. Mass Percentage mass of A in solution Mass % of A =
total mass of solution
Mass % of A =
 100
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40. Parts per Million and Parts per Billion
Parts per Million (ppm) = mg solute/kg solution
mass of A in solution
total mass of solution
ppm =
 106
Parts per Billion (ppb) = mg solute/kg solution
mass of A in solution
total mass of solution
ppb =
 109
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41. Sample Exercise 13.4 (p. 542)
A solution is made by dissolving 13.5 g glucose (C6H12O6) in kg of water. What is the mass percentage of solute in this solution?
(11.9%)
A 2.5-g sample of groundwater was found to contain 5.4 mg of Zn2+. What is the concentration of Zn2+ in parts per million?
(2.2 ppm)
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42. Practice Exercise 13.4
Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water.
(2.91%)
A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution?
(90.5 g NaOCl)
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43. Mole Fraction, Molarity and Molality
Common expressions of concentration based on the number of moles of one or more components
Remember: mass can be converted to moles using MM
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44. total moles in solution
Mole Fraction (X)
moles of A
total moles in solution
XA =
In some applications, one needs the mole fraction of solvent, not solute — make sure you find the quantity you need!
Note: mole fraction has no units.
Note: mole fractions range from 0 to 1
Particularly useful for gases
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45. Molarity (M) mol of solute M = L of solution
You will recall this concentration measure from Chapter 4.
Since volume is temperature-dependent, molarity can change with temperature.
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46. Molality (m) mol of solute m = kg of solvent
Since both moles and mass do not change with temperature, molality (unlike molarity) is not temperature-dependent.
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47. Sample Exercise 13.5 (p. 544)
A solution is made by dissolving 4.35 g glucose (C6H12O6) in 25.0 mL of water. Calculate the molality of glucose in the solution. (Assume DH2O = 1.00 g/mL)
(0.964 m)
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48. Practice Exercise 13.5
What is the molality of a solution made by dissolving 36.5 g of naphthalene (C10H8) in 425 g of toluene (C7H8)?
(0.671 m)
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49. Where we are going next
Problems that involve conversions between different types of units and other applications that require preparatory steps
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50. Changing Molarity to Molality
If we know the density of the solution, we can calculate the molality from the molarity and vice versa.
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51. Sample Exercise 13.6 (p. 544)
A solution of hydrochloric acid contains 36% HCl by mass.
a) Calculate the mole fraction of HCl in
the solution.
(0.22)
b) Calculate the molality of HCl in the
solution.
(15 m)
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52. Practice Exercise 13.6
A commercial bleach solution contains 3.62 mass % NaOCl in water.
Calculate
the mole fraction of NaOCl in the
solution and (9.00 x 10-3)
b) the molality of NaOCl in the solution (0.505 m)
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53. Sample Exercise 13.7 (p. 546)
A solution with a density of g/mL contains 5.0 g of toluene (C7H8) and 225 g of benzene.
Calculate the molarity of the solution. (0.21 M)
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54. Practice Exercise 13.7
A solution containing equal masses of glycerol (C3H8O3) and water has a density of 1.10 g/mL. Calculate
a) the molality of glycerol; (10.9 m)
b) the mole fraction of glycerol; (0.163)
c) the molarity of glycerol in the solution. (5.97 M)
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55. Warmup
Caffeine (C8H10N4O2) is a stimulant found in coffee and tea. If a solution of caffeine in chloroform (CHCl3) as a solvent has a concentration of m, calculate
a) the percent caffeine by mass (1.44%)
b) the mole fraction of caffeine ( )
(p. 567, Q# 13.54)
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56. Overview
Solutions
Colligative properties, especially to find MM, including Friday’s lab
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57. Colligative Properties
Changes in colligative properties depend only on the number of solute particles present, not on the identity of the solute particles.
Physical properties
“depending on the collection”, i.e. collective effects of the solute particles
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58. Colligative Properties
Application of molality
Examples:
C12H22O11  C12H22O11 one particle
NaCl  Na+ + Cl- two particles
CaCl2  Ca Cl- three particles
This will have the greatest effect (of these examples)
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59. Colligative Properties
Among colligative properties are
Vapor pressure lowering
Boiling point elevation*
Melting point depression* - Friday’s lab
Osmotic pressure*
*quantitative
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60. Vapor Pressure
Because of solute-solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase.
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61. Vapor Pressure Focus on the solvent
e.g. NaCl dissolved in H2O – lower water VP than pure H2O
Lower VP because the Na+ and Cl- particles (nonvolatile solutes) get surrounded by H2O  fewer “free” H2O particles left to vaporize at surface
Stronger IMFs  harder for solvent molecules to leave
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62. Vapor Pressure
The vapor pressure of a solution is lower than that of the pure solvent.
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63. Raoult’s Law For an ideal solution: PA = XAPA where
XA is the mole fraction of compound A, and
PA is the normal vapor pressure of A at that temperature.
NOTE: This is one of those times when you want to make sure you have the vapor pressure of the solvent.
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64. Raoult’s Law Real solutions show approximately ideal behavior when:
The solute concentration is low
The solute and solvent have similarly sized molecules
The solute and solvent have similar types of IMFs
Raoult’s Law breaks down when the solvent-solvent and solute-solute IMFs are >>> or <<< solute-solvent IMFs.
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65. Sample Exercise 13.8 (p. 547)
Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25oC. Calculate the vapor pressure at 25oC of a solution made by adding 50.0 mL of glycerin to mL of water. The vapor pressure of pure water at 25oC is 23.8 torr.
(23.2 torr)
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66. Practice Exercise 13.8
The vapor pressure of pure water at 110oC is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110oC. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution?
(rearrange equation to find mole fraction)
(0.290)
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67. Boiling Point Elevation and Freezing Point Depression
Nonvolatile solute-solvent interactions also cause solutions to have higher boiling points and lower freezing points than the pure solvent.
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68. Boiling Point Elevation
# particles of nonvolatile solute  ¯ VP
 ­ boiling point
Since bp = VP at atm P, if the VP is lower,
more energy must be used to overcome atm P
= higher T at boiling
It takes more energy to break apart solvent/solute bonds to allow the solvent to vaporize.
For H2O: 1 mole particles = oC (molal bp elevation constant)
1 kg H2O
Note: 1 m solution of NaCl is 2 m in total solute particles
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69. Boiling Point Elevation
The change in boiling point is proportional to the molality of the solution:
Tb = Kb  m
where Kb is the molal boiling point elevation constant, a property of the solvent.
m = molality
Tb is added to the normal boiling point of the solvent.
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70. Boiling Point Elevation
The change in freezing point can be found similarly:
Tf = Kf  m
Here Kf is the molal freezing point depression constant of the solvent.
Tf is subtracted from the normal boiling point of the solvent.
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71. Freezing Point Depression
# particles of nonvolatile solute  ¯ freezing point
When a solution freezes, crystals of almost pure solvent form first
Solute molecules are usually not soluble in the solid phase of the solvent, so become more concentrated in the liquid phase
 ¯ VP
 triple point occurs at a lower T because of the lower VP for the solution.
The melting-point (freezing-point) curve is a vertical line from the triple point
 the solution freezes at a lower temperature than the pure solvent
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72. Boiling Point Elevation and Freezing Point Depression
Note that in both equations, T does not depend on what the solute is, but only on how many particles are dissolved.
Tb = Kb  m
Tf = Kf  m
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73. Boiling Point Elevation and Freezing Point Depression
e.g. What is the boiling point of a solution that contains 1.25 mol CaCl2 in g of water?
mol CaCl2 = 1.25 mol CaCl2 x g = m x 3 = 2.68 m
1000. g H2O g H2O 1 kg
= # particles when CaCl dissociates in H2O
Tb = Kb  m
= 0.512oC x 2.68 m = 1.37oC oC = oC
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74. Sample Exercise 13.9 (p. 550)
Automotive antifreeze consists of ethylene glycol (C2H6O2), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0 mass % solution of ethylene glycol in water.
(102.7oC; oC)
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75. Practice Exercise 13.9
Calculate the freezing point of a solution containing kg of chloroform (CHCl3) and 42.0 g of eucalyptol (C10H18O), a fragrant substance found in the leaves of eucalyptus trees.
(See Table 13.4)
(-65.6oC)
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76. Sample Exercise (p. 551)
List the following aqueous solutions in order of their expected freezing points:
0.050 m CaCl2; m HCl; m HC2H3O2; m C12H22O11.
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77. Practice Exercise 13.10
Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water:
1 mol Co(NO3)2, 2 mol of KCl, 3 mol of ethylene glycol (C2H6O2)?
(2 mol KCl)
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78. Check in End of last semester
colligative properties, esp. freezing point depression and boiling point elevation
Calculations: predict new bp or fp
Next: Colligative props of electrolytes
Osmotic pressure
How to determine molar mass using colligative properties
- freezing point depression lab
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79. Colligative Properties of Electrolytes
Since these properties depend on the number of particles dissolved, solutions of electrolytes (which dissociate in solution) should show greater changes than those of nonelectrolytes.
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80. Colligative Properties of Electrolytes
However, a 1M solution of NaCl does not show twice the change in freezing point that a 1M solution of methanol does.
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81. van’t Hoff Factor
One mole of NaCl in water does not really give rise to two moles of ions.
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82. van’t Hoff Factor
Some Na+ and Cl- reassociate for a short time, so the true concentration of particles is somewhat less than two times the concentration of NaCl.
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83. van’t Hoff Factor
Reassociation is more likely at higher concentration.
Therefore, the number of particles present is concentration-dependent.
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84. van’t Hoff Factor Tf = Kf  m  i
We modify the previous equations by multiplying by the van’t Hoff factor, i.
Tf = Kf  m  i
(is is used incorrectly in your handout on p. 18)
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85. Osmosis
Some substances form semipermeable membranes, allowing some smaller particles to pass through, but blocking other larger particles.
In biological systems, most semipermeable membranes allow water to pass through, but solutes are not free to do so.
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86. Osmosis
In osmosis, there is net movement of solvent from the area of higher solvent concentration (lower solute concentration) to the are of lower solvent concentration (higher solute concentration).
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87. Osmotic Pressure n  = ( )RT = MRT V
The pressure required to stop osmosis, known as osmotic pressure, , is n V
 = ( )RT = MRT
where M is the molarity of the solution.
If the osmotic pressure is the same on both sides of a membrane (i.e., the concentrations are the same), the solutions are isotonic.
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88. Osmosis in Blood Cells
If the solute concentration outside the cell is greater than that inside the cell, the solution is hypertonic.
Water will flow out of the cell, and crenation results.
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89. Osmosis in Cells
If the solute concentration outside the cell is less than that inside the cell, the solution is hypotonic.
Water will flow into the cell, and hemolysis results.
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90. Sample Exercise 13.11 (p. 553) (on your own)
The average osmotic pressure of blood is 7.7 atm at 25oC. What concentration of glucose (C6H12O6) will be isotonic with blood?
(0.31 M)
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91. Practice Exercise 13.11 (on your own)
What is the osmotic pressure at 20oC of a M sucrose (C12H22O11) solution?
(0.048 atm or 37 torr)
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92. Using Colligative Properties to Determine Molar Mass
Any of the four colligative properties may be used to determine molar mass
Thursday’s lab
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93. Sample Exercise (p. 555)
A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving g of the substance in 40.0 g of CCl4. The boiling point of the resultant solution was 0.357oC higher than that of the pure solvent. Calculate the molar mass of the solute.
(88.0 g/mol)
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94. Practice Exercise 13.12
Camphor (C10H16O) melts at 179.8oC, and it has a particularly large freezing-point-depression constant, Kf = 40.0oC/m. When g of an organic substance of unknown molar mass is dissolved in g of liquid camphor, the freezing point of the mixture is found to be 176.7oC. What is the molar mass of the solute?
(110 g/mol)
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95. Sample Exercise (p. 556)
The osmotic pressure of an aqueous solution of a certain protein was measured in order to determine its molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution at 25oC was found to be 1.54 torr. Calculate the molar mass of the protein.
(8.45 x 103 g/mol)
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96. Practice Exercise 13.13
A sample of 2.05 g of the plastic polystyrene was dissolved in enough toluene to form L of solution. The osmotic pressure of this solution was found to be 1.21 kPa at 25oC.
Calculate the molar mass of the polystyrene.
(4.20 x 104 g/mol)
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98. Colloids (review on your own)
Suspensions of particles larger than individual ions or molecules, but too small to be settled out by gravity are called colloids.
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99. Tyndall Effect Colloidal suspensions can scatter rays of light.
This phenomenon is known as the Tyndall effect.
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100. Colloids in Biological Systems
Some molecules have a polar, hydrophilic (water-loving) end and a non-polar, hydrophobic (water-hating) end.
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101. Colloids in Biological Systems
Sodium stearate is one example of such a molecule.
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102. Colloids in Biological Systems
These molecules can aid in the emulsification of fats and oils in aqueous solutions.
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103. Sample Integrative Exercise 13 (p. 560)
A L solution is made by dissolving g of CaCl2(s) in water.
a) Calculate the osmotic pressure of this solution at 27oC, assuming that it is completely dissociated into its component ions (2.93 atm)
The measured osmotic pressure of this solution is 2.56 atm at 27oC. Explain why it is less than the value calculated in (a), and calculate the van’t Hoff factor, i, for the solute in this solution (2.62)
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104. Sample Integrative Exercise 13 (p. 560)
c) The enthalpy of solution for CaCl2 is DH = kJ/mol. If the final temperature of the solution was 27.0oC, what was its initial temperature? (Assume that the density of the solution is 1.00 g/mL, that its specific heat is 4.18 J/g-K, and that the solution loses no heat to its surroundings.)
(26.2oC)
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