chapter outline - wou.edu
TRANSCRIPT
342019
1
Chapter 11 Properties of Solutions -
Their Concentrations and Colligative Properties
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
342019
2
Lattice energy (E) increases as Q
increases andor as r decreases
Lattice energy (U) is the energy required to completely separate one mole of
a solid ionic compound into gaseous ions It is always endothermic
eg MgF2(s) Mg2+(g) + 2F-(g)
Calculating Lattice Energies Using the Born-Haber Cycle
Q1 is the charge on the cation
Q2 is the charge on the anion
d is the distance between the ions
CompoundLattice
Energy (U)
MgF2 2957 Q= +2-1
MgO 3938 Q= +2-2
LiF
LiCl
1036
853
radius F lt
radius Cl
119864119890119897 = 231 119909 10minus19119869 ∙ 11989911989811987611199091198762119889
Hessrsquo Law is used to calculate Ulattice
The calculation is broken down into a series of steps (cycles)
Steps
1 sublimation of 1 mole of the metal = ΔHsub
2 if necessary breaking bond of a diatomic = f x ΔHBE
3 ionization of the gas-phase metal = IE1 + IE2 +
4 electron affinity of the nonmetal = EA1 + EA2 +
formation of 1 mole of the salt from ions(g) = ΔHlast = -Ulattice
Born-Haber Cycles ndash Calculating Ulattice
342019
3
M(s) + n X2(g)
M(g) +
M(g) + X(g)
f X2(g)
X(g)M+(g) + X(g)M+(g) +
X-(g)M+(g) +
MX(s)
ΔHsub
f ΔHBE
IE1 + IE2 etcEA1 + EA2 etc
ΔHlast = -Ulattice
ΔHf
ΔHf = ΔHsub + f x ΔHBE + IE + EA + ΔHlast
Calculating Ulattice for NaCl(s)
342019
4
ΔHlast = ΔHf minus ΔHsub minus frac12 ΔHBE minus IE1 minus EA1
342019
5
Sample Exercise 111
Calculating Lattice Energy
Calcium fluoride occurs in nature as the mineral fluorite which is the
principle source of the worldrsquos supply of fluorine Use the following data to
calculate the lattice energy of CaF2
ΔHsubCa = 168 kJmol
BEF2 = 155 kJmol
EAF = -328 kJmol
IE1Ca = 590 kJmol
IE2Ca = 1145 kJmol
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
342019
6
Vapor PressurePressure exerted by a gas in equilibrium with its liquid
Where are we going with this
Colligative Properties ndash boiling
point elevation freezing point
depression osmosis
Vapor Pressure of Solutions
bull Vapor Pressurendash Pressure exerted by a gas
in equilibrium with liquid
ndash Rates of
evaporationcondensation
are equal
342019
7
The tendency for a liquid to evaporate
increases as -
1 the temperature rises
2 the surface area
increases
3 the intermolecular
forces decrease
Boiling Point
ldquoIf you have a beaker of water open to the
atmosphere the mass of the atmosphere is
pressing down on the surface As heat is
added more and more water evaporates
pushing the molecules of the atmosphere
aside (wsystem lt 0) If enough heat is added
a temperature is eventually reached at which
the vapor pressure of the liquid equals the
atmospheric pressure and the liquid boilsrdquo
342019
8
the temperature at which the vapor pressure of a liquid is equal to the
external atmospheric pressure of 1 atm
Increasing the external atmospheric pressure increases the boiling point
Decreasing the external atmospheric pressure decreases the boiling point
Normal boiling point
Diethyl ether - dipole-dipole
interactions
Water - hydrogen
bonding (stronger)
Intermolecular Forces and Vapor Pressure
342019
9
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
Pre
ssu
re
Temperature
solid
liquid
gas
Plot of ln(P) vs 1T yields
straight line
bull Slope = minusΔHvapR
bull Intercept = constant
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
342019
10
Clausius-Clapeyron Equation
CTR
HP
1ln
vap
vap
How to use when given (P1 T1) and (P2T2)
Practice Exercise p 473
Pentane (C5H12) gas is used to blow the bubbles in molten polystyrene that turn it into
Styrofoam which is used in coffee cups and other products that have good thermal
insulation properties The normal boiling point of pentane is 36 oC its vapor pressure
at 25 oC is 505 torr What is the enthalpy of vaporization of pentane
342019
11
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
Mixtures of Volatile Substances
Raoultrsquos Law
bull Total vapor pressure of an ideal solution
depends on how much the partial pressure of
each volatile component contributes to total
vapor pressure of solution
bull Ptotal = 1P1 + 2P2 + 3P3 + hellip
where i = mole fraction of component i and
Pi = equilibrium vapor pressure of pure
volatile component at a given temperature
342019
12
Sample Exercise 113 (simplified)
Calculating the Vapor Pressure of a Solution
What is the vapor pressure of a solution prepared by dissolving 13 g
of heptane (C7H16) in 87 g of octane (C8H18) at 25 oC
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
13
Colligative Properties of Solutions
bull Colligative Properties
ndash Set of properties of a solution relative to the pure
solvent
ndash Due to solutendashsolvent interactions
ndash Depend on concentration of solute particles not
the identity of particles
ndash Include lowering of vapor pressure boiling point
elevation freezing point depression osmosis
and osmotic pressure
Vapor Pressure of Solutions
bull Raoultrsquos Law for Solutions
bull Vapor pressure of solution is equal to the vapor
pressure of the pure solvent multiplied by the
mole fraction of solvent in the solution
bull Psolution = solventP solvent
bull Vapor pressure lowering
bull A colligative property of solutions (Section 115)
bull Ideal Solution
bull One that obeys Raoultrsquos law
342019
14
Homework Problem 1138
Calculating the Vapor Pressure of a Solution of One
or More Nonvolatile Substances
A solution contains 45 moles of H2O 030 moles of sucrose (C12H22O11) and
020 moles of glucose Sucrose and glucose are nonvolatile What is the mole
fraction of water in the solution What is the vapor pressure of the solution at
35 oC given that the vapor pressure of pure water at 35 oC is 422 Torr
Psolution = solventP solvent
χ H2O =mol H2O
total moles
Psolution = H2OP H2O
Psolution = (090)(422 Torr) = 38 Torr
χ H2O =45 moles
45 + 030 + 020 moles= 090
Colligative Properties of Solutions
bull Colligative Properties
Consequences
in bp and fp of
solution relative to
pure solvent
Vapor pressure
lowering
342019
15
Solute Concentration Molality
bull Changes in boiling pointfreezing point
of solutions depends on molality
ndash Preferred concentration unit for properties
involving temperature changes because it
is independent of temperature
ndash For typical solutions molality gt molarity
solute
kg of solvent
nm
Calculating Molality
Starting with
a) Mass of solute
b) Volume of solvent or
c) Volume of solution
342019
16
Practice Exercise p 483
Calculating the Molality of a Solution
What is the molality of a solution prepared by dissolving 782 g of ethylene glycol
HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water
is 100 gmL
120002 =mol solute
kg solvent
mol solute = 782 g x1 mol
6207 g= 12599 mol
kg solvent = 150 L x1000 mL
Lx10 g
mLx
10 kg
1000 g= 150 kg
120002 =12599 mol
150 kg= 0840120002
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
2
Lattice energy (E) increases as Q
increases andor as r decreases
Lattice energy (U) is the energy required to completely separate one mole of
a solid ionic compound into gaseous ions It is always endothermic
eg MgF2(s) Mg2+(g) + 2F-(g)
Calculating Lattice Energies Using the Born-Haber Cycle
Q1 is the charge on the cation
Q2 is the charge on the anion
d is the distance between the ions
CompoundLattice
Energy (U)
MgF2 2957 Q= +2-1
MgO 3938 Q= +2-2
LiF
LiCl
1036
853
radius F lt
radius Cl
119864119890119897 = 231 119909 10minus19119869 ∙ 11989911989811987611199091198762119889
Hessrsquo Law is used to calculate Ulattice
The calculation is broken down into a series of steps (cycles)
Steps
1 sublimation of 1 mole of the metal = ΔHsub
2 if necessary breaking bond of a diatomic = f x ΔHBE
3 ionization of the gas-phase metal = IE1 + IE2 +
4 electron affinity of the nonmetal = EA1 + EA2 +
formation of 1 mole of the salt from ions(g) = ΔHlast = -Ulattice
Born-Haber Cycles ndash Calculating Ulattice
342019
3
M(s) + n X2(g)
M(g) +
M(g) + X(g)
f X2(g)
X(g)M+(g) + X(g)M+(g) +
X-(g)M+(g) +
MX(s)
ΔHsub
f ΔHBE
IE1 + IE2 etcEA1 + EA2 etc
ΔHlast = -Ulattice
ΔHf
ΔHf = ΔHsub + f x ΔHBE + IE + EA + ΔHlast
Calculating Ulattice for NaCl(s)
342019
4
ΔHlast = ΔHf minus ΔHsub minus frac12 ΔHBE minus IE1 minus EA1
342019
5
Sample Exercise 111
Calculating Lattice Energy
Calcium fluoride occurs in nature as the mineral fluorite which is the
principle source of the worldrsquos supply of fluorine Use the following data to
calculate the lattice energy of CaF2
ΔHsubCa = 168 kJmol
BEF2 = 155 kJmol
EAF = -328 kJmol
IE1Ca = 590 kJmol
IE2Ca = 1145 kJmol
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
342019
6
Vapor PressurePressure exerted by a gas in equilibrium with its liquid
Where are we going with this
Colligative Properties ndash boiling
point elevation freezing point
depression osmosis
Vapor Pressure of Solutions
bull Vapor Pressurendash Pressure exerted by a gas
in equilibrium with liquid
ndash Rates of
evaporationcondensation
are equal
342019
7
The tendency for a liquid to evaporate
increases as -
1 the temperature rises
2 the surface area
increases
3 the intermolecular
forces decrease
Boiling Point
ldquoIf you have a beaker of water open to the
atmosphere the mass of the atmosphere is
pressing down on the surface As heat is
added more and more water evaporates
pushing the molecules of the atmosphere
aside (wsystem lt 0) If enough heat is added
a temperature is eventually reached at which
the vapor pressure of the liquid equals the
atmospheric pressure and the liquid boilsrdquo
342019
8
the temperature at which the vapor pressure of a liquid is equal to the
external atmospheric pressure of 1 atm
Increasing the external atmospheric pressure increases the boiling point
Decreasing the external atmospheric pressure decreases the boiling point
Normal boiling point
Diethyl ether - dipole-dipole
interactions
Water - hydrogen
bonding (stronger)
Intermolecular Forces and Vapor Pressure
342019
9
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
Pre
ssu
re
Temperature
solid
liquid
gas
Plot of ln(P) vs 1T yields
straight line
bull Slope = minusΔHvapR
bull Intercept = constant
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
342019
10
Clausius-Clapeyron Equation
CTR
HP
1ln
vap
vap
How to use when given (P1 T1) and (P2T2)
Practice Exercise p 473
Pentane (C5H12) gas is used to blow the bubbles in molten polystyrene that turn it into
Styrofoam which is used in coffee cups and other products that have good thermal
insulation properties The normal boiling point of pentane is 36 oC its vapor pressure
at 25 oC is 505 torr What is the enthalpy of vaporization of pentane
342019
11
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
Mixtures of Volatile Substances
Raoultrsquos Law
bull Total vapor pressure of an ideal solution
depends on how much the partial pressure of
each volatile component contributes to total
vapor pressure of solution
bull Ptotal = 1P1 + 2P2 + 3P3 + hellip
where i = mole fraction of component i and
Pi = equilibrium vapor pressure of pure
volatile component at a given temperature
342019
12
Sample Exercise 113 (simplified)
Calculating the Vapor Pressure of a Solution
What is the vapor pressure of a solution prepared by dissolving 13 g
of heptane (C7H16) in 87 g of octane (C8H18) at 25 oC
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
13
Colligative Properties of Solutions
bull Colligative Properties
ndash Set of properties of a solution relative to the pure
solvent
ndash Due to solutendashsolvent interactions
ndash Depend on concentration of solute particles not
the identity of particles
ndash Include lowering of vapor pressure boiling point
elevation freezing point depression osmosis
and osmotic pressure
Vapor Pressure of Solutions
bull Raoultrsquos Law for Solutions
bull Vapor pressure of solution is equal to the vapor
pressure of the pure solvent multiplied by the
mole fraction of solvent in the solution
bull Psolution = solventP solvent
bull Vapor pressure lowering
bull A colligative property of solutions (Section 115)
bull Ideal Solution
bull One that obeys Raoultrsquos law
342019
14
Homework Problem 1138
Calculating the Vapor Pressure of a Solution of One
or More Nonvolatile Substances
A solution contains 45 moles of H2O 030 moles of sucrose (C12H22O11) and
020 moles of glucose Sucrose and glucose are nonvolatile What is the mole
fraction of water in the solution What is the vapor pressure of the solution at
35 oC given that the vapor pressure of pure water at 35 oC is 422 Torr
Psolution = solventP solvent
χ H2O =mol H2O
total moles
Psolution = H2OP H2O
Psolution = (090)(422 Torr) = 38 Torr
χ H2O =45 moles
45 + 030 + 020 moles= 090
Colligative Properties of Solutions
bull Colligative Properties
Consequences
in bp and fp of
solution relative to
pure solvent
Vapor pressure
lowering
342019
15
Solute Concentration Molality
bull Changes in boiling pointfreezing point
of solutions depends on molality
ndash Preferred concentration unit for properties
involving temperature changes because it
is independent of temperature
ndash For typical solutions molality gt molarity
solute
kg of solvent
nm
Calculating Molality
Starting with
a) Mass of solute
b) Volume of solvent or
c) Volume of solution
342019
16
Practice Exercise p 483
Calculating the Molality of a Solution
What is the molality of a solution prepared by dissolving 782 g of ethylene glycol
HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water
is 100 gmL
120002 =mol solute
kg solvent
mol solute = 782 g x1 mol
6207 g= 12599 mol
kg solvent = 150 L x1000 mL
Lx10 g
mLx
10 kg
1000 g= 150 kg
120002 =12599 mol
150 kg= 0840120002
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
3
M(s) + n X2(g)
M(g) +
M(g) + X(g)
f X2(g)
X(g)M+(g) + X(g)M+(g) +
X-(g)M+(g) +
MX(s)
ΔHsub
f ΔHBE
IE1 + IE2 etcEA1 + EA2 etc
ΔHlast = -Ulattice
ΔHf
ΔHf = ΔHsub + f x ΔHBE + IE + EA + ΔHlast
Calculating Ulattice for NaCl(s)
342019
4
ΔHlast = ΔHf minus ΔHsub minus frac12 ΔHBE minus IE1 minus EA1
342019
5
Sample Exercise 111
Calculating Lattice Energy
Calcium fluoride occurs in nature as the mineral fluorite which is the
principle source of the worldrsquos supply of fluorine Use the following data to
calculate the lattice energy of CaF2
ΔHsubCa = 168 kJmol
BEF2 = 155 kJmol
EAF = -328 kJmol
IE1Ca = 590 kJmol
IE2Ca = 1145 kJmol
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
342019
6
Vapor PressurePressure exerted by a gas in equilibrium with its liquid
Where are we going with this
Colligative Properties ndash boiling
point elevation freezing point
depression osmosis
Vapor Pressure of Solutions
bull Vapor Pressurendash Pressure exerted by a gas
in equilibrium with liquid
ndash Rates of
evaporationcondensation
are equal
342019
7
The tendency for a liquid to evaporate
increases as -
1 the temperature rises
2 the surface area
increases
3 the intermolecular
forces decrease
Boiling Point
ldquoIf you have a beaker of water open to the
atmosphere the mass of the atmosphere is
pressing down on the surface As heat is
added more and more water evaporates
pushing the molecules of the atmosphere
aside (wsystem lt 0) If enough heat is added
a temperature is eventually reached at which
the vapor pressure of the liquid equals the
atmospheric pressure and the liquid boilsrdquo
342019
8
the temperature at which the vapor pressure of a liquid is equal to the
external atmospheric pressure of 1 atm
Increasing the external atmospheric pressure increases the boiling point
Decreasing the external atmospheric pressure decreases the boiling point
Normal boiling point
Diethyl ether - dipole-dipole
interactions
Water - hydrogen
bonding (stronger)
Intermolecular Forces and Vapor Pressure
342019
9
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
Pre
ssu
re
Temperature
solid
liquid
gas
Plot of ln(P) vs 1T yields
straight line
bull Slope = minusΔHvapR
bull Intercept = constant
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
342019
10
Clausius-Clapeyron Equation
CTR
HP
1ln
vap
vap
How to use when given (P1 T1) and (P2T2)
Practice Exercise p 473
Pentane (C5H12) gas is used to blow the bubbles in molten polystyrene that turn it into
Styrofoam which is used in coffee cups and other products that have good thermal
insulation properties The normal boiling point of pentane is 36 oC its vapor pressure
at 25 oC is 505 torr What is the enthalpy of vaporization of pentane
342019
11
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
Mixtures of Volatile Substances
Raoultrsquos Law
bull Total vapor pressure of an ideal solution
depends on how much the partial pressure of
each volatile component contributes to total
vapor pressure of solution
bull Ptotal = 1P1 + 2P2 + 3P3 + hellip
where i = mole fraction of component i and
Pi = equilibrium vapor pressure of pure
volatile component at a given temperature
342019
12
Sample Exercise 113 (simplified)
Calculating the Vapor Pressure of a Solution
What is the vapor pressure of a solution prepared by dissolving 13 g
of heptane (C7H16) in 87 g of octane (C8H18) at 25 oC
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
13
Colligative Properties of Solutions
bull Colligative Properties
ndash Set of properties of a solution relative to the pure
solvent
ndash Due to solutendashsolvent interactions
ndash Depend on concentration of solute particles not
the identity of particles
ndash Include lowering of vapor pressure boiling point
elevation freezing point depression osmosis
and osmotic pressure
Vapor Pressure of Solutions
bull Raoultrsquos Law for Solutions
bull Vapor pressure of solution is equal to the vapor
pressure of the pure solvent multiplied by the
mole fraction of solvent in the solution
bull Psolution = solventP solvent
bull Vapor pressure lowering
bull A colligative property of solutions (Section 115)
bull Ideal Solution
bull One that obeys Raoultrsquos law
342019
14
Homework Problem 1138
Calculating the Vapor Pressure of a Solution of One
or More Nonvolatile Substances
A solution contains 45 moles of H2O 030 moles of sucrose (C12H22O11) and
020 moles of glucose Sucrose and glucose are nonvolatile What is the mole
fraction of water in the solution What is the vapor pressure of the solution at
35 oC given that the vapor pressure of pure water at 35 oC is 422 Torr
Psolution = solventP solvent
χ H2O =mol H2O
total moles
Psolution = H2OP H2O
Psolution = (090)(422 Torr) = 38 Torr
χ H2O =45 moles
45 + 030 + 020 moles= 090
Colligative Properties of Solutions
bull Colligative Properties
Consequences
in bp and fp of
solution relative to
pure solvent
Vapor pressure
lowering
342019
15
Solute Concentration Molality
bull Changes in boiling pointfreezing point
of solutions depends on molality
ndash Preferred concentration unit for properties
involving temperature changes because it
is independent of temperature
ndash For typical solutions molality gt molarity
solute
kg of solvent
nm
Calculating Molality
Starting with
a) Mass of solute
b) Volume of solvent or
c) Volume of solution
342019
16
Practice Exercise p 483
Calculating the Molality of a Solution
What is the molality of a solution prepared by dissolving 782 g of ethylene glycol
HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water
is 100 gmL
120002 =mol solute
kg solvent
mol solute = 782 g x1 mol
6207 g= 12599 mol
kg solvent = 150 L x1000 mL
Lx10 g
mLx
10 kg
1000 g= 150 kg
120002 =12599 mol
150 kg= 0840120002
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
4
ΔHlast = ΔHf minus ΔHsub minus frac12 ΔHBE minus IE1 minus EA1
342019
5
Sample Exercise 111
Calculating Lattice Energy
Calcium fluoride occurs in nature as the mineral fluorite which is the
principle source of the worldrsquos supply of fluorine Use the following data to
calculate the lattice energy of CaF2
ΔHsubCa = 168 kJmol
BEF2 = 155 kJmol
EAF = -328 kJmol
IE1Ca = 590 kJmol
IE2Ca = 1145 kJmol
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
342019
6
Vapor PressurePressure exerted by a gas in equilibrium with its liquid
Where are we going with this
Colligative Properties ndash boiling
point elevation freezing point
depression osmosis
Vapor Pressure of Solutions
bull Vapor Pressurendash Pressure exerted by a gas
in equilibrium with liquid
ndash Rates of
evaporationcondensation
are equal
342019
7
The tendency for a liquid to evaporate
increases as -
1 the temperature rises
2 the surface area
increases
3 the intermolecular
forces decrease
Boiling Point
ldquoIf you have a beaker of water open to the
atmosphere the mass of the atmosphere is
pressing down on the surface As heat is
added more and more water evaporates
pushing the molecules of the atmosphere
aside (wsystem lt 0) If enough heat is added
a temperature is eventually reached at which
the vapor pressure of the liquid equals the
atmospheric pressure and the liquid boilsrdquo
342019
8
the temperature at which the vapor pressure of a liquid is equal to the
external atmospheric pressure of 1 atm
Increasing the external atmospheric pressure increases the boiling point
Decreasing the external atmospheric pressure decreases the boiling point
Normal boiling point
Diethyl ether - dipole-dipole
interactions
Water - hydrogen
bonding (stronger)
Intermolecular Forces and Vapor Pressure
342019
9
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
Pre
ssu
re
Temperature
solid
liquid
gas
Plot of ln(P) vs 1T yields
straight line
bull Slope = minusΔHvapR
bull Intercept = constant
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
342019
10
Clausius-Clapeyron Equation
CTR
HP
1ln
vap
vap
How to use when given (P1 T1) and (P2T2)
Practice Exercise p 473
Pentane (C5H12) gas is used to blow the bubbles in molten polystyrene that turn it into
Styrofoam which is used in coffee cups and other products that have good thermal
insulation properties The normal boiling point of pentane is 36 oC its vapor pressure
at 25 oC is 505 torr What is the enthalpy of vaporization of pentane
342019
11
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
Mixtures of Volatile Substances
Raoultrsquos Law
bull Total vapor pressure of an ideal solution
depends on how much the partial pressure of
each volatile component contributes to total
vapor pressure of solution
bull Ptotal = 1P1 + 2P2 + 3P3 + hellip
where i = mole fraction of component i and
Pi = equilibrium vapor pressure of pure
volatile component at a given temperature
342019
12
Sample Exercise 113 (simplified)
Calculating the Vapor Pressure of a Solution
What is the vapor pressure of a solution prepared by dissolving 13 g
of heptane (C7H16) in 87 g of octane (C8H18) at 25 oC
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
13
Colligative Properties of Solutions
bull Colligative Properties
ndash Set of properties of a solution relative to the pure
solvent
ndash Due to solutendashsolvent interactions
ndash Depend on concentration of solute particles not
the identity of particles
ndash Include lowering of vapor pressure boiling point
elevation freezing point depression osmosis
and osmotic pressure
Vapor Pressure of Solutions
bull Raoultrsquos Law for Solutions
bull Vapor pressure of solution is equal to the vapor
pressure of the pure solvent multiplied by the
mole fraction of solvent in the solution
bull Psolution = solventP solvent
bull Vapor pressure lowering
bull A colligative property of solutions (Section 115)
bull Ideal Solution
bull One that obeys Raoultrsquos law
342019
14
Homework Problem 1138
Calculating the Vapor Pressure of a Solution of One
or More Nonvolatile Substances
A solution contains 45 moles of H2O 030 moles of sucrose (C12H22O11) and
020 moles of glucose Sucrose and glucose are nonvolatile What is the mole
fraction of water in the solution What is the vapor pressure of the solution at
35 oC given that the vapor pressure of pure water at 35 oC is 422 Torr
Psolution = solventP solvent
χ H2O =mol H2O
total moles
Psolution = H2OP H2O
Psolution = (090)(422 Torr) = 38 Torr
χ H2O =45 moles
45 + 030 + 020 moles= 090
Colligative Properties of Solutions
bull Colligative Properties
Consequences
in bp and fp of
solution relative to
pure solvent
Vapor pressure
lowering
342019
15
Solute Concentration Molality
bull Changes in boiling pointfreezing point
of solutions depends on molality
ndash Preferred concentration unit for properties
involving temperature changes because it
is independent of temperature
ndash For typical solutions molality gt molarity
solute
kg of solvent
nm
Calculating Molality
Starting with
a) Mass of solute
b) Volume of solvent or
c) Volume of solution
342019
16
Practice Exercise p 483
Calculating the Molality of a Solution
What is the molality of a solution prepared by dissolving 782 g of ethylene glycol
HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water
is 100 gmL
120002 =mol solute
kg solvent
mol solute = 782 g x1 mol
6207 g= 12599 mol
kg solvent = 150 L x1000 mL
Lx10 g
mLx
10 kg
1000 g= 150 kg
120002 =12599 mol
150 kg= 0840120002
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
5
Sample Exercise 111
Calculating Lattice Energy
Calcium fluoride occurs in nature as the mineral fluorite which is the
principle source of the worldrsquos supply of fluorine Use the following data to
calculate the lattice energy of CaF2
ΔHsubCa = 168 kJmol
BEF2 = 155 kJmol
EAF = -328 kJmol
IE1Ca = 590 kJmol
IE2Ca = 1145 kJmol
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
342019
6
Vapor PressurePressure exerted by a gas in equilibrium with its liquid
Where are we going with this
Colligative Properties ndash boiling
point elevation freezing point
depression osmosis
Vapor Pressure of Solutions
bull Vapor Pressurendash Pressure exerted by a gas
in equilibrium with liquid
ndash Rates of
evaporationcondensation
are equal
342019
7
The tendency for a liquid to evaporate
increases as -
1 the temperature rises
2 the surface area
increases
3 the intermolecular
forces decrease
Boiling Point
ldquoIf you have a beaker of water open to the
atmosphere the mass of the atmosphere is
pressing down on the surface As heat is
added more and more water evaporates
pushing the molecules of the atmosphere
aside (wsystem lt 0) If enough heat is added
a temperature is eventually reached at which
the vapor pressure of the liquid equals the
atmospheric pressure and the liquid boilsrdquo
342019
8
the temperature at which the vapor pressure of a liquid is equal to the
external atmospheric pressure of 1 atm
Increasing the external atmospheric pressure increases the boiling point
Decreasing the external atmospheric pressure decreases the boiling point
Normal boiling point
Diethyl ether - dipole-dipole
interactions
Water - hydrogen
bonding (stronger)
Intermolecular Forces and Vapor Pressure
342019
9
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
Pre
ssu
re
Temperature
solid
liquid
gas
Plot of ln(P) vs 1T yields
straight line
bull Slope = minusΔHvapR
bull Intercept = constant
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
342019
10
Clausius-Clapeyron Equation
CTR
HP
1ln
vap
vap
How to use when given (P1 T1) and (P2T2)
Practice Exercise p 473
Pentane (C5H12) gas is used to blow the bubbles in molten polystyrene that turn it into
Styrofoam which is used in coffee cups and other products that have good thermal
insulation properties The normal boiling point of pentane is 36 oC its vapor pressure
at 25 oC is 505 torr What is the enthalpy of vaporization of pentane
342019
11
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
Mixtures of Volatile Substances
Raoultrsquos Law
bull Total vapor pressure of an ideal solution
depends on how much the partial pressure of
each volatile component contributes to total
vapor pressure of solution
bull Ptotal = 1P1 + 2P2 + 3P3 + hellip
where i = mole fraction of component i and
Pi = equilibrium vapor pressure of pure
volatile component at a given temperature
342019
12
Sample Exercise 113 (simplified)
Calculating the Vapor Pressure of a Solution
What is the vapor pressure of a solution prepared by dissolving 13 g
of heptane (C7H16) in 87 g of octane (C8H18) at 25 oC
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
13
Colligative Properties of Solutions
bull Colligative Properties
ndash Set of properties of a solution relative to the pure
solvent
ndash Due to solutendashsolvent interactions
ndash Depend on concentration of solute particles not
the identity of particles
ndash Include lowering of vapor pressure boiling point
elevation freezing point depression osmosis
and osmotic pressure
Vapor Pressure of Solutions
bull Raoultrsquos Law for Solutions
bull Vapor pressure of solution is equal to the vapor
pressure of the pure solvent multiplied by the
mole fraction of solvent in the solution
bull Psolution = solventP solvent
bull Vapor pressure lowering
bull A colligative property of solutions (Section 115)
bull Ideal Solution
bull One that obeys Raoultrsquos law
342019
14
Homework Problem 1138
Calculating the Vapor Pressure of a Solution of One
or More Nonvolatile Substances
A solution contains 45 moles of H2O 030 moles of sucrose (C12H22O11) and
020 moles of glucose Sucrose and glucose are nonvolatile What is the mole
fraction of water in the solution What is the vapor pressure of the solution at
35 oC given that the vapor pressure of pure water at 35 oC is 422 Torr
Psolution = solventP solvent
χ H2O =mol H2O
total moles
Psolution = H2OP H2O
Psolution = (090)(422 Torr) = 38 Torr
χ H2O =45 moles
45 + 030 + 020 moles= 090
Colligative Properties of Solutions
bull Colligative Properties
Consequences
in bp and fp of
solution relative to
pure solvent
Vapor pressure
lowering
342019
15
Solute Concentration Molality
bull Changes in boiling pointfreezing point
of solutions depends on molality
ndash Preferred concentration unit for properties
involving temperature changes because it
is independent of temperature
ndash For typical solutions molality gt molarity
solute
kg of solvent
nm
Calculating Molality
Starting with
a) Mass of solute
b) Volume of solvent or
c) Volume of solution
342019
16
Practice Exercise p 483
Calculating the Molality of a Solution
What is the molality of a solution prepared by dissolving 782 g of ethylene glycol
HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water
is 100 gmL
120002 =mol solute
kg solvent
mol solute = 782 g x1 mol
6207 g= 12599 mol
kg solvent = 150 L x1000 mL
Lx10 g
mLx
10 kg
1000 g= 150 kg
120002 =12599 mol
150 kg= 0840120002
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
6
Vapor PressurePressure exerted by a gas in equilibrium with its liquid
Where are we going with this
Colligative Properties ndash boiling
point elevation freezing point
depression osmosis
Vapor Pressure of Solutions
bull Vapor Pressurendash Pressure exerted by a gas
in equilibrium with liquid
ndash Rates of
evaporationcondensation
are equal
342019
7
The tendency for a liquid to evaporate
increases as -
1 the temperature rises
2 the surface area
increases
3 the intermolecular
forces decrease
Boiling Point
ldquoIf you have a beaker of water open to the
atmosphere the mass of the atmosphere is
pressing down on the surface As heat is
added more and more water evaporates
pushing the molecules of the atmosphere
aside (wsystem lt 0) If enough heat is added
a temperature is eventually reached at which
the vapor pressure of the liquid equals the
atmospheric pressure and the liquid boilsrdquo
342019
8
the temperature at which the vapor pressure of a liquid is equal to the
external atmospheric pressure of 1 atm
Increasing the external atmospheric pressure increases the boiling point
Decreasing the external atmospheric pressure decreases the boiling point
Normal boiling point
Diethyl ether - dipole-dipole
interactions
Water - hydrogen
bonding (stronger)
Intermolecular Forces and Vapor Pressure
342019
9
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
Pre
ssu
re
Temperature
solid
liquid
gas
Plot of ln(P) vs 1T yields
straight line
bull Slope = minusΔHvapR
bull Intercept = constant
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
342019
10
Clausius-Clapeyron Equation
CTR
HP
1ln
vap
vap
How to use when given (P1 T1) and (P2T2)
Practice Exercise p 473
Pentane (C5H12) gas is used to blow the bubbles in molten polystyrene that turn it into
Styrofoam which is used in coffee cups and other products that have good thermal
insulation properties The normal boiling point of pentane is 36 oC its vapor pressure
at 25 oC is 505 torr What is the enthalpy of vaporization of pentane
342019
11
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
Mixtures of Volatile Substances
Raoultrsquos Law
bull Total vapor pressure of an ideal solution
depends on how much the partial pressure of
each volatile component contributes to total
vapor pressure of solution
bull Ptotal = 1P1 + 2P2 + 3P3 + hellip
where i = mole fraction of component i and
Pi = equilibrium vapor pressure of pure
volatile component at a given temperature
342019
12
Sample Exercise 113 (simplified)
Calculating the Vapor Pressure of a Solution
What is the vapor pressure of a solution prepared by dissolving 13 g
of heptane (C7H16) in 87 g of octane (C8H18) at 25 oC
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
13
Colligative Properties of Solutions
bull Colligative Properties
ndash Set of properties of a solution relative to the pure
solvent
ndash Due to solutendashsolvent interactions
ndash Depend on concentration of solute particles not
the identity of particles
ndash Include lowering of vapor pressure boiling point
elevation freezing point depression osmosis
and osmotic pressure
Vapor Pressure of Solutions
bull Raoultrsquos Law for Solutions
bull Vapor pressure of solution is equal to the vapor
pressure of the pure solvent multiplied by the
mole fraction of solvent in the solution
bull Psolution = solventP solvent
bull Vapor pressure lowering
bull A colligative property of solutions (Section 115)
bull Ideal Solution
bull One that obeys Raoultrsquos law
342019
14
Homework Problem 1138
Calculating the Vapor Pressure of a Solution of One
or More Nonvolatile Substances
A solution contains 45 moles of H2O 030 moles of sucrose (C12H22O11) and
020 moles of glucose Sucrose and glucose are nonvolatile What is the mole
fraction of water in the solution What is the vapor pressure of the solution at
35 oC given that the vapor pressure of pure water at 35 oC is 422 Torr
Psolution = solventP solvent
χ H2O =mol H2O
total moles
Psolution = H2OP H2O
Psolution = (090)(422 Torr) = 38 Torr
χ H2O =45 moles
45 + 030 + 020 moles= 090
Colligative Properties of Solutions
bull Colligative Properties
Consequences
in bp and fp of
solution relative to
pure solvent
Vapor pressure
lowering
342019
15
Solute Concentration Molality
bull Changes in boiling pointfreezing point
of solutions depends on molality
ndash Preferred concentration unit for properties
involving temperature changes because it
is independent of temperature
ndash For typical solutions molality gt molarity
solute
kg of solvent
nm
Calculating Molality
Starting with
a) Mass of solute
b) Volume of solvent or
c) Volume of solution
342019
16
Practice Exercise p 483
Calculating the Molality of a Solution
What is the molality of a solution prepared by dissolving 782 g of ethylene glycol
HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water
is 100 gmL
120002 =mol solute
kg solvent
mol solute = 782 g x1 mol
6207 g= 12599 mol
kg solvent = 150 L x1000 mL
Lx10 g
mLx
10 kg
1000 g= 150 kg
120002 =12599 mol
150 kg= 0840120002
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
7
The tendency for a liquid to evaporate
increases as -
1 the temperature rises
2 the surface area
increases
3 the intermolecular
forces decrease
Boiling Point
ldquoIf you have a beaker of water open to the
atmosphere the mass of the atmosphere is
pressing down on the surface As heat is
added more and more water evaporates
pushing the molecules of the atmosphere
aside (wsystem lt 0) If enough heat is added
a temperature is eventually reached at which
the vapor pressure of the liquid equals the
atmospheric pressure and the liquid boilsrdquo
342019
8
the temperature at which the vapor pressure of a liquid is equal to the
external atmospheric pressure of 1 atm
Increasing the external atmospheric pressure increases the boiling point
Decreasing the external atmospheric pressure decreases the boiling point
Normal boiling point
Diethyl ether - dipole-dipole
interactions
Water - hydrogen
bonding (stronger)
Intermolecular Forces and Vapor Pressure
342019
9
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
Pre
ssu
re
Temperature
solid
liquid
gas
Plot of ln(P) vs 1T yields
straight line
bull Slope = minusΔHvapR
bull Intercept = constant
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
342019
10
Clausius-Clapeyron Equation
CTR
HP
1ln
vap
vap
How to use when given (P1 T1) and (P2T2)
Practice Exercise p 473
Pentane (C5H12) gas is used to blow the bubbles in molten polystyrene that turn it into
Styrofoam which is used in coffee cups and other products that have good thermal
insulation properties The normal boiling point of pentane is 36 oC its vapor pressure
at 25 oC is 505 torr What is the enthalpy of vaporization of pentane
342019
11
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
Mixtures of Volatile Substances
Raoultrsquos Law
bull Total vapor pressure of an ideal solution
depends on how much the partial pressure of
each volatile component contributes to total
vapor pressure of solution
bull Ptotal = 1P1 + 2P2 + 3P3 + hellip
where i = mole fraction of component i and
Pi = equilibrium vapor pressure of pure
volatile component at a given temperature
342019
12
Sample Exercise 113 (simplified)
Calculating the Vapor Pressure of a Solution
What is the vapor pressure of a solution prepared by dissolving 13 g
of heptane (C7H16) in 87 g of octane (C8H18) at 25 oC
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
13
Colligative Properties of Solutions
bull Colligative Properties
ndash Set of properties of a solution relative to the pure
solvent
ndash Due to solutendashsolvent interactions
ndash Depend on concentration of solute particles not
the identity of particles
ndash Include lowering of vapor pressure boiling point
elevation freezing point depression osmosis
and osmotic pressure
Vapor Pressure of Solutions
bull Raoultrsquos Law for Solutions
bull Vapor pressure of solution is equal to the vapor
pressure of the pure solvent multiplied by the
mole fraction of solvent in the solution
bull Psolution = solventP solvent
bull Vapor pressure lowering
bull A colligative property of solutions (Section 115)
bull Ideal Solution
bull One that obeys Raoultrsquos law
342019
14
Homework Problem 1138
Calculating the Vapor Pressure of a Solution of One
or More Nonvolatile Substances
A solution contains 45 moles of H2O 030 moles of sucrose (C12H22O11) and
020 moles of glucose Sucrose and glucose are nonvolatile What is the mole
fraction of water in the solution What is the vapor pressure of the solution at
35 oC given that the vapor pressure of pure water at 35 oC is 422 Torr
Psolution = solventP solvent
χ H2O =mol H2O
total moles
Psolution = H2OP H2O
Psolution = (090)(422 Torr) = 38 Torr
χ H2O =45 moles
45 + 030 + 020 moles= 090
Colligative Properties of Solutions
bull Colligative Properties
Consequences
in bp and fp of
solution relative to
pure solvent
Vapor pressure
lowering
342019
15
Solute Concentration Molality
bull Changes in boiling pointfreezing point
of solutions depends on molality
ndash Preferred concentration unit for properties
involving temperature changes because it
is independent of temperature
ndash For typical solutions molality gt molarity
solute
kg of solvent
nm
Calculating Molality
Starting with
a) Mass of solute
b) Volume of solvent or
c) Volume of solution
342019
16
Practice Exercise p 483
Calculating the Molality of a Solution
What is the molality of a solution prepared by dissolving 782 g of ethylene glycol
HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water
is 100 gmL
120002 =mol solute
kg solvent
mol solute = 782 g x1 mol
6207 g= 12599 mol
kg solvent = 150 L x1000 mL
Lx10 g
mLx
10 kg
1000 g= 150 kg
120002 =12599 mol
150 kg= 0840120002
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
8
the temperature at which the vapor pressure of a liquid is equal to the
external atmospheric pressure of 1 atm
Increasing the external atmospheric pressure increases the boiling point
Decreasing the external atmospheric pressure decreases the boiling point
Normal boiling point
Diethyl ether - dipole-dipole
interactions
Water - hydrogen
bonding (stronger)
Intermolecular Forces and Vapor Pressure
342019
9
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
Pre
ssu
re
Temperature
solid
liquid
gas
Plot of ln(P) vs 1T yields
straight line
bull Slope = minusΔHvapR
bull Intercept = constant
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
342019
10
Clausius-Clapeyron Equation
CTR
HP
1ln
vap
vap
How to use when given (P1 T1) and (P2T2)
Practice Exercise p 473
Pentane (C5H12) gas is used to blow the bubbles in molten polystyrene that turn it into
Styrofoam which is used in coffee cups and other products that have good thermal
insulation properties The normal boiling point of pentane is 36 oC its vapor pressure
at 25 oC is 505 torr What is the enthalpy of vaporization of pentane
342019
11
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
Mixtures of Volatile Substances
Raoultrsquos Law
bull Total vapor pressure of an ideal solution
depends on how much the partial pressure of
each volatile component contributes to total
vapor pressure of solution
bull Ptotal = 1P1 + 2P2 + 3P3 + hellip
where i = mole fraction of component i and
Pi = equilibrium vapor pressure of pure
volatile component at a given temperature
342019
12
Sample Exercise 113 (simplified)
Calculating the Vapor Pressure of a Solution
What is the vapor pressure of a solution prepared by dissolving 13 g
of heptane (C7H16) in 87 g of octane (C8H18) at 25 oC
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
13
Colligative Properties of Solutions
bull Colligative Properties
ndash Set of properties of a solution relative to the pure
solvent
ndash Due to solutendashsolvent interactions
ndash Depend on concentration of solute particles not
the identity of particles
ndash Include lowering of vapor pressure boiling point
elevation freezing point depression osmosis
and osmotic pressure
Vapor Pressure of Solutions
bull Raoultrsquos Law for Solutions
bull Vapor pressure of solution is equal to the vapor
pressure of the pure solvent multiplied by the
mole fraction of solvent in the solution
bull Psolution = solventP solvent
bull Vapor pressure lowering
bull A colligative property of solutions (Section 115)
bull Ideal Solution
bull One that obeys Raoultrsquos law
342019
14
Homework Problem 1138
Calculating the Vapor Pressure of a Solution of One
or More Nonvolatile Substances
A solution contains 45 moles of H2O 030 moles of sucrose (C12H22O11) and
020 moles of glucose Sucrose and glucose are nonvolatile What is the mole
fraction of water in the solution What is the vapor pressure of the solution at
35 oC given that the vapor pressure of pure water at 35 oC is 422 Torr
Psolution = solventP solvent
χ H2O =mol H2O
total moles
Psolution = H2OP H2O
Psolution = (090)(422 Torr) = 38 Torr
χ H2O =45 moles
45 + 030 + 020 moles= 090
Colligative Properties of Solutions
bull Colligative Properties
Consequences
in bp and fp of
solution relative to
pure solvent
Vapor pressure
lowering
342019
15
Solute Concentration Molality
bull Changes in boiling pointfreezing point
of solutions depends on molality
ndash Preferred concentration unit for properties
involving temperature changes because it
is independent of temperature
ndash For typical solutions molality gt molarity
solute
kg of solvent
nm
Calculating Molality
Starting with
a) Mass of solute
b) Volume of solvent or
c) Volume of solution
342019
16
Practice Exercise p 483
Calculating the Molality of a Solution
What is the molality of a solution prepared by dissolving 782 g of ethylene glycol
HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water
is 100 gmL
120002 =mol solute
kg solvent
mol solute = 782 g x1 mol
6207 g= 12599 mol
kg solvent = 150 L x1000 mL
Lx10 g
mLx
10 kg
1000 g= 150 kg
120002 =12599 mol
150 kg= 0840120002
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
9
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
Pre
ssu
re
Temperature
solid
liquid
gas
Plot of ln(P) vs 1T yields
straight line
bull Slope = minusΔHvapR
bull Intercept = constant
Clausius-Clapeyron EquationVapor Pressure vs Temperature
CTR
HP
1ln
vap
vap
342019
10
Clausius-Clapeyron Equation
CTR
HP
1ln
vap
vap
How to use when given (P1 T1) and (P2T2)
Practice Exercise p 473
Pentane (C5H12) gas is used to blow the bubbles in molten polystyrene that turn it into
Styrofoam which is used in coffee cups and other products that have good thermal
insulation properties The normal boiling point of pentane is 36 oC its vapor pressure
at 25 oC is 505 torr What is the enthalpy of vaporization of pentane
342019
11
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
Mixtures of Volatile Substances
Raoultrsquos Law
bull Total vapor pressure of an ideal solution
depends on how much the partial pressure of
each volatile component contributes to total
vapor pressure of solution
bull Ptotal = 1P1 + 2P2 + 3P3 + hellip
where i = mole fraction of component i and
Pi = equilibrium vapor pressure of pure
volatile component at a given temperature
342019
12
Sample Exercise 113 (simplified)
Calculating the Vapor Pressure of a Solution
What is the vapor pressure of a solution prepared by dissolving 13 g
of heptane (C7H16) in 87 g of octane (C8H18) at 25 oC
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
13
Colligative Properties of Solutions
bull Colligative Properties
ndash Set of properties of a solution relative to the pure
solvent
ndash Due to solutendashsolvent interactions
ndash Depend on concentration of solute particles not
the identity of particles
ndash Include lowering of vapor pressure boiling point
elevation freezing point depression osmosis
and osmotic pressure
Vapor Pressure of Solutions
bull Raoultrsquos Law for Solutions
bull Vapor pressure of solution is equal to the vapor
pressure of the pure solvent multiplied by the
mole fraction of solvent in the solution
bull Psolution = solventP solvent
bull Vapor pressure lowering
bull A colligative property of solutions (Section 115)
bull Ideal Solution
bull One that obeys Raoultrsquos law
342019
14
Homework Problem 1138
Calculating the Vapor Pressure of a Solution of One
or More Nonvolatile Substances
A solution contains 45 moles of H2O 030 moles of sucrose (C12H22O11) and
020 moles of glucose Sucrose and glucose are nonvolatile What is the mole
fraction of water in the solution What is the vapor pressure of the solution at
35 oC given that the vapor pressure of pure water at 35 oC is 422 Torr
Psolution = solventP solvent
χ H2O =mol H2O
total moles
Psolution = H2OP H2O
Psolution = (090)(422 Torr) = 38 Torr
χ H2O =45 moles
45 + 030 + 020 moles= 090
Colligative Properties of Solutions
bull Colligative Properties
Consequences
in bp and fp of
solution relative to
pure solvent
Vapor pressure
lowering
342019
15
Solute Concentration Molality
bull Changes in boiling pointfreezing point
of solutions depends on molality
ndash Preferred concentration unit for properties
involving temperature changes because it
is independent of temperature
ndash For typical solutions molality gt molarity
solute
kg of solvent
nm
Calculating Molality
Starting with
a) Mass of solute
b) Volume of solvent or
c) Volume of solution
342019
16
Practice Exercise p 483
Calculating the Molality of a Solution
What is the molality of a solution prepared by dissolving 782 g of ethylene glycol
HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water
is 100 gmL
120002 =mol solute
kg solvent
mol solute = 782 g x1 mol
6207 g= 12599 mol
kg solvent = 150 L x1000 mL
Lx10 g
mLx
10 kg
1000 g= 150 kg
120002 =12599 mol
150 kg= 0840120002
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
10
Clausius-Clapeyron Equation
CTR
HP
1ln
vap
vap
How to use when given (P1 T1) and (P2T2)
Practice Exercise p 473
Pentane (C5H12) gas is used to blow the bubbles in molten polystyrene that turn it into
Styrofoam which is used in coffee cups and other products that have good thermal
insulation properties The normal boiling point of pentane is 36 oC its vapor pressure
at 25 oC is 505 torr What is the enthalpy of vaporization of pentane
342019
11
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
Mixtures of Volatile Substances
Raoultrsquos Law
bull Total vapor pressure of an ideal solution
depends on how much the partial pressure of
each volatile component contributes to total
vapor pressure of solution
bull Ptotal = 1P1 + 2P2 + 3P3 + hellip
where i = mole fraction of component i and
Pi = equilibrium vapor pressure of pure
volatile component at a given temperature
342019
12
Sample Exercise 113 (simplified)
Calculating the Vapor Pressure of a Solution
What is the vapor pressure of a solution prepared by dissolving 13 g
of heptane (C7H16) in 87 g of octane (C8H18) at 25 oC
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
13
Colligative Properties of Solutions
bull Colligative Properties
ndash Set of properties of a solution relative to the pure
solvent
ndash Due to solutendashsolvent interactions
ndash Depend on concentration of solute particles not
the identity of particles
ndash Include lowering of vapor pressure boiling point
elevation freezing point depression osmosis
and osmotic pressure
Vapor Pressure of Solutions
bull Raoultrsquos Law for Solutions
bull Vapor pressure of solution is equal to the vapor
pressure of the pure solvent multiplied by the
mole fraction of solvent in the solution
bull Psolution = solventP solvent
bull Vapor pressure lowering
bull A colligative property of solutions (Section 115)
bull Ideal Solution
bull One that obeys Raoultrsquos law
342019
14
Homework Problem 1138
Calculating the Vapor Pressure of a Solution of One
or More Nonvolatile Substances
A solution contains 45 moles of H2O 030 moles of sucrose (C12H22O11) and
020 moles of glucose Sucrose and glucose are nonvolatile What is the mole
fraction of water in the solution What is the vapor pressure of the solution at
35 oC given that the vapor pressure of pure water at 35 oC is 422 Torr
Psolution = solventP solvent
χ H2O =mol H2O
total moles
Psolution = H2OP H2O
Psolution = (090)(422 Torr) = 38 Torr
χ H2O =45 moles
45 + 030 + 020 moles= 090
Colligative Properties of Solutions
bull Colligative Properties
Consequences
in bp and fp of
solution relative to
pure solvent
Vapor pressure
lowering
342019
15
Solute Concentration Molality
bull Changes in boiling pointfreezing point
of solutions depends on molality
ndash Preferred concentration unit for properties
involving temperature changes because it
is independent of temperature
ndash For typical solutions molality gt molarity
solute
kg of solvent
nm
Calculating Molality
Starting with
a) Mass of solute
b) Volume of solvent or
c) Volume of solution
342019
16
Practice Exercise p 483
Calculating the Molality of a Solution
What is the molality of a solution prepared by dissolving 782 g of ethylene glycol
HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water
is 100 gmL
120002 =mol solute
kg solvent
mol solute = 782 g x1 mol
6207 g= 12599 mol
kg solvent = 150 L x1000 mL
Lx10 g
mLx
10 kg
1000 g= 150 kg
120002 =12599 mol
150 kg= 0840120002
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
11
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar Mass
Chapter Outline
Mixtures of Volatile Substances
Raoultrsquos Law
bull Total vapor pressure of an ideal solution
depends on how much the partial pressure of
each volatile component contributes to total
vapor pressure of solution
bull Ptotal = 1P1 + 2P2 + 3P3 + hellip
where i = mole fraction of component i and
Pi = equilibrium vapor pressure of pure
volatile component at a given temperature
342019
12
Sample Exercise 113 (simplified)
Calculating the Vapor Pressure of a Solution
What is the vapor pressure of a solution prepared by dissolving 13 g
of heptane (C7H16) in 87 g of octane (C8H18) at 25 oC
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
13
Colligative Properties of Solutions
bull Colligative Properties
ndash Set of properties of a solution relative to the pure
solvent
ndash Due to solutendashsolvent interactions
ndash Depend on concentration of solute particles not
the identity of particles
ndash Include lowering of vapor pressure boiling point
elevation freezing point depression osmosis
and osmotic pressure
Vapor Pressure of Solutions
bull Raoultrsquos Law for Solutions
bull Vapor pressure of solution is equal to the vapor
pressure of the pure solvent multiplied by the
mole fraction of solvent in the solution
bull Psolution = solventP solvent
bull Vapor pressure lowering
bull A colligative property of solutions (Section 115)
bull Ideal Solution
bull One that obeys Raoultrsquos law
342019
14
Homework Problem 1138
Calculating the Vapor Pressure of a Solution of One
or More Nonvolatile Substances
A solution contains 45 moles of H2O 030 moles of sucrose (C12H22O11) and
020 moles of glucose Sucrose and glucose are nonvolatile What is the mole
fraction of water in the solution What is the vapor pressure of the solution at
35 oC given that the vapor pressure of pure water at 35 oC is 422 Torr
Psolution = solventP solvent
χ H2O =mol H2O
total moles
Psolution = H2OP H2O
Psolution = (090)(422 Torr) = 38 Torr
χ H2O =45 moles
45 + 030 + 020 moles= 090
Colligative Properties of Solutions
bull Colligative Properties
Consequences
in bp and fp of
solution relative to
pure solvent
Vapor pressure
lowering
342019
15
Solute Concentration Molality
bull Changes in boiling pointfreezing point
of solutions depends on molality
ndash Preferred concentration unit for properties
involving temperature changes because it
is independent of temperature
ndash For typical solutions molality gt molarity
solute
kg of solvent
nm
Calculating Molality
Starting with
a) Mass of solute
b) Volume of solvent or
c) Volume of solution
342019
16
Practice Exercise p 483
Calculating the Molality of a Solution
What is the molality of a solution prepared by dissolving 782 g of ethylene glycol
HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water
is 100 gmL
120002 =mol solute
kg solvent
mol solute = 782 g x1 mol
6207 g= 12599 mol
kg solvent = 150 L x1000 mL
Lx10 g
mLx
10 kg
1000 g= 150 kg
120002 =12599 mol
150 kg= 0840120002
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
12
Sample Exercise 113 (simplified)
Calculating the Vapor Pressure of a Solution
What is the vapor pressure of a solution prepared by dissolving 13 g
of heptane (C7H16) in 87 g of octane (C8H18) at 25 oC
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
13
Colligative Properties of Solutions
bull Colligative Properties
ndash Set of properties of a solution relative to the pure
solvent
ndash Due to solutendashsolvent interactions
ndash Depend on concentration of solute particles not
the identity of particles
ndash Include lowering of vapor pressure boiling point
elevation freezing point depression osmosis
and osmotic pressure
Vapor Pressure of Solutions
bull Raoultrsquos Law for Solutions
bull Vapor pressure of solution is equal to the vapor
pressure of the pure solvent multiplied by the
mole fraction of solvent in the solution
bull Psolution = solventP solvent
bull Vapor pressure lowering
bull A colligative property of solutions (Section 115)
bull Ideal Solution
bull One that obeys Raoultrsquos law
342019
14
Homework Problem 1138
Calculating the Vapor Pressure of a Solution of One
or More Nonvolatile Substances
A solution contains 45 moles of H2O 030 moles of sucrose (C12H22O11) and
020 moles of glucose Sucrose and glucose are nonvolatile What is the mole
fraction of water in the solution What is the vapor pressure of the solution at
35 oC given that the vapor pressure of pure water at 35 oC is 422 Torr
Psolution = solventP solvent
χ H2O =mol H2O
total moles
Psolution = H2OP H2O
Psolution = (090)(422 Torr) = 38 Torr
χ H2O =45 moles
45 + 030 + 020 moles= 090
Colligative Properties of Solutions
bull Colligative Properties
Consequences
in bp and fp of
solution relative to
pure solvent
Vapor pressure
lowering
342019
15
Solute Concentration Molality
bull Changes in boiling pointfreezing point
of solutions depends on molality
ndash Preferred concentration unit for properties
involving temperature changes because it
is independent of temperature
ndash For typical solutions molality gt molarity
solute
kg of solvent
nm
Calculating Molality
Starting with
a) Mass of solute
b) Volume of solvent or
c) Volume of solution
342019
16
Practice Exercise p 483
Calculating the Molality of a Solution
What is the molality of a solution prepared by dissolving 782 g of ethylene glycol
HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water
is 100 gmL
120002 =mol solute
kg solvent
mol solute = 782 g x1 mol
6207 g= 12599 mol
kg solvent = 150 L x1000 mL
Lx10 g
mLx
10 kg
1000 g= 150 kg
120002 =12599 mol
150 kg= 0840120002
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
13
Colligative Properties of Solutions
bull Colligative Properties
ndash Set of properties of a solution relative to the pure
solvent
ndash Due to solutendashsolvent interactions
ndash Depend on concentration of solute particles not
the identity of particles
ndash Include lowering of vapor pressure boiling point
elevation freezing point depression osmosis
and osmotic pressure
Vapor Pressure of Solutions
bull Raoultrsquos Law for Solutions
bull Vapor pressure of solution is equal to the vapor
pressure of the pure solvent multiplied by the
mole fraction of solvent in the solution
bull Psolution = solventP solvent
bull Vapor pressure lowering
bull A colligative property of solutions (Section 115)
bull Ideal Solution
bull One that obeys Raoultrsquos law
342019
14
Homework Problem 1138
Calculating the Vapor Pressure of a Solution of One
or More Nonvolatile Substances
A solution contains 45 moles of H2O 030 moles of sucrose (C12H22O11) and
020 moles of glucose Sucrose and glucose are nonvolatile What is the mole
fraction of water in the solution What is the vapor pressure of the solution at
35 oC given that the vapor pressure of pure water at 35 oC is 422 Torr
Psolution = solventP solvent
χ H2O =mol H2O
total moles
Psolution = H2OP H2O
Psolution = (090)(422 Torr) = 38 Torr
χ H2O =45 moles
45 + 030 + 020 moles= 090
Colligative Properties of Solutions
bull Colligative Properties
Consequences
in bp and fp of
solution relative to
pure solvent
Vapor pressure
lowering
342019
15
Solute Concentration Molality
bull Changes in boiling pointfreezing point
of solutions depends on molality
ndash Preferred concentration unit for properties
involving temperature changes because it
is independent of temperature
ndash For typical solutions molality gt molarity
solute
kg of solvent
nm
Calculating Molality
Starting with
a) Mass of solute
b) Volume of solvent or
c) Volume of solution
342019
16
Practice Exercise p 483
Calculating the Molality of a Solution
What is the molality of a solution prepared by dissolving 782 g of ethylene glycol
HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water
is 100 gmL
120002 =mol solute
kg solvent
mol solute = 782 g x1 mol
6207 g= 12599 mol
kg solvent = 150 L x1000 mL
Lx10 g
mLx
10 kg
1000 g= 150 kg
120002 =12599 mol
150 kg= 0840120002
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
14
Homework Problem 1138
Calculating the Vapor Pressure of a Solution of One
or More Nonvolatile Substances
A solution contains 45 moles of H2O 030 moles of sucrose (C12H22O11) and
020 moles of glucose Sucrose and glucose are nonvolatile What is the mole
fraction of water in the solution What is the vapor pressure of the solution at
35 oC given that the vapor pressure of pure water at 35 oC is 422 Torr
Psolution = solventP solvent
χ H2O =mol H2O
total moles
Psolution = H2OP H2O
Psolution = (090)(422 Torr) = 38 Torr
χ H2O =45 moles
45 + 030 + 020 moles= 090
Colligative Properties of Solutions
bull Colligative Properties
Consequences
in bp and fp of
solution relative to
pure solvent
Vapor pressure
lowering
342019
15
Solute Concentration Molality
bull Changes in boiling pointfreezing point
of solutions depends on molality
ndash Preferred concentration unit for properties
involving temperature changes because it
is independent of temperature
ndash For typical solutions molality gt molarity
solute
kg of solvent
nm
Calculating Molality
Starting with
a) Mass of solute
b) Volume of solvent or
c) Volume of solution
342019
16
Practice Exercise p 483
Calculating the Molality of a Solution
What is the molality of a solution prepared by dissolving 782 g of ethylene glycol
HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water
is 100 gmL
120002 =mol solute
kg solvent
mol solute = 782 g x1 mol
6207 g= 12599 mol
kg solvent = 150 L x1000 mL
Lx10 g
mLx
10 kg
1000 g= 150 kg
120002 =12599 mol
150 kg= 0840120002
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
15
Solute Concentration Molality
bull Changes in boiling pointfreezing point
of solutions depends on molality
ndash Preferred concentration unit for properties
involving temperature changes because it
is independent of temperature
ndash For typical solutions molality gt molarity
solute
kg of solvent
nm
Calculating Molality
Starting with
a) Mass of solute
b) Volume of solvent or
c) Volume of solution
342019
16
Practice Exercise p 483
Calculating the Molality of a Solution
What is the molality of a solution prepared by dissolving 782 g of ethylene glycol
HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water
is 100 gmL
120002 =mol solute
kg solvent
mol solute = 782 g x1 mol
6207 g= 12599 mol
kg solvent = 150 L x1000 mL
Lx10 g
mLx
10 kg
1000 g= 150 kg
120002 =12599 mol
150 kg= 0840120002
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
16
Practice Exercise p 483
Calculating the Molality of a Solution
What is the molality of a solution prepared by dissolving 782 g of ethylene glycol
HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water
is 100 gmL
120002 =mol solute
kg solvent
mol solute = 782 g x1 mol
6207 g= 12599 mol
kg solvent = 150 L x1000 mL
Lx10 g
mLx
10 kg
1000 g= 150 kg
120002 =12599 mol
150 kg= 0840120002
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine Molar
Mass
Chapter Outline
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
17
copy 2012 by W W Norton amp Company
Colligative Properties of Solutions
Colligative Properties
bull Solution properties that depend on concentration
of solute particles not the identity of particles
Previous example
vapor pressure
lowering
Consequences
change in bp
and fp of solution
copy 2012 by W W Norton amp Company
Boiling-Point Elevation
and Freezing-Point Depression
Boiling Point Elevation (ΔTb)
bull ΔTb = Kb∙m
bull Kb = boiling point elevation
constant of solvent m = molality
Freezing Point Depression (ΔTf)
bull ΔTf = Kf∙m
bull Kf = freezing-point depression
constant m = molality
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
18
Sample Exercise 117 Calculating the
Freezing Point Depression of a Solution
What is the freezing point of an automobile radiator fluid prepared by mixing equal
volumes of ethylene glycol (MW=6207) and water at a temperature where the
density of ethylene glycol is 1114 gmL and the density of water is 1000 gmL
The freezing point depression constant of water Kf = 186 oCm
Tf = Kf m Assume 100 L of each = 1000 mL
kg H2O (solvent) = 1000 mL x100 g
mLx
1 kg1000 g
=100 kg H2O
mol EG (solute) = 1000 mL x1114 g x
ml1 mol6207 g
=1795 mol EG
Tf = Kf m = (186 oCm)(1795 m) = 334 oC
m = 1795 mol100 kg
= 1795 m
New fp = 00 oC ndash 334 oC = -334 oC
The vanrsquot Hoff Factor
Solutions of Electrolytes
bull Need to correct for number of particles formed when ionic substance dissolves
vanrsquot Hoff Factor (i)
bull number of ions in formula unit
bull eg NaCl i = 2
ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m
Deviations from theoretical value due to ion pair formation
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
19
copy 2012 by W W Norton amp Company
Values of vanrsquot Hoff Factors
Sample Exercise 119
Using the vanrsquot Hoff FactorSome Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to roasting them to help the birds
retain moisture and to season the meat Brining involves completely immersing a turkey for
about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl
MW=5844) in 76 L of water (dH2O = 100 gmL) Suppose a turkey soaking in such a
solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC
Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for
this NaCl solution
Tf = i Kf m
kg H2O (solvent) = 76 x 103mL x100 gmL
1199091 kg
1000 g= 76 kg H2O
mol NaCl (solute) = 910 g x1 mol5844 g
= 156 mol NaCl
m = 156 mol76 kg
= 205 m
Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC
New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
20
111 Energy Changes when Substances Dissolve
112 Vapor Pressure
113 Mixtures of Volatile Substances
114 Colligative Properties of Solutions
115 Osmosis and Osmotic Pressure
116 Using Colligative Properties to Determine
Molar Mass
Chapter Outline
Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
ΔTf or ΔTb
moles
MW = g samplemoles
Step 1
solve for molalitym = ΔTfKf
Step 2
solve for molesmolality
kg of solvent
Step 3
calc MW
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf
342019
21
Sample Exercise 1112 Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to
enhance recovery of petroleum and natural gas a process sometimes called ldquofrackingrdquo The
freezing point of a solution prepared by dissolving 100 mg of eicosene in 100 g of benzene is
175 oC lower that the freezing point of pure benzene What is the molar mass of eicosene
(Kf for benzene is 490 oCm)
Step 1
solve for molality=
175 oC
490 oCm= 0357 m
Step 2
solve for moles
moles solute = 0357 mol
kg benzeneX 000100 kg benzene
= 357 x 10-4 mol eicosene
Step 3
calc MWMW =
100 x 10minus3g eicosene
357 x 10minus4 mol= 280 gmol
m =∆T
Kf