Chapter 5
Mixtures at the Molecular
Level: Properties of Solutions
Chemistry: The Molecular Nature
of Matter, 6E
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Chapter 5: Solutions
Solution
 Homogeneous mixture
 Composed of solvent and solute(s)
Solvent
 More abundant component of mixture
Solute(s)
 Less abundant or other component(s) of mixture
Ex. Lactated Ringer’s solution
 NaCl, KCl, CaCl2, NaC3H5O3 in water
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Rule of Thumb
 “Like dissolves Like”
 Use polar solvent for polar solute
 Use Nonpolar solvent for nonpolar
solute
 When strengths of intermolecular
attractions are similar in solute and
solvent, solutions form because net
energy exchange is about the same
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Chemistry: The Molecular Nature of Matter, 6E
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Process of Dissolution
 Polar solutes interact with
and dissolve in polar
solvents
 H-bonding solutes interact
with and dissolve in Hbonding solvents
Ex. Ethanol in water
H3C
CH 2
O
H



O
H

H

 Both are polar molecules
 Both form hydrogen bonds
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Chemistry: The Molecular Nature of Matter, 6E
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Ex. of Miscible Solution
 When ethanol dissolves in water,
 Solvent and solute are “similar” (IMF strength)
 Solution will form
H
H
H
H3C
C
H2
O
O
O
H
H
H
O
H2C
O
H
CH3
H
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Chemistry: The Molecular Nature of Matter, 6E
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Ex. Benzene in CCl4
 CCl4
 Nonpolar
 Benzene, C6H6
 Nonpolar
 Similar in strength to CCl4
 Does dissolve, solution forms
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Chemistry: The Molecular Nature of Matter, 6E
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Ex. of Immiscible Solution
Benzene in water
 Solvent and solute are very “different”
 No solution forms
 2 layers, Don’t Mix
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Chemistry: The Molecular Nature of Matter, 6E
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Learning Check
Which of the following are miscible in water?
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Chemistry: The Molecular Nature of Matter, 6E
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Your Turn!
Which of the following molecules is soluble in
C6H6?
A. NH3
B. CH3NH2
C. CH3OH
D. CH3CH3
E. CH3Cl
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Solutions of Solids in Liquids
 Basic principles remain the same when
solutes are solids
 Sodium chloride (NaCl)
 Ionic bonding
 Strong intermolecular forces
 Ions dissolve in water because ion-dipole forces
of water with ions strong enough to overcome
ion-ion attractions
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Hydration of Solid Solute
 At edges, fewer
oppositely charged ions
around
 H2O can come in
 Ion-dipole forces
 Remove ion
 New ion at surface
 Process continues until
all ions in solution
 Hydration of ions
 Completely surrounded
by solvent
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Hydration vs. Solvation
 Hydration
 Ions surrounded by water molecules
 Solvation
 General term for surrounding solute particle by
solvent molecules
 Do polar molecules dissolve in H2O?
 Yes
 Attractions between solvent and solute dipoles
(dipole-dipole interactions) dislodge molecules
from solid
 Bring into solution
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Polar Molecule in Water
 H2O reorients so
 Positive Hs are near negative ends of solute
 Negative Os are near positive ends of solute
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Solvation in Nonpolar Solvents?
 Wax and benzene
 Weak London dispersion forces in both
 Wax molecules
 Easily slip from solid
 Slide between benzene molecules
 Form new London forces between solvent and
solute
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Solubility
 Mass of solute that forms saturated solution with
given mass of solvent at specified temperature
g solute
solubility 
100g solvent
 If extra solute added to saturated solution, extra
solute will remain as separate phase

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Effect of T on Solubility of Solids and
Liquids in Liquid Solvent
 If heat is absorbed when solute dissolves,
solubility  when T 
soluteundissolve d  energy


solutedissolved
 If energy is released when solute dissolves,
solubility  when T 
soluteundissolve d


solutedissolved  energy
 When apply “stress” to equilibrium (by  T),
equilibrium will shift so as to relieve (absorbs or
minimizes) stress
 LeChâtelier’s Principle
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Solubility of Most Substances
Increases with Temperature
 Most substances
become more
soluble as T 
 Amount solubility 
 Varies considerably
 Depends on
substance
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Effect of T on Gas Solubility in Liquids
 Solubility of gases usually  as T 
Table 13.2 Solubilities of Common Gases in Water
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Effect of Pressure on Gas Solubility
 Solubility  as P 
Why?
  P means  V above
solution for gas
 Gas goes into
solution
 Relieves stress on
system
 Conversely, solubility
 as P 
 Soda in can
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Chemistry: The Molecular Nature of Matter, 6E
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Effect of Pressure on Gas Solubility
A. At some P, equilibrium exists between vapor phase and
solution
 ratein = rateout
B.  in P puts stress on equilibrium
  frequency of collisions so ratein > rateout
 More gas molecules dissolve than are leaving solution
C. More gas dissolved
 Rateout will  until Rateout = Ratein and equilibrium
restored
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Henry’s Law
 Pressure-Solubility Law
 “Concentration of gas in liquid at any given
temperature is directly proportional to partial
pressure of gas over solution”
Cgas = kHPgas (T is constant)
Cgas = concentration of gas
Pgas = partial pressure of gas
kH = Henry's Law constant
 Unique to each gas
 Tabulated
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Henry’s Law
 True only at low concentrations and
pressures where gases do NOT react with
solvent
 Alternate form
C1 C2

P1
P2
 C1 and P1 refer to an initial set of conditions
 C2 and P2 refer to a final set of conditions
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Ex. 1 Using Henry’s Law
Calculate the concentration of CO2 in a soft drink that
is bottled with a partial pressure of CO2 of 5 atm over
the liquid at 25 °C. The Henry’s Law constant for CO2
in water at this temperature is 3.12  102 mol/L·atm.
C CO2  k H (CO2 )PCO2
= 3.12  102 mol/L·atm * 5.0 atm
= 0.156 mol/L  0.16 mol/L
When under 5.0 atm pressure
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Chemistry: The Molecular Nature of Matter, 6E
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Ex. 1 Using Henry’s Law
Calculate the concentration of CO2 in a soft drink
after the bottle is opened and equilibrates at 25 °C
under a partial pressure of CO2 of 4.0  104 ·atm.
C1 C 2

P1
P2
C2
P2 C1
C2 
P1

0.156 mol/L4.0  10 4 atm 

5.0atm
C2 = 1.2  104 · mol/L
When open to air
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Chemistry: The Molecular Nature of Matter, 6E
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Learning Check
What is the concentration of dissolved nitrogen
in a solution that is saturated in N2 at 2.0 atm?
kH= 8.42×107 (M / atm)
• Cg=kHPg
• Cg= 8.42×107 (M / atm) × 2.0 atm
• Cg=1.7 ×10 6 M
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Your Turn!
 How many grams of oxygen gas at 1.0 atm
will dissolve in 10.0 L of water at 25 oC if
Henry’s constant is 1.3 x 10-3 M atm-1 at this
temperature ?
A. 0.42 g
B. 0.013 g
C. 0.042 g
D. 0.21 g
E. 2.4 g
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Solubility of Polar vs. Nonpolar Gases
 Gas molecules with polar bonds are much more
soluble in water than nonpolar molecules like oxygen
and nitrogen
 CO2, SO2, NH3 >> O2, N2, Ar
 Form H-bonds with H2O
 Some gases have increased solubility because they
react with H2O to some extent
Ex. CO2(aq) + H2O
H2CO3(aq)
H+(aq) +
HCO3–(aq)
SO2(aq) + H2O
H2SO3(aq)
H+(aq) + HSO3–(aq)
NH3(aq) + H2O
NH4+(aq) + HO–(aq)
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Case Study
When you open a bottle
of seltzer, it fizzes. How
should you store it to
increase the time before
it goes flat?
Gases are more soluble at
low temperature and high
pressure. Cap it and cool it.
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Ex. Percent by Mass
What is the percent by mass of NaCl in a
solution consisting of 12.5 g of NaCl and 75.0
g water?
mass of solute
percent by mass 
 100%
mass of solution
wt %NaCl
12.5 g

 100
(12.5  75.0)g
wt %NaCl  14.3% NaCl
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Colligative Properties
 Physical properties of solutions
 Depend mostly on relative populations of
particles in mixtures
 Don’t depend on their chemical identities
Effects of Solute on Vapor Pressure of Solvents
 Solutes that can’t evaporate from solution are
called nonvolatile solutes
Fact: All solutions of nonvolatile solutes have
lower vapor pressures than their pure solvents
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Raoult's Law
 Vapor pressure of solution, Psoln, equals
product of mole fraction of solvent, Xsolvent,
and its vapor pressure when pure, Psolvent
 Applies for dilute solutions
Psolution  X solvent Psolvent

Psolution  vapor pressure of the solution
X solvent  mole fraction of the solvent

Psolvent
 vapor pressure of pure solvent
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Alternate form of Raoult’s Law
 Plot of Psoln vs. Xsolvent
should be linear
 Slope =

Psolvent
 Intercept = 0
 Change in vapor
pressure can be
expressed as

P  change in P  (Psolvent  Psolution )
 Usually more interested in how solute’s mole fraction
changes the vapor pressure of solvent
P 

X solutePsolvent
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Ex. Glycerin (using Raoult’s Law)
Glycerin, C3H8O3, is a nonvolatile nonelectrolyte with
a density of 1.26 g/mL at 25 °C. Calculate the
change in vapor pressure as 25 °C of a solution made
by adding 50.0 mL of glycerin to 500.0 mL of water.
The vapor pressure of pure water at 25 °C is 23.8
torr.
To solve use:

P  X solute Psolvent
First we need Xsolute, so we need mole glycerin
and mole H2O.
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Ex. Glycerin (cont.)
Mole glycerin
1mol C 3H8 O 3
g
50.0mL C 3H8 O 3  1.26

mL 92.1g C 3H8 O 3
 0.684mol C 3H8 O 3
Mole water
1mol H2 O
g
500.0mL H2 O  1.00

 27.75mol H2 O
mL 18.02g H2 O
Mole fraction glycerin
0.684mol
X C3H8O3 
 2.406  10  2
27.75  0.684mol

P  X solute Psolvent  2.406  10

2
 23.8torr
= 0.573 torr
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Ex. Glycerin (cont.)
What is the final pressure?
 Can solve two ways:
P  Psolvent  Psolution Psolution  Psolvent  P


Psoln  23.8torr  0.573torr  23.2torr
 Or
X H2O
27.75mol

 0.9759
27.75  0.684mol
Psolution 

X solventPsolvent
 0.9759  23.8torr  23.2torr
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Learning Check
The vapor pressure of 2-methylhexane is 37.986 torr at
15°C. What would be the pressure of the mixture of 78.0
g 2-methylhexane and 15 g naphthalene, which is nearly
non-volatile at this temperature?
Psolution = XsolventPosolvent
naphthalene
C10H8
MM 128.17
2-methylhexane
C7H16
MM 100.2
15 g
mole naphthalen e 
 0.117 mol
128.17 g/mol
78.0 g
mole 2 - methylhexa ne 
 0.7784 mol
100.2 g/mol
0.7784 mol
X 2  methylhexane 
 0.869
0.7784 mol  0.117 mol
P  0.869  37.986 torr 
= 33.02 torr = 33 torr
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Why Nonvolatile Solute Lowers
Vapor Pressure
 To evaporate, molecule must have enough
Kinetic Energy to escape surface—say 1%
 Only those molecules escape
 Set up equilibrium between liquid and vapor
 Add solute to solvent to get 20% (w/w) solution
 Now only 1% of 80% solvent can escape or 0.8%
of all molecules
 So vapor pressure  because fraction of
solvent molecules capable of leaving
solution 
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Why Nonvolatile Solute Lowers Vapor
Pressure
A. Lots of solvent molecules in liquid phase
 Rate of evaporation and condensation high
B. Fewer solvent molecules in liquid
 Rate of evaporation lower
 At equilibrium, fewer molecules in gas phase
 Vapor pressure lower
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Solutions That Contain Two or
More Volatile Components
 Now vapor contains molecules of both
components
 Partial pressure of each component A and B is
given by Raoult’s Law
PA  X APA
PB  X BPB
 Total pressure of solution of components A and
B given by Dalton’s Law of Partial Pressures
Ptotal  PA  PB  X A PA  X B PB
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For Ideal, Two Component Solution
of Volatile Components
P Total  PA  PB 

X APA


X BPB
PA  X APA
PB  X BPB
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Ex. Benzene and Toluene
 Consider a mixture of benzene, C6H6, and
toluene, C7H8, containing 1.0 mol benzene
and 2.0 mol toluene. At 20 °C, the vapor
pressures of the pure substances are:
P°benzene = 75 torr
P°toluene = 22 torr
 Assuming the mixture obeys Raoult’s law,
what is the total pressure above this
solution?
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Ex. Benzene and Toluene (cont.)
1. Calculate mole fractions of A and B
X benzene
1.0mol

 0.33 benzene
1.0  2.0 mol
X toluene
2.0mol

 0.67 toluene
1.0  2.0 mol
2. Calculate partial pressures of A and B

Pbenzene  X benzene  Pbenzene
 0.33  75torr  25torr

Ptoluene  X toluene  Ptoluene
 0.67  22torr  15torr
3. Calculate total pressure
Ptotal  Pbenzene  Ptoluene
 (25  15)torr  40torr
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Learning Check
The vapor pressure of 2-methylheptane is
233.95 torr at 55°C. 3-ethylpentane has a
vapor pressure of 207.68 at the same
temperature. What would be the pressure of
the mixture of 78.0g 2-methylheptane and 15
g 3-ethylpentane?
2-methylheptane
C8H18
MM 114.23 g/mol
3-ethylpentane
C7H16
MM 100.2 g/mol
Psolution = XAPoA + XBPoB
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Chemistry: The Molecular Nature of Matter, 6E
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Learning Check
78.0 g
mole 2 - methylheptane 
 0.6828 mol
114.23 g/mol
15 g
mole 3 - ethylpenta ne 
 0.1497 mol
100.2 g/mol
X 2 - methylpentane
X 3 - ethylpentane
0.68283 mol

 0.827
(0.68283 mol  0.14 97 mol)
0.1497 mol

 0.173
(0.68283 mol  0.1497 mol)
P  0.827  233.95 torr   0.173  207.68 torr 
P = 230 torr
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Your Turn!
n-hexane and n-heptane are miscible in a large degree
and both volatile. If the vapor pressure of pure
hexane is 151.28 mm Hg, and heptane is 45.67 at
25º, which equation can be used to determine the
mole fraction of hexane in the mixture if the mixture’s
vapor pressure is 145.5 mm Hg?
A. X(151.28 mmHg) = 145.5 mmHg
B. X(151.28 mmHg) + (X)(45.67 mm Hg) = 145.5
mmHg
C. X(151.28 mmHg) + (1 – X)(45.67 mm Hg)
= 145.5 mm Hg
D. None of these
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Solutes also Affect Freezing and
Boiling Points of Solutions
Facts:
 Freezing Point of solution always Lower
than pure solvent
 Boiling Point of solution always Higher
than pure solvent
Why?
 Consider the phase diagram of H2O
 Solid, liquid, gas phases in equilibrium
 Blue lines
 P vs. T
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Pure Water
 Triple Point (TP)
 All 3 phases exist in
equilibrium
simultaneously
760
 Pure H2O
 Dashed lines at 760 torr
(1 atm) that intersect
solid/liquid and
liquid/gas curves
 Give T for Freezing
Point (FP) and Boiling
Point (BP)
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TP
TFP
Chemistry: The Molecular Nature of Matter, 6E
TBP
47
Solution—Effect of Solute
 Solute molecules stay
in solution only
 None in vapor
 None in solid
 Crystal structure prevents
from entering
 Liquid/vapor
  number solvent
molecules entering vapor
 Need higher T to get all
liquid to gas
 Line at higher T along
phase border (red)
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Solution—Effect of Solute
 Triple point lower
and to left ()
 Solid/liquid
 Solid/liquid line to left
(red)
 Lower T all along
phase boundary
 Solute keeps solvent
in solution longer
 Must go to lower T to
form crystal
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Freezing Point Depression and
Boiling Point Elevation
Solution
 Observe  BP and FP over pure solvent
 Presence of solute, depresses FP and elevates BP
 Both Tf and Tb depend on relative amounts of
solvent and solute
Colligative properties
 Boiling Point Elevation (Tb)
  in boiling point of solution vs. pure solvent
 Freezing Point Depression (Tf )
  in freezing point of solution vs. pure solvent
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Freezing Point Depression (Tf)
Tf = Kf m
where
Tf = (Tfp  Tsoln)
m = concentration in Molality
Kf = molal freezing point depression constant
Units of °C/molal
Depend on solvent, see Table 13.3
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Boiling Point Elevation (Tb)
where
Tb = Kb m
Tb = (Tsoln  Tbp)
m = concentration in Molality
Kb = molal boiling point elevation constant
Units of °C/m
Depend on solvent, see Table 13.3
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Table 13.3 Kf and Kb
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