Chapter 12
Solutions
From Chapter 1: Classification of matter
Homogeneous (Solutions)
(visibly indistinguishable)
Mixtures
(multiple components)
Heterogeneous
(visibly distinguishable)
Matter
Elements
Pure Substances
(one component)
Compounds
Solution = Solute + Solvent
Vodka = ethanol + water
Brass = copper + zinc
If solvent is water, the solution is called an aqueous solution.
Beer
Wine
Ethanol Concentration
Liquor
Four Concentrations
mass of solute
(1) Mass percent
 100%
mass of solution
Unit: none
molesof solute
(2) Molarity(M) 
liters of solution
Unit: mol/L
Four Concentrations
(3)
moles of solute
Mole fraction of solute 
moles of solution
χA 
nA
nA  nB
Mole fraction of solvent 
χB 
nB
nA  nB
Unit: none
moles of solvent
moles of solution
Unit: none
χA  χB  1
Four Concentrations
moles of solute
(4) Molaliy (m) 
kilograms of solvent
Unit: mol/kg
A solution contains 5.0 g of toluene (C7H8) and 225 g of
benzene (C6H6) and has a density of 0.876 g/mL.
Calculate the mass percent and mole fraction of C7H8, and
the molarity and molality of the solution.
Practice on Example 12.4 on page 533 and
compare your results with the answers.
Electrical Conductivity of Aqueous Solutions
salts
strong electrolyte
strong acids
strong bases
weak acids
solute
weak electrolyte
weak bases
nonelectrolyte
many organic compounds
van’t Hoff factor
moles of particles from solute
i
moles of solute dissolved
nonelectrolyte:
i=1
Unit: none
weak electrolyte: depends on degree of dissociation
strong electrolyte: depends on chemical formula
moles of particles from solute
i
moles of solute dissolved
MgBr2
Mg3(PO4)2
Hexane
MgSO4
NaOH
Glucose
FeCl3
Four properties of solutions
(1) Boiling point elevation
water = solvent
water + sugar = solution
Boiling point = 100 °C
Boiling point > 100 °C
Solution compared to pure solvent
Sugar Dissolved in
Water to Make Candy
Causes the Boiling
Point to be Elevated
∆Tb = Tb,solution − Tb,solvent = i Kb m
i: van’t Hoff factor
m: molality
Kb: boiling-point elevation constant
Units
Kb is characteristic of the solvent. Does not
depend on solute.
Boiling point elevation can be used to find molar mass of solute.
ΔTb
ΔTb  i Kb m  m 
iKb
∆Tb ― experiments
i ― electrolyte or nonelectrolyte
Kb ― table or reference book
molesof solute
ΔTb
m

kilogramof solvent i K b
mass of solute
molar mass of solute 
molesof solute
A solution was prepared by dissolving 18.00 g glucose in 150.0 g
water. The resulting solution was found to have a boiling point of
100.34 °C. Calculate the molar mass of glucose. Glucose is
molecular solid that is present as individual molecules in solution.
180 g/mol
molesof solute
ΔTb
m

kilogramof solvent i K b
mass of solute
molar mass of solute 
molesof solute
Four properties of solutions
(1) Boiling point elevation
(2) Freezing point depression
water = solvent
freezing point = 0 °C
water + salt = solution
freezing point < 0 °C
Solution compared to pure solvent
∆Tf = Tf,solvent − Tf,solution = i Kf m
i: van’t Hoff factor
m: molality
Kf: freezing-point depression constant
Units
Kf is characteristic of the solvent. Does not
depend on solute.
Freezing point depression can be used to find molar mass of solute
ΔTf
ΔTf  i Kf m  m 
i Kf
∆Tf ― experiments
i ― electrolyte or nonelectrolyte
Kf ― table or reference book
molesof solute
ΔTf
m

kilogramof solvent i K f
mass of solute
molar mass of solute 
molesof solute
A chemist is trying to identify a human hormone that controls
metabolism by determining its molar mass. A sample weighing
0.546 g was dissolved in 15.0 g benzene, and the freezing-point
depression was determined to be 0.240 °C. Calculate the molar
mass of the hormone.
776 g/mol
molesof solute
ΔTf
m

kilogramof solvent i K f
mass of solute
molar mass of solute 
molesof solute
The Addition of Antifreeze Lowers the Freezing Point
of Water in a Car's Radiator
water
0 °C
< 0 °C
100 °C
> 100 °C
antifreeze = water + ethylene glycol
Four properties of solutions
(1) Boiling point elevation
(2) Freezing point depression
(3) Osmotic pressure
Osmotic Pressure
Π = iMRT
Π ― osmotic pressuue
i ― van’t Hoff factor
M ― molarity
R ― ideal gas constant
T ― temperature
Π = iMRT
Units
Π ― atm
i ― none
M ― mol/L
R ― atm·L·K−1·mol−1
T―K
Osmotic pressure can be used to find molar mass of solute.

Π  iMRT  M 
iRT
Π ― experiments
i ― electrolyte or nonelectrolyte
R ― constant
T ― experiments
molesof solute

M

litersof solution i RT
mass of solute
molar mass of solute 
molesof solute
To determine the molar mass of a certain protein, 1.00 x 10−3 g
of it was dissolved in enough water to make 1.00 mL of solution.
The osmotic pressure of this solution was found to be 1.12 torr
at 25.0 °C. Calculate the molar mass of the protein.
1.66 x 104 g/mol
molesof solute

M

litersof solution i RT
mass of solute
molar mass of solute 
molesof solute
Practice on Example 12.10 on page 546 and
compare your results with the answers.
What concentration of NaCl in water is needed to produce an
aqueous solution isotonic with blood ( Π = 7.70 atm at 25 °C)?
0.158 mol/L
Four properties of solutions
(1) Boiling point elevation
(2) Freezing point depression
(3) Osmotic pressure
(4) Lowering the vapor pressure
Nonvolatile solute to volatile solvent
Lowering Vapor Pressure
The Presence of a Nonvolatile Solute Lowers the
Vapor Pressure of the Solvent
Surface Molecules
Liquid Surface
pure solvent
Liquid Surface
solvent + solute
When you count the number of solute particles, use
van’t Hoff factor i.
Four Concentrations
(3)
moles of solute
Mole fraction of solute 
moles of solution
χA 
nA
nA  nB
Mole fraction of solvent 
χB 
nB
nA  nB
Unit: none
moles of solvent
moles of solution
Unit: none
χA  χB  1
Raoult’s Law: Case 1
Nonvolatile solute in a Volatile solvent
Psolution  P
0
solvent
Psolution
0
solvent
solvent
― vapor pressure of solution
P
― vapor pressure of pure solvent
solvent
― mole fraction of solvent
For a Solution
that Obeys
Raoult's Law, a
Plot of Psoln
Versus Xsolvent,
Give a Straight
Line
Example 12.6, page 537
Calculate the vapor pressure at 25 °C of a solution containing
99.5 g of sucrose (C12H22O11) and 300 mL of water. The vapor
pressure of pure water at 25 °C is 23.8 torr. Assume the
density of water to be 1.00 g/mL.
23.4 torr
Predict the vapor pressure of a solution prepared by mixing
35.0 g solid Na2SO4 (molar mass = 142 g/mol) with 175 g
water at 25 °C. The vapor pressure of pure water at 25 °C is
23.76 torr.
Liquid Surface
solvent + solute
When you count the number of solute particles, use
van’t Hoff factor i.
Predict the vapor pressure of a solution prepared by mixing
35.0 g solid Na2SO4 (molar mass = 142 g/mol) with 175 g
water at 25 °C. The vapor pressure of pure water at 25 °C is
23.76 torr.
22.1 torr
Raoult’s Law: Case 2
Volatile solute in a Volatile solvent
Recall Dalton’s law of partial pressures
Psolution  PA  PB
 P A  P B
0
A
0
B
Vapor Pressure for a Solution of Two Volatile Liquids
XA + XB = 1
A mixture of benzene (C6H6) and toluene (C7H8) containing
1.0 mol of benzene and 2.0 mol of toluene. At 20 °C the vapor
pressures of pure benzene and toluene are 75 torr and 22 torr,
respectively. What is the vapor pressure of the mixture?
What is the mole fraction of benzene in the vapor?
Lowering vapor pressure can be used to find molar mass of solute.
Psolution  P
0
solvent
χ solvent  χ solvent
0
solvent
Psolution and P
χ solvent
Psolution
 0
Psolvent
― experiments
nsolvent
Psolution

 0
nsolvent  nsolute Psolvent
mass of solute
molar mass of solute 
molesof solute
Modified Example 12.6, page 537
At 25 °C of a solution is prepared by dissolving 99.5 g of
sucrose (nonelectrolyte, nonvolatile) into 300 mL of water. The
vapor pressure of the solution and pure water are 23.4 torr and
23.8 torr, respectively. Assume the density of water to be
1.00 g/mL. Calculate the molar mass of sucrose.
A solution that obeys Raoult’s Law is called an
ideal solution.
A solution is prepared by mixing 5.81 g acetone (molar mass =
58.1 g/mol) and 11.0 g chloroform (molar mass = 119.4 g/mol).
At 35 °C, this solution has a total vapor pressure of 260. torr.
Is this an ideal solution? The vapor pressure of pure acetone
and pure chloroform at 35 °C are 345 torr and 293 torr,
respectively.
What kind of solution is ideal?
pure solvent
10%
P0
χ
# of
molecules in vapor = 100 x 1 x 10% = 10
15%
solvent + solute
5%
10%
χ
Raoult’s law:
Deviate from
Raoult’s law
P0
Psln
# of
molecules in vapor = 100 x 0.8 x 10% = 8
# of
molecules in vapor = 100 x 0.8 x 5% = 4
# of
molecules in vapor = 100 x 0.8 x 15% = 12
What kind of solution is ideal?
Solute-solute, solvent-solvent, and solute-solvent
interactions are very similar.
Comparison to ideal gas.
Example 12.7, page 540
A solution contains 3.95 g of carbon disulfide (CS2) and 2.43
g of acetone (CH3COCH3). The vapor pressures at 35 C of
pure carbon disulfide and pure acetone are 515 torr and 332
torr, respectively. Assuming ideal behavior, calculate the
vapor pressures of each of the components and the total
vapor pressure above the solution. The experimentally
measured total vapor pressure of the solution at 35 C is 645
torr. Is the solution ideal? If not, what can you say about the
relative strength of carbon disulfide–acetone interactions
compared to the acetone–acetone and carbon disulfide–
carbon disulfide interactions?
Four Colligative properties of solutions
(1) Boiling point elevation:
∆Tb = i Kb m
(2) Freezing point depression:
∆Tf = i Kf m
(3) Osmotic pressure: Π = iMRT
0
P

P
(4) Lowering the vapor pressure: solution solvent solvent
Colligative: depend on the quantity (number of particles,
concentration) but not the kind or identity of the solute particles.