1
Chapter 12
Solutions
Copyright © Cengage Learning. All rights reserved. 12 | 2
Solution Formation
1. Types of Solutions
2. Solubility and the S...
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Colligative Properties
4. Ways of Expressing Concentration
5. Va...
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Solution Formation
1. Types of Solutions
a. Define solute and so...
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2. Solubility and the Solution Process
a. List the conditions th...
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3. Effects of Temperature and Pressure on
Solubility
a. State th...
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Colligative Properties
4. Ways of Expressing Concentration
a. De...
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i. Convert molality to mole fraction.
j. Convert mole fraction t...
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6. Boiling-Point Elevation and Freezing-Point
Depression
a. Defi...
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7. Osmosis
a. Describe a system where osmosis will take
place.
...
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Colloid Formation
9. Colloids
a. Define colloid.
b. Explain the...
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A solution is composed of two parts: the solute and
the solvent...
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Fluids that mix with or
dissolve in each other in
all proportio...
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a.The 5 g of Mo is the solute; the 80 g of Cr is the
solvent.
b...
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A saturated solution is
in equilibrium with
respect to the amou...
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The solubility of a solute is the amount that
dissolves in a gi...
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A supersaturated solution is a solution that
contains more diss...
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Solubility can be understood in terms of two
factors:
1. The na...
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For molecular solutions,
this can be summarized
as “Like dissol...
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Concept Check 12.2
You have two molecular compounds, X and Y.
C...
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When considering ionic solutes in water, we need
to examine the...
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The stronger ion-dipole
force between the ion
and the solvent—t...
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The force of attraction
between water and
both a cation and an
...
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The process of
dissolving occurs at
the surfaces of the
solid. ...
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The hydration energy for AB2 must be greater than
the hydration...
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In general, solubility depends on temperature.
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In most cases, solubility increases with increasing
temperature...
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When dissolving absorbs heat (is endothermic),
the temperature ...
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Hot packs use an
exothermic solution
process.
Cold packs use an...
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Henry’s law describes the effect of pressure on
gas solubility:...
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In general, pressure has
little or no effect on the
solubility ...
?
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Helium–oxygen mixtures are sometimes
used as the breathing ga...
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P1 = 1.0 atm P2 = 46 atm
S1 = 0.94 g/mL S2 = ?
1
2
12
P
P
SS ×=...
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At high elevations, the partial pressure of oxygen
decreases, d...
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The concentration of a solute can be quantitatively
expressed i...
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Molarity is the moles of solute per liter of solution.
It is ab...
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Mass percentage of solute is the percentage by
mass of solute i...
?
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An experiment calls for 36.0 g of a
5.00% aqueous solution of...
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A 5.00% aqueous solution of KBr has 5.00 g KBr
per 100. g solut...
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Molality is the moles of solute per kilogram of
solvent. It is ...
?
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Iodine dissolves in a variety of organic
solvents. For exampl...
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Mass of solute = 5.00 g I2
Mass of solvent = 30.0 g CH2Cl2
kg1
...
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Mole fraction is the moles of component over the
total moles of...
?
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A solution of iodine, I2, in methylene
chloride, CH2Cl2, cont...
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Mass of solute = 5.00 g I2
Mass of solvent = 56.0 g CH2Cl2
mol0...
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0.02902I =Χ
totalmol0.6791
Imol0.01970 2
I2
=Χ
0.97122ClCH =Χ
2...
?
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A bottle of bourbon is labeled 94 proof,
meaning that it is 4...
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mol8.16
g46.08
mol1
mL1
g0.80
OHHCmL470 52
=××
0.22ethanol =Χ
m...
?
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A 3.6 m solution of calcium chloride,
CaCl2, is used in tract...
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Moles solute = 3.6 mol CaCl2
0.0612CaCl =Χ
OHmol55.5
g18.02
mol...
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Converting between molality and mole fraction is
relatively sim...
?
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A solution contains 8.89 × 10-3
mole
fraction of I2 dissolved...
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Moles solute = 0.00889 mol I2
kg
mol
0.106=m
22
3
ClCHkg0.08417...
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Converting between molality and molarity requires
knowing the d...
?
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Citric acid, HC6H7O7, is often used in
fruit beverages to add...
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Moles solute = 2.331 mol HC6H7O7
L
mol
1.827=
g447.88
mol1
g192...
?
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An aqueous solution of ethanol is 14.1
M C2H5OH. The density ...
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g649.7
mol1
g46.08
mol14.1soluteofMass =×=
g853
mL
g0.853
L10
m...
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Colligative properties of solutions are properties
that depend ...
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The vapor pressure of a solution, P, is less than
the vapor pre...
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We can explain this relationship by recalling that
evaporation ...
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To establish an equilibrium, the gaseous
solvent will condense ...
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To find the vapor-pressure lowering (a colligative
property), w...
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Note that vapor-pressure lowering is directly
proportional to t...
?
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Eugenol, C10H12O2, is the chief
constituent of oil of clove. ...
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mol0.0521
g164.22
mol1
g8.56soluteMoles =×=
mmHg2.04Δ =P
mol1.0...
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When the solute is nonvolatile, it has no
appreciable vapor pre...
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The phase diagram on the next slide shows the
changes in freezi...
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The boiling point of a solution is higher than the
boiling poin...
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The freezing point of a solution is lower than the
freezing poi...
?
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A solution is made up of eugenol,
C10H12O2, in diethyl ether ...
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m = 0.575 m
Kf = 1.79°C/m Kb = 2.02°C/m
Tf° = –116.3°C Tb° = 34...
?
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In a freezing-point depression
experiment, the molality of a ...
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Solute mass = 58.1 mg
Solvent mass = 5.00 g
m = 0.0784 mol/kg
m...
?
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An 11.2-g sample of sulfur was
dissolved in 40.0 g of carbon ...
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Solute mass = 11.2 g
Solvent mass = 40.0 g
∆Tb = 2.63°C
Kb = 2....
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m
m
K
T
m 1.096
C
2.40
C2.63Δ
f
b
=
°
°
==
fbΔ KmT ×=
mol0.0438...
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.8SformulamolecularThe
g/mol255.5
mol0.04383
g11.2
massMolar ==...
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Osmosis is the phenomenon of solvent flow
through a semipermeab...
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Water is the
solvent.
Water will flow
from the left to
the righ...
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Water is the solvent.
It flows from the beaker into
the thistle...
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Osmotic pressure, π, is equal to the pressure that,
when applie...
?
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Dextran, a polymer of glucose units, is
produced by bacteria ...
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T = 21°C + 273 = 294 K
Mass solute = 1.50 g
Liters solution = 1...
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8 mol L atm
3.75 10 0.08206 294 K
L mol K
π − •
= × × ×
•
MRTπ ...
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One can show that 760.0 mmHg is equivalent to
the pressure exer...
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Ionic solutes dissolve to form more than one
particle per formu...
?
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What is the osmotic pressure at 25.0°C
of an isotonic saline ...
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T = 25°C + 273 = 298 K
Mass solute = 0.900 g NaCl
Volume soluti...
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K298
Kmol
atmL
0.08206
L
mol
0.15402 ×
•
•
××=π
iMRTπ =
atm7.53...
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A colloid is a dispersion of particles of one
substance (the di...
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Colloids exhibit the Tyndall effect.
The path of the light is v...
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Colloid Solution
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Colloids are characterized according to the state
(solid, liqui...
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Colloids in which the continuous phase is water
are categorized...
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Coagulation is the process by which the
dispersed phase of a c...
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When molecules or ions that have both a
hydrophobic end and a ...
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Soap consists of compounds such as sodium
stearate. Sodium ste...
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In water solution, the stearate ions associate to
form micelle...
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Far left:
Vegetable oil floating
on water (dyed
green).
Left:
...
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Synthetic detergents also form association
colloids. Sodium la...
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The detergent molecules we have discussed so far
are classifie...
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Many cationic
detergents
also have
germicidal
properties and
a...
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  • Chapter12new

    1. 1. 1 Chapter 12 Solutions
    2. 2. Copyright © Cengage Learning. All rights reserved. 12 | 2 Solution Formation 1. Types of Solutions 2. Solubility and the Solution Process 3. Effects of Temperature and Pressure on Solubility Contents and Concepts
    3. 3. Copyright © Cengage Learning. All rights reserved. 12 | 3 Colligative Properties 4. Ways of Expressing Concentration 5. Vapor Pressure of a Solution 6. Boiling-Point Elevation and Freezing-Point Depression 7. Osmosis 8. Colligative Properties of Ionic Solutions Colloid Formation 9. Colloids
    4. 4. Copyright © Cengage Learning. All rights reserved. 12 | 4 Solution Formation 1. Types of Solutions a. Define solute and solvent. b. Define miscible fluid. c. Provide examples of gaseous solutions, liquid solutions, and solid solutions. Learning Objectives
    5. 5. Copyright © Cengage Learning. All rights reserved. 12 | 5 2. Solubility and the Solution Process a. List the conditions that must be present to have a saturated solution, to have an unsaturated solution, and to have a supersaturated solution. b. Describe the factors that make one substance soluble in another. c. Determine when a molecular solution will form when substances are mixed. d. Learn which conditions must be met to create an ionic solution.
    6. 6. Copyright © Cengage Learning. All rights reserved. 12 | 6 3. Effects of Temperature and Pressure on Solubility a. State the general trends for the solubility of gases and solids with temperature. b. Explain how the solubility of a gas changes with temperature. c. Apply Henry’s law.
    7. 7. Copyright © Cengage Learning. All rights reserved. 12 | 7 Colligative Properties 4. Ways of Expressing Concentration a. Define colligative property. b. Define molarity. c. Define mass percentage of solute. d. Perform calculations with the mass percentage of a solute. e. Define molality. f. Calculate the molality of a solute. g. Define mole fraction. h. Calculate the mole fractions of components.
    8. 8. Copyright © Cengage Learning. All rights reserved. 12 | 8 i. Convert molality to mole fraction. j. Convert mole fraction to molality. k. Convert molality to molarity. l. Convert molarity to molality. 5. Vapor Pressure of a Solution a. Explain vapor-pressure lowering of a solvent. b. State Raoult’s law. c. Calculate vapor-pressure lowering. d. Describe an ideal solution.
    9. 9. Copyright © Cengage Learning. All rights reserved. 12 | 9 6. Boiling-Point Elevation and Freezing-Point Depression a. Define boiling-point elevation and freezing- point depression. b. Calculate boiling-point elevation and freezing-point depression. c. Calculate the molecular mass of a solute from molality. d. Calculate the molecular mass from freezing-point depression.
    10. 10. Copyright © Cengage Learning. All rights reserved. 12 | 10 7. Osmosis a. Describe a system where osmosis will take place. b. Calculate osmotic pressure. 8. Colligative Properties of Ionic Solutions a. Determine the colligative properties of ionic solutions.
    11. 11. Copyright © Cengage Learning. All rights reserved. 12 | 11 Colloid Formation 9. Colloids a. Define colloid. b. Explain the Tyndall effect. c. Give examples of hydrophilic colloids and hydrophobic colloids. e. Describe coagulation. f. Explain how micelles can form an association colloid.
    12. 12. Copyright © Cengage Learning. All rights reserved. 12 | 12 A solution is composed of two parts: the solute and the solvent. Solute The gas (or solid) in a solution of gases (or solids), or the component present in the smaller amount. Solvent The liquid in the case of a solution of gases or solids, or the component present in the larger amount.
    13. 13. Copyright © Cengage Learning. All rights reserved. 12 | 13 Fluids that mix with or dissolve in each other in all proportions are said to be miscible (left). Fluids that do not dissolve in each other are said to be immiscible (right).
    14. 14. Copyright © Cengage Learning. All rights reserved. 12 | 14 a.The 5 g of Mo is the solute; the 80 g of Cr is the solvent. b.MgCl2 is the solute; H2O is the solvent. c. O2 and N2 are the solutes; Ar is the solvent. Concept Check 12.1 Identify the solute(s) and solvent(s) in the following solutions. a. 80 g of Cr and 5 g of Mo b. 5 g of MgCl2 dissolved in 1000 g of H2O c. 39% N2, 41% Ar, and the rest O2
    15. 15. Copyright © Cengage Learning. All rights reserved. 12 | 15 A saturated solution is in equilibrium with respect to the amount of dissolved solute. The rate at which the solute leaves the solid state equals the rate at which the solute returns to the solid state.
    16. 16. Copyright © Cengage Learning. All rights reserved. 12 | 16 The solubility of a solute is the amount that dissolves in a given quantity of solvent at a given temperature. An unsaturated solution is a solution not in equilibrium with respect to a given dissolved substance and in which more of the substance can be dissolved.
    17. 17. Copyright © Cengage Learning. All rights reserved. 12 | 17
    18. 18. Copyright © Cengage Learning. All rights reserved. 12 | 18 A supersaturated solution is a solution that contains more dissolved substance than a saturated solution does. This occurs when a solution is prepared at a higher temperature and is then slowly cooled. This is a very unstable situation, so any disturbance causes precipitation.
    19. 19. Copyright © Cengage Learning. All rights reserved. 12 | 19 Solubility can be understood in terms of two factors: 1. The natural tendency toward disorder favors dissolving. 2. The relative forces between and within species must be considered. Stronger forces within solute species oppose dissolving. Stronger forces between species favor dissolving.
    20. 20. Copyright © Cengage Learning. All rights reserved. 12 | 20 For molecular solutions, this can be summarized as “Like dissolves like.” In other words, solutes dissolve in solvents that have the same type of intermolecular forces. An immiscible solute and solvent are illustrated at right.
    21. 21. Copyright © Cengage Learning. All rights reserved. 12 | 21 Concept Check 12.2 You have two molecular compounds, X and Y. Compound X has stronger intermolecular forces than compound Y. X has no dipole-dipole bonding of any kind, and Y exhibits hydrogen bonding. Predict the relative solubility of compounds X and Y in water and in a nonpolar solvent. Explain your reasoning.
    22. 22. Copyright © Cengage Learning. All rights reserved. 12 | 22 When considering ionic solutes in water, we need to examine the hydration energy and the lattice energy.
    23. 23. Copyright © Cengage Learning. All rights reserved. 12 | 23 The stronger ion-dipole force between the ion and the solvent—that is, hydration energy— favors dissolving. A stronger force between ions—that is, lattice energy—opposes dissolving.
    24. 24. Copyright © Cengage Learning. All rights reserved. 12 | 24 The force of attraction between water and both a cation and an anion is illustrated to the left with lithium fluoride, LiF.
    25. 25. Copyright © Cengage Learning. All rights reserved. 12 | 25 The process of dissolving occurs at the surfaces of the solid. Here we see water hydrating (dissolving) ions.
    26. 26. Copyright © Cengage Learning. All rights reserved. 12 | 26 The hydration energy for AB2 must be greater than the hydration energy for CB2. Concept Check 12.3 The hypothetical ionic compound AB2 is very soluble in water. Another hypothetical ionic compound, CB2, is only slightly soluble in water. The lattice energies for these compounds are about the same. Provide an explanation for the solubility difference between these compounds.
    27. 27. Copyright © Cengage Learning. All rights reserved. 12 | 27 In general, solubility depends on temperature.
    28. 28. Copyright © Cengage Learning. All rights reserved. 12 | 28 In most cases, solubility increases with increasing temperature. However, for a number of compounds, solubility decreases with increasing temperature. The difference is explained by differences in the heat of solution.
    29. 29. Copyright © Cengage Learning. All rights reserved. 12 | 29 When dissolving absorbs heat (is endothermic), the temperature of the solution decreases as the solute dissolves. The solubility will increase as temperature increases. When dissolving releases heat (is exothermic), the temperature of the solution increases as the solute dissolves. The solubility will decrease as temperature increases.
    30. 30. Copyright © Cengage Learning. All rights reserved. 12 | 30 Hot packs use an exothermic solution process. Cold packs use an endothermic solution process.
    31. 31. Copyright © Cengage Learning. All rights reserved. 12 | 31 Henry’s law describes the effect of pressure on gas solubility: The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the solution. This is expressed mathematically in the equation S = kHP where S = gas solubility kH = Henry’s law constant for the gas P = partial pressure of the gas over the solution
    32. 32. Copyright © Cengage Learning. All rights reserved. 12 | 32 In general, pressure has little or no effect on the solubility of solids or liquids in water. The solubility of a gas increases as pressure increases, as illustrated at right.
    33. 33. ? Copyright © Cengage Learning. All rights reserved. 12 | 33 Helium–oxygen mixtures are sometimes used as the breathing gas in deep-sea diving. At sea level (where the pressure is 1.0 atm), the solubility of pure helium in blood is 0.94 g/mL. What is the solubility of pure helium at a depth of 1500 feet? Pressure increases by 1.0 atm for every 33 feet of depth, so at 1500 feet the pressure is 46 atm. (For a helium–oxygen mixture, the solubility of helium will depend on its initial partial pressure, which will be less than 1.0 atm.)
    34. 34. Copyright © Cengage Learning. All rights reserved. 12 | 34 P1 = 1.0 atm P2 = 46 atm S1 = 0.94 g/mL S2 = ? 1 2 12 P P SS ×= atm1.0 atm46 mL g0.94 2 ×=S mL g43 2 =S
    35. 35. Copyright © Cengage Learning. All rights reserved. 12 | 35 At high elevations, the partial pressure of oxygen decreases, decreasing the solubility of oxygen in water. Fish require a certain minimum level of dissolved oxygen to survive. Concept Check 12.4 Most fish have a very difficult time surviving at elevations much above 3500 m. Howcould Henry’s law be used to account for this fact?
    36. 36. Copyright © Cengage Learning. All rights reserved. 12 | 36 The concentration of a solute can be quantitatively expressed in several ways: 1. Molarity 2. Mass percentage of solute 3. Molality 4. Mole fraction
    37. 37. Copyright © Cengage Learning. All rights reserved. 12 | 37 Molarity is the moles of solute per liter of solution. It is abbreviated as M. solutionofliters soluteofmoles =M
    38. 38. Copyright © Cengage Learning. All rights reserved. 12 | 38 Mass percentage of solute is the percentage by mass of solute in a solution. 100% solutionofgrams soluteofgrams soluteofpercentageMass × =
    39. 39. ? Copyright © Cengage Learning. All rights reserved. 12 | 39 An experiment calls for 36.0 g of a 5.00% aqueous solution of potassium bromide. Describe how you would make up such a solution.
    40. 40. Copyright © Cengage Learning. All rights reserved. 12 | 40 A 5.00% aqueous solution of KBr has 5.00 g KBr per 100. g solution. The remainder of the 100. g is water: 95 g. We can use this ratio to determine the mass of KBr in 36.0 g solution: Since 1.80 g KBr is required for 36.0 g of solution, the remainder consists of 34.2 g water. We make the solution by mixing 1.8 g KBr in 34.2 g water. KBrg1.80 solution100.g KBrg5.00 solutiong36.0 =×
    41. 41. Copyright © Cengage Learning. All rights reserved. 12 | 41 Molality is the moles of solute per kilogram of solvent. It is abbreviated as m. solventofkilograms soluteofmoles =m
    42. 42. ? Copyright © Cengage Learning. All rights reserved. 12 | 42 Iodine dissolves in a variety of organic solvents. For example, in methylene chloride, it forms an orange solution. What is the molality of a solution of 5.00 g iodine, I2, in 30.0 g of methylene chloride, CH2Cl2?
    43. 43. Copyright © Cengage Learning. All rights reserved. 12 | 43 Mass of solute = 5.00 g I2 Mass of solvent = 30.0 g CH2Cl2 kg1 g10 Ig253.8 Imol1 solventg30.0 Ig5.00 3 2 22 ××=m kg mol 0.657=m
    44. 44. Copyright © Cengage Learning. All rights reserved. 12 | 44 Mole fraction is the moles of component over the total moles of solution. It is abbreviated Χ. solutionofmolestotal soluteofmoles =Χ
    45. 45. ? Copyright © Cengage Learning. All rights reserved. 12 | 45 A solution of iodine, I2, in methylene chloride, CH2Cl2, contains 5.00 g I2 and 56.0 g CH2Cl2. What is the mole fraction of each component in this solution?
    46. 46. Copyright © Cengage Learning. All rights reserved. 12 | 46 Mass of solute = 5.00 g I2 Mass of solvent = 56.0 g CH2Cl2 mol0.01970 Ig253.8 Imol1 Ig5.00 soluteMoles 2 2 2 =× = mol0.6594 ClCHg84.93 ClCHmol1 Ig56.0 solventMoles 22 22 2 =× = mol0.6791 mol0.6594mol0.01970molesTotal = +=
    47. 47. Copyright © Cengage Learning. All rights reserved. 12 | 47 0.02902I =Χ totalmol0.6791 Imol0.01970 2 I2 =Χ 0.97122ClCH =Χ 2 2 2 2 CH Cl 0.6594 mol CH Cl 0.6791mol total Χ =
    48. 48. ? Copyright © Cengage Learning. All rights reserved. 12 | 48 A bottle of bourbon is labeled 94 proof, meaning that it is 47% by volume of alcohol in water. What is the mole fraction of ethyl alcohol, C2H5OH, in the bourbon? The density of ethyl alcohol is 0.80 g/mL. One liter of bourbon contains 470 mL of alcohol and 530 mL of water. To solve this problem, we will convert the volume of ethyl alcohol to mass using density, and then convert to moles using molar mass.
    49. 49. Copyright © Cengage Learning. All rights reserved. 12 | 49 mol8.16 g46.08 mol1 mL1 g0.80 OHHCmL470 52 =×× 0.22ethanol =Χ mol29.4 g18.02 mol1 mL1 g1.00 OHmL530 2 =×× mol37.6mol29.4mol8.16molesTotal =+= totalmol37.6 OHHCmol8.16 52 ethanol =Χ
    50. 50. ? Copyright © Cengage Learning. All rights reserved. 12 | 50 A 3.6 m solution of calcium chloride, CaCl2, is used in tractor tires to give them weight. The addition of CaCl2 also prevents water in the tires from freezing at temperatures above –20°C. What are the mole fractions of CaCl2 and water in such a solution? The 3.6 m solution contains 3.6 mol CaCl2 in 1.0 kg of water. To solve this problem, we will convert the mass of water to moles, and then compute the mole fractions.
    51. 51. Copyright © Cengage Learning. All rights reserved. 12 | 51 Moles solute = 3.6 mol CaCl2 0.0612CaCl =Χ OHmol55.5 g18.02 mol1 kg1 g10 OHkg1.0 solventMoles 2 3 2 =×× = mol59.1mol55.5mol3.6molesTotal =+= totalmol59.1 mol3.6 2CaCl =Χ 0.94OH2 =Χ totalmol59.1 mol55.5 2H =OΧ
    52. 52. Copyright © Cengage Learning. All rights reserved. 12 | 52 Converting between molality and mole fraction is relatively simple, because you know the masses or moles of both the solute and the solvent.
    53. 53. ? Copyright © Cengage Learning. All rights reserved. 12 | 53 A solution contains 8.89 × 10-3 mole fraction of I2 dissolved in 0.9911 mole fraction of CH2Cl2 (methylene chloride). What is the molality of I2 in this solution? We will assume we have 1 mole of the solution, so we begin with 8.89 × 10-3 mol I2 and 0.9911 mol CH2Cl2. Next, we will convert the moles of solvent into grams, and then into kilograms. Finally, we will compute the molality of the solution.
    54. 54. Copyright © Cengage Learning. All rights reserved. 12 | 54 Moles solute = 0.00889 mol I2 kg mol 0.106=m 22 3 ClCHkg0.08417 g10 kg1 mol1 g84.93 mol0.9911solventKilograms = ××= kg0.08417 mol0.00889 =m
    55. 55. Copyright © Cengage Learning. All rights reserved. 12 | 55 Converting between molality and molarity requires knowing the density of the solution. This enables you to calculate the mass or volume of the solution. You can then distinguish the amount of solute from the amount of solvent, or combine them to find the volume of solution. These types of conversions are illustrated in the following problems.
    56. 56. ? Copyright © Cengage Learning. All rights reserved. 12 | 56 Citric acid, HC6H7O7, is often used in fruit beverages to add tartness. An aqueous solution of citric acid is 2.331 m HC6H7O7. What is the molarity of the solution? The density of the solution is 1.1346 g/mL. A 2.331 m solution contains 2.331 mol solute in 1.000 kg solvent. We will use this relationship first to convert the moles of citric acid to grams and then to find the mass of solution. Using density, we can then find the volume of solution. Finally, we will compute the molarity of the solution.
    57. 57. Copyright © Cengage Learning. All rights reserved. 12 | 57 Moles solute = 2.331 mol HC6H7O7 L mol 1.827= g447.88 mol1 g192.14 mol2.331soluteMass =×= L1.2761 mL L10 g1.1346 mL1 g1447.88solutionLiters 3 = ××= − g1447.88g1000.00g447.88solutionofMass =+= L1.2761 mol2.331 =M
    58. 58. ? Copyright © Cengage Learning. All rights reserved. 12 | 58 An aqueous solution of ethanol is 14.1 M C2H5OH. The density of the solution is 0.853 g/cm3 . What is the molality of ethanol in the solution? We will work with 1.00 L of solution. First, we will convert volume to mass using the density. Then, we will find the masses of the solute and the solvent. Finally, we will compute the molality.
    59. 59. Copyright © Cengage Learning. All rights reserved. 12 | 59 g649.7 mol1 g46.08 mol14.1soluteofMass =×= g853 mL g0.853 L10 mL1 L1.000solutionofMass 3- = ××= kg mol 69.4= kg0.2033 g10 kg1 g203.3 g649.7g853solventofMass 3 =×= −= kg0.2033 mol14.1 =m
    60. 60. Copyright © Cengage Learning. All rights reserved. 12 | 60 Colligative properties of solutions are properties that depend on the concentration of the solute molecules or ions in solution but not on the chemical identity of the solute. 1. Vapor-pressure lowering 2. Boiling-point elevation 3. Freezing-point lowering 4. Osmotic pressure
    61. 61. Copyright © Cengage Learning. All rights reserved. 12 | 61 The vapor pressure of a solution, P, is less than the vapor pressure of the pure solvent, P°. When the solute is nonvolatile, the vapor pressure of a solution is the mole fraction of the solvent times the vapor pressure of pure solvent. o solventsolventsolution PXP = o PP solvent<solution
    62. 62. Copyright © Cengage Learning. All rights reserved. 12 | 62 We can explain this relationship by recalling that evaporation occurs at the surface of the solution. When a solute is present, less of the surface is occupied by solvent molecules.
    63. 63. Copyright © Cengage Learning. All rights reserved. 12 | 63 To establish an equilibrium, the gaseous solvent will condense in the more concentrated solution until the vapor pressures and concentrations are equal.
    64. 64. Copyright © Cengage Learning. All rights reserved. 12 | 64 To find the vapor-pressure lowering (a colligative property), we rearrange the following equation: solution o solvent PPP −=∆ ( ) o solventsolute solventsolute 1-Since PXP =∆ = XX o solventsolvent o solvent PPP X−=∆ ( )solvent o solvent 1- XPP =∆
    65. 65. Copyright © Cengage Learning. All rights reserved. 12 | 65 Note that vapor-pressure lowering is directly proportional to the solute concentration, the definition of a colligative property. o solventsolutesolvent PXP =∆
    66. 66. ? Copyright © Cengage Learning. All rights reserved. 12 | 66 Eugenol, C10H12O2, is the chief constituent of oil of clove. This pale yellow liquid dissolves in ethanol, C2H5OH; it has a boiling point of 255°C. As a result, we know eugenol’s vapor pressure is very low; it can be considered nonvolatile. What is the vapor-pressure lowering at 20.0°C of a solution containing 8.56 g of eugenol in 50.0 g of ethanol? The vapor pressure of ethanol at 20.0°C is 44.6 mmHg.
    67. 67. Copyright © Cengage Learning. All rights reserved. 12 | 67 mol0.0521 g164.22 mol1 g8.56soluteMoles =×= mmHg2.04Δ =P mol1.085 g46.08 mol1 g50.0solventMoles =×= mmHg44.60.04581Δ Δ 0 solventsolvent ×= = P PΧP mol1.137 mol1.085mol0.0521molesTotal = += 0.04581 mol1.137 mol0.0521 solute ==Χ
    68. 68. Copyright © Cengage Learning. All rights reserved. 12 | 68 When the solute is nonvolatile, it has no appreciable vapor pressure itself and forms an ideal solution. When the solute is volatile, a nonideal solution results.
    69. 69. Copyright © Cengage Learning. All rights reserved. 12 | 69 The phase diagram on the next slide shows the changes in freezing and boiling points when a nonvolatile solute is added to a solvent. The blue line shows the pure solvent; the purple line shows the solution.
    70. 70. Copyright © Cengage Learning. All rights reserved. 12 | 70
    71. 71. Copyright © Cengage Learning. All rights reserved. 12 | 71 The boiling point of a solution is higher than the boiling point of pure solvent. The boiling-point elevation, ∆Tb, is given by the following equation: ∆Tb = mKb
    72. 72. Copyright © Cengage Learning. All rights reserved. 12 | 72 The freezing point of a solution is lower than the freezing point of pure solvent. The freezing-point depression, ∆Tf, is given by the following equation: ∆Tf = mKf
    73. 73. ? Copyright © Cengage Learning. All rights reserved. 12 | 73 A solution is made up of eugenol, C10H12O2, in diethyl ether (“ether”). If the solution is 0.575 m eugenol in ether, what are the freezing point and the boiling point of the solution? The freezing point and the boiling point of pure ether are –116.3°C and 34.6°C, respectively. The freezing-point depression and boiling-point elevation constants are 1.79°C/m and 2.02°C/m, respectively.
    74. 74. Copyright © Cengage Learning. All rights reserved. 12 | 74 m = 0.575 m Kf = 1.79°C/m Kb = 2.02°C/m Tf° = –116.3°C Tb° = 34.6°C ∆Tf = m × Kf ∆Tb = m × Kb ∆Tf = 0.575 m × 1.79°C/m ∆Tf = 1.03°C This is the freezing-point depression. Tf = –116.3 – 1.03 = –117.3°C ∆Tb = 0.575 m × 2.02°C/m ∆Tb = 1.16°C This is the boiling-point elevation. Tb = 34.6 + 1.16 = 35.8°C
    75. 75. ? Copyright © Cengage Learning. All rights reserved. 12 | 75 In a freezing-point depression experiment, the molality of a solution of 58.1 mg anethole in 5.00 g benzene was determined to be 0.0784 m. What is the molar mass of anethole? First, we will use the freezing-point depression data to find the molality of the solution. Next, we will use the molality and mass of solvent to find the moles of solute. Finally, we will use the moles of solute and mass of solute to find the molar mass.
    76. 76. Copyright © Cengage Learning. All rights reserved. 12 | 76 Solute mass = 58.1 mg Solvent mass = 5.00 g m = 0.0784 mol/kg mol103.92 kg0.00500m0.0784 solventkgsolutemol 4- ×= ×= ×= m solventkg solutemol =m g/mol.148ismassmolarThe g/mol148 mol103.92 g1058.1 massMolar 4- -3 = × × =
    77. 77. ? Copyright © Cengage Learning. All rights reserved. 12 | 77 An 11.2-g sample of sulfur was dissolved in 40.0 g of carbon disulfide. The boiling-point elevation of carbon disulfide was found to be 2.63°C. What is the molar mass of the sulfur in the solution? What is the formula of molecular sulfur? The boiling-point elevation constant, Kb, for carbon disulfide is 2.40°C/m.
    78. 78. Copyright © Cengage Learning. All rights reserved. 12 | 78 Solute mass = 11.2 g Solvent mass = 40.0 g ∆Tb = 2.63°C Kb = 2.40°C/m First, we will use the definition of the colligative property to calculate the molality. Next, we will use the definition of molality to calculate the moles of solute. Finally, using the mass and moles of solute, we will find the molar mass and the molecular formula of the sulfur.
    79. 79. Copyright © Cengage Learning. All rights reserved. 12 | 79 m m K T m 1.096 C 2.40 C2.63Δ f b = ° ° == fbΔ KmT ×= mol0.04383 kg0.04001.096 solventkgsolutemol = ×= ×= m m solventkg solutemol =m
    80. 80. Copyright © Cengage Learning. All rights reserved. 12 | 80 .8SformulamolecularThe g/mol255.5 mol0.04383 g11.2 massMolar == g/mol.32.065ismassformulaThe S.issulfurforformulaempiricalThe 8 32.065 255.5 ≈=n
    81. 81. Copyright © Cengage Learning. All rights reserved. 12 | 81 Osmosis is the phenomenon of solvent flow through a semipermeable membrane to equalize the solute concentration on both sides of the membrane. A semipermeable membrane allows solvent molecules to pass through but not solute molecules.
    82. 82. Copyright © Cengage Learning. All rights reserved. 12 | 82 Water is the solvent. Water will flow from the left to the right.
    83. 83. Copyright © Cengage Learning. All rights reserved. 12 | 83 Water is the solvent. It flows from the beaker into the thistle tube.
    84. 84. Copyright © Cengage Learning. All rights reserved. 12 | 84
    85. 85. Copyright © Cengage Learning. All rights reserved. 12 | 85 Osmotic pressure, π, is equal to the pressure that, when applied, just stops osmosis. Osmotic pressure is a colligative property of a solution. π = MRT
    86. 86. ? Copyright © Cengage Learning. All rights reserved. 12 | 86 Dextran, a polymer of glucose units, is produced by bacteria growing in sucrose solutions. Solutions of dextran in water have been used as a blood plasma substitute. At 21°C, what is the osmotic pressure (in mmHg) of a solution containing 1.50 g of dextran dissolved in 100.0 mL of aqueous solution, if the average molecular mass of the dextran is 4.0 × 108 amu?
    87. 87. Copyright © Cengage Learning. All rights reserved. 12 | 87 T = 21°C + 273 = 294 K Mass solute = 1.50 g Liters solution = 100.0 × 10-3 L Molecular mass = 4.0 × 108 amu Molar mass = 4.0 × 108 g/mol mol103.75 g104.0 mol1 g1.50soluteMoles 8 9− ×= × ×= mol/L103.75 L10100.0 mol103.75 Molarity 8- 3- -9 ×= × × =
    88. 88. Copyright © Cengage Learning. All rights reserved. 12 | 88 8 mol L atm 3.75 10 0.08206 294 K L mol K π − • = × × × • MRTπ = 4 6.9 10 mmHgπ − = × atm1 mmHg760 atm109.05 7 ××= −
    89. 89. Copyright © Cengage Learning. All rights reserved. 12 | 89 One can show that 760.0 mmHg is equivalent to the pressure exerted by a column of water 10.334 m high. Thus each 1.00 mmHg of pressure is equivalent to the pressure of a 1.36-cm column of water. If the density of this dextran solution is equal to that of water, what height of solution would exert a pressure equal to its osmotic pressure? 4 9.4 10 cmπ − = × 4 1.36 cm 6.9 10 mmHg 1.00 mmHg π − = × ×
    90. 90. Copyright © Cengage Learning. All rights reserved. 12 | 90
    91. 91. Copyright © Cengage Learning. All rights reserved. 12 | 91 Ionic solutes dissolve to form more than one particle per formula unit. We alter the colligative property equations to account for this fact by including i, the number of ions per formula unit: ∆P = iP°AXB ∆Tb = iKbm ∆Tf = iKfm π = iMRT
    92. 92. ? Copyright © Cengage Learning. All rights reserved. 12 | 92 What is the osmotic pressure at 25.0°C of an isotonic saline solution (a solution having an osmotic pressure equal to that of blood) that contains 0.900 g NaCl in 100.0 mL of aqueous solution? Assume that i has the ideal value (based on the formula).
    93. 93. Copyright © Cengage Learning. All rights reserved. 12 | 93 T = 25°C + 273 = 298 K Mass solute = 0.900 g NaCl Volume solution = 100.0 mL = 100.0 × 10−3 L Molar mass solute = 58.44 g/mol i = 2 mol0.01540 g58.44 mol1 g0.900soluteMoles =×= 2 3 1.540 10 mol Molarity 0.1540 mol/L 100.0 10 L − − × = = ×
    94. 94. Copyright © Cengage Learning. All rights reserved. 12 | 94 K298 Kmol atmL 0.08206 L mol 0.15402 × • • ××=π iMRTπ = atm7.53=π
    95. 95. Copyright © Cengage Learning. All rights reserved. 12 | 95 A colloid is a dispersion of particles of one substance (the dispersed phase) throughout another substance or solution (the continuous phase). The dispersed particles range from 1000 pm to 200,000 pm in size—much larger than single molecules or single ions. Fog is an example of a colloid. In fog, water droplets are dispersed through air.
    96. 96. Copyright © Cengage Learning. All rights reserved. 12 | 96 Colloids exhibit the Tyndall effect. The path of the light is visible through a colloid because the light is reflected by the relatively larger-sized particles in the dispersed phase.
    97. 97. Copyright © Cengage Learning. All rights reserved. 12 | 97 Colloid Solution
    98. 98. Copyright © Cengage Learning. All rights reserved. 12 | 98 Colloids are characterized according to the state (solid, liquid, or gas) of the dispersed phase and the state of the continuous phase. • Fog and smoke are aerosols, which are liquid droplets or solid particles dispersed throughout a gas. • An emulsion consists of liquid droplets dispersed throughout another liquid (for example, particles of butterfat dispersed through homogenized milk). • A sol consists of solid particles dispersed in a liquid.
    99. 99. Copyright © Cengage Learning. All rights reserved. 12 | 99 Colloids in which the continuous phase is water are categorized into two major classes: hydrophilic colloids and hydrophobic colloids. Hydrophilic colloid A colloid in which there is a strong attraction between the dispersed phase and the continuous phase (water). Hydrophobic colloid A colloid in which there is a lack of attraction between the dispersed phase and the continuous phase (water).
    100. 100. Copyright © Cengage Learning. All rights reserved. 12 | 100 Coagulation is the process by which the dispersed phase of a colloid is made to aggregate and thereby separate from the continuous phase. It is analogous to precipitation from a solution. Curdled milk is an example of coagulation.
    101. 101. Copyright © Cengage Learning. All rights reserved. 12 | 101 When molecules or ions that have both a hydrophobic end and a hydrophilic end are dispersed in water, they associate, or aggregate, to form colloidal-sized particles called micelles. A colloid in which the dispersed phase consists of micelles is called an association colloid.
    102. 102. Copyright © Cengage Learning. All rights reserved. 12 | 102 Soap consists of compounds such as sodium stearate. Sodium stearate is an example of a molecule with hydrophobic and hydrophilic ends. Hydrophobic end Hydrophilic end
    103. 103. Copyright © Cengage Learning. All rights reserved. 12 | 103 In water solution, the stearate ions associate to form micelles in which the hydrocarbon ends point inward toward one another and away from the water, and ionic carboxyl groups are on the outside of the micelle facing the water. The cleansing action of soap occurs because oil and grease can be absorbed into the hydrophobic centers of soap micelles and washed away.
    104. 104. Copyright © Cengage Learning. All rights reserved. 12 | 104
    105. 105. Copyright © Cengage Learning. All rights reserved. 12 | 105
    106. 106. Copyright © Cengage Learning. All rights reserved. 12 | 106 Far left: Vegetable oil floating on water (dyed green). Left: When the mixture is shaken with soap, an emulsion forms as the oil droplets are absorbed into soap micelles.
    107. 107. Copyright © Cengage Learning. All rights reserved. 12 | 107 Synthetic detergents also form association colloids. Sodium lauryl sulfate is a synthetic detergent present in toothpastes and shampoos.
    108. 108. Copyright © Cengage Learning. All rights reserved. 12 | 108 The detergent molecules we have discussed so far are classified in the trade as anionics, because they have a negative charge at the hydrophilic end. Other detergent molecules are classified as cationics, because they have a positive charge at the hydrophilic end.
    109. 109. Copyright © Cengage Learning. All rights reserved. 12 | 109
    110. 110. Copyright © Cengage Learning. All rights reserved. 12 | 110 Many cationic detergents also have germicidal properties and are used in hospital disinfectants and in mouthwashes.

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