Chemistry, The Central Science, 11th edition
Theodore L. Brown, H. Eugene LeMay, Jr.,
and Bruce E. Bursten
Chapter 13
Properties of Solutions
John D. Bookstaver
St. Charles Community College
Cottleville, MO
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Overview of Chapter 13 Solutions
Review/On your own:
•
describe the three factors involved in solubility (solute-solute, solvent-solvent
and solute-solvent interactions)
•
explain what factors affect solubility and predict solubilities of solutes in solvents
•
define and apply the terms miscible and immiscible, saturated, unsaturated and
supersaturated solutions
•
use Henry’s Law to calculate the solubility or pressure of a gas in liquid solution
•
calculate solubilities (g/100 g H2O)
•
interpret solubility graphs
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Overview of Chapter 13 Solutions
New Material
• calculate concentration using mass percentage, ppm, ppb, mole
fraction, molarity and molality and convert amongst them
• define colligative properties
• use Raoult’s Law to calculate vapor pressures and mole
fractions of solutions
• calculate freezing point depression, boiling point elevation and
osmotic pressure
• calculate the molar mass from any of the four colligative
properties
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Solutions
• Solutions are homogeneous mixtures of two
or more pure substances.
• In a solution, the solute is dispersed uniformly
throughout the solvent.
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The intermolecular
forces between solute
and solvent particles
must be strong enough
to compete with those
between solute particles
and those between
solvent particles.
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How Does a Solution Form?
As a solution forms, the solvent pulls solute
particles apart and surrounds, or solvates,
them.
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How Does a Solution Form
If an ionic salt is
soluble in water, it is
because the iondipole interactions
are strong enough
to overcome the
lattice energy of the
salt crystal.
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Energy Changes in Solution
• Simply put, three
processes affect the
energetics of solution:
– separation of solute
particles,
– separation of solvent
particles,
– new interactions
between solute and
solvent.
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Energy Changes in Solution
The enthalpy
change of the
overall process
depends on H for
each of these steps.
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Why Do Endothermic
Processes Occur?
Things do not tend to
occur spontaneously
(i.e., without outside
intervention) unless
the energy of the
system is lowered.
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Why Do Endothermic
Processes Occur?
Yet we know the in
some processes,
like the dissolution
of NH4NO3 in water,
heat is absorbed,
not released.
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Enthalpy Is Only Part of the Picture
The reason is that
increasing the disorder
or randomness (known
as entropy) of a system
tends to lower the
energy of the system.
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Enthalpy Is Only Part of the Picture
So even though
enthalpy may increase,
the overall energy of
the system can still
decrease if the system
becomes more
disordered.
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Student, Beware!
Just because a substance disappears when it
comes in contact with a solvent, it doesn’t
mean the substance dissolved.
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Student, Beware!
• Dissolution is a physical change — you can get back the
original solute by evaporating the solvent.
• If you can’t, the substance didn’t dissolve, it reacted.
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Sample Exercise 13.1 (p. 534)
By the process illustrated below, water vapor
reacts with excess solid sodium sulfate to form
the hydrated form of the salt. The chemical
reaction is
Na2SO4(s) + 10 H2O(g)  Na2SO4• 10 H2O(s)
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Sample Exercise 13.1 (p. 534)
Essentially all of the water vapor in the
closed container is consumed in this
reaction. If we consider our system to
consist initially of Na2SO4(s) and 10 H2O(g)
a) does the system become more or less
ordered in this process, and
b) does the entropy of the system increase or
decrease?
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Practice Exercise 13.1
Does the entropy of the system
increase or decrease when the
stopcock is opened to allow mixing of
the two gases in the apparatus?
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Types of Solutions
• Saturated
– In a saturated solution,
the solvent holds as
much solute as is
possible at that
temperature.
– Dissolved solute is in
dynamic equilibrium
with solid solute
particles.
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Types of Solutions
• Unsaturated
– If a solution is
unsaturated, less
solute than can
dissolve in the
solvent at that
temperature is
dissolved in the
solvent.
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Types of Solutions
• Supersaturated
– In supersaturated solutions, the solvent holds
more solute than is normally possible at that
temperature.
– These solutions are unstable; crystallization can
usually be stimulated by adding a “seed crystal” or Solutions
scratching the side of the flask.
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Factors Affecting Solubility
• Chemists use the axiom “like dissolves like."
– Polar substances tend to dissolve in polar solvents.
– Nonpolar substances tend to dissolve in nonpolar
solvents.
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Factors Affecting Solubility
The more similar the
intermolecular
attractions, the more
likely one substance
is to be soluble in
another.
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Factors Affecting Solubility
Glucose (which has
hydrogen bonding)
is very soluble in
water, while
cyclohexane (which
only has dispersion
forces) is not.
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Factors Affecting Solubility
• Vitamin A is soluble in nonpolar compounds
(like fats).
• Vitamin C is soluble in water.
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Sample Exercise 13.2 (p. 539)
Predict whether each of the following
substances is more likely to dissolve in
carbon tetrachloride (CCl4) or in water:
C7H16, Na2SO4, HCl, and I2.
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Practice Exercise 13.2
Arrange the following substances in
order of increasing solubility in water:
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Gases in Solution
• In general, the
solubility of gases in
water increases with
increasing mass.
• Larger molecules
have stronger
dispersion forces.
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Gases in Solution
• The solubility of
liquids and solids
does not change
appreciably with
pressure.
• The solubility of a
gas in a liquid is
directly proportional
to its pressure.
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Henry’s Law
Sg = kPg
where
• Sg is the solubility of
the gas,
• k is the Henry’s Law
constant for that gas in
that solvent, and
• Pg is the partial
pressure of the gas
above the liquid.
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Sample Exercise 13.3 (p. 540)
Calculate the concentration of CO2 in a
soft drink that is bottled with a partial
pressure of CO2 of 4.0 atm over the
liquid at 25oC. The Henry’s Law
constant for CO2 in water at this
temperature is 3.1 x 10-2 mol/L-atm.
(0.12 M)
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Practice Exercise 13.3
Calculate the concentration of CO2 in a
soft drink after the bottle is opened and
equilibrates at 25oC under a CO2 partial
pressure of 3.0 x 10-4 atm.
(9.3 x 10-6 M)
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Temperature
Generally, the
solubility of solid
solutes in liquid
solvents increases
with increasing
temperature.
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Temperature
• The opposite is true
of gases.
– Carbonated soft
drinks are more
“bubbly” if stored in
the refrigerator.
– Warm lakes have
less O2 dissolved in
them than cool lakes.
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Ways of
Expressing
Concentrations
of Solutions
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Ways of Expressing
Concentration
• All methods involve quantifying the
amount of solute per amount of solvent,
or solution.
• Dilute and concentrated are qualitative
ways to describe concentration.
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Quantitative Expressions of
Concentration
• Require specific information regarding
such quantities as
• Masses
• Moles
• Liters of the solute, solvent, or solution
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Quantitative Expressions of
Concentration
• Mass percentage, which includes ppm,
ppb
• Mole fraction
• Molarity
• Molality
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Mass Percentage
mass of A in solution
 100
Mass % of A =
total mass of solution
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Parts per Million and
Parts per Billion
Parts per Million (ppm) = mg solute/kg solution
mass of A in solution
 106
ppm =
total mass of solution
Parts per Billion (ppb) = mg solute/kg solution
mass of A in solution
 109
ppb =
total mass of solution
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Sample Exercise 13.4 (p. 542)
a) A solution is made by dissolving 13.5 g
glucose (C6H12O6) in 0.100 kg of water.
What is the mass percentage of solute in
this solution?
(11.9%)
b) A 2.5-g sample of groundwater was found to
contain 5.4 mg of Zn2+. What is the
concentration of Zn2+ in parts per million?
(2.2 ppm)
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Practice Exercise 13.4
a) Calculate the mass percentage of NaCl in a
solution containing 1.50 g of NaCl in 50.0 g
of water.
(2.91%)
b) A commercial bleaching solution contains
3.62 mass % sodium hypochlorite, NaOCl.
What is the mass of NaOCl in a bottle
containing 2500 g of bleaching solution?
(90.5 g NaOCl)
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Mole Fraction, Molarity and
Molality
• Common expressions of concentration
based on the number of moles of one or
more components
• Remember: mass can be converted to
moles using MM
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Mole Fraction (X)
moles of A
XA =
total moles in solution
• In some applications, one needs the mole
fraction of solvent, not solute — make sure
you find the quantity you need!
• Note: mole fraction has no units.
• Note: mole fractions range from 0 to 1
• Particularly useful for gases
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Molarity (M)
M=
mol of solute
L of solution
• You will recall this concentration
measure from Chapter 4.
• Since volume is temperaturedependent, molarity can change with
temperature.
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Molality (m)
m=
mol of solute
kg of solvent
Since both moles and mass do not
change with temperature, molality
(unlike molarity) is not temperaturedependent.
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Sample Exercise 13.5 (p. 544)
A solution is made by dissolving 4.35 g
glucose (C6H12O6) in 25.0 mL of water.
Calculate the molality of glucose in the
solution. (Assume DH2O = 1.00 g/mL)
(0.964 m)
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Practice Exercise 13.5
What is the molality of a solution made
by dissolving 36.5 g of naphthalene
(C10H8) in 425 g of toluene (C7H8)?
(0.671 m)
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Where we are going next
• Problems that involve conversions
between different types of units and
other applications that require
preparatory steps
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Changing Molarity to Molality
If we know the
density of the
solution, we can
calculate the
molality from the
molarity and vice
versa.
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Sample Exercise 13.6 (p. 544)
A solution of hydrochloric acid contains 36%
HCl by mass.
a) Calculate the mole fraction of HCl in
the solution.
(0.22)
b) Calculate the molality of HCl in the
solution.
(15 m)
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Practice Exercise 13.6
A commercial bleach solution contains
3.62 mass % NaOCl in water.
Calculate
a) the mole fraction of NaOCl in the
solution and
(9.00 x 10-3)
b) the molality of NaOCl in the solution
(0.505 m)
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Sample Exercise 13.7 (p. 546)
A solution with a density of 0.876 g/mL
contains 5.0 g of toluene (C7H8) and
225 g of benzene.
Calculate the molarity of the solution.
(0.21 M)
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Practice Exercise 13.7
A solution containing equal masses of
glycerol (C3H8O3) and water has a
density of 1.10 g/mL. Calculate
a) the molality of glycerol; (10.9 m)
b) the mole fraction of glycerol; (0.163)
c) the molarity of glycerol in the solution.
(5.97 M)
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Warmup
Caffeine (C8H10N4O2) is a stimulant found in
coffee and tea. If a solution of caffeine in
chloroform (CHCl3) as a solvent has a
concentration of 0.0750 m, calculate
a) the percent caffeine by mass (1.44%)
b) the mole fraction of caffeine (0.00887)
(p. 567, Q# 13.54)
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Overview
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• Colligative properties, especially to find
MM, including Friday’s lab
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Colligative Properties
• Changes in colligative properties depend only
on the number of solute particles present, not
on the identity of the solute particles.
• Physical properties
• “depending on the collection”, i.e.
collective effects of the solute particles
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Colligative Properties
Application of molality
Examples:
C12H22O11  C12H22O11
NaCl  Na+ + ClCaCl2  Ca2+ + 2 Cl-
one particle
two particles
three particles
This will have the greatest effect
(of these examples)
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Colligative Properties
• Among colligative properties are
– Vapor pressure lowering
– Boiling point elevation*
– Melting point depression* - Friday’s lab
– Osmotic pressure*
*quantitative
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Vapor Pressure
Because of solutesolvent intermolecular
attraction, higher
concentrations of
nonvolatile solutes
make it harder for
solvent to escape to
the vapor phase.
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Vapor Pressure
Focus on the solvent
e.g. NaCl dissolved in H2O –
lower water VP than pure
H2O
Lower VP because the Na+
and Cl- particles
(nonvolatile solutes) get
surrounded by H2O 
fewer “free” H2O particles
left to vaporize at surface
Stronger IMFs  harder for
solvent molecules to leave
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Vapor Pressure
The vapor pressure of
a solution is lower
than that of the pure
solvent.
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Raoult’s Law
For an ideal solution:
PA = XAPA
where
– XA is the mole fraction of compound A, and
– PA is the normal vapor pressure of A at that
temperature.
NOTE: This is one of those times when you want to
make sure you have the vapor pressure of the
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Raoult’s Law
Real solutions show approximately ideal
behavior when:
- The solute concentration is low
- The solute and solvent have similarly sized
molecules
- The solute and solvent have similar types of
IMFs
- Raoult’s Law breaks down when the solventsolvent and solute-solute IMFs are >>> or
<<< solute-solvent IMFs.
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Sample Exercise 13.8 (p. 547)
Glycerin (C3H8O3) is a nonvolatile
nonelectrolyte with a density of 1.26 g/mL at
25oC. Calculate the vapor pressure at 25oC
of a solution made by adding 50.0 mL of
glycerin to 500.0 mL of water. The vapor
pressure of pure water at 25oC is 23.8 torr.
(23.2 torr)
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Practice Exercise 13.8
The vapor pressure of pure water at 110oC is
1070 torr. A solution of ethylene glycol and
water has a vapor pressure of 1.00 atm at
110oC. Assuming that Raoult’s law is obeyed,
what is the mole fraction of ethylene glycol in
the solution?
(rearrange equation to find mole fraction)
(0.290)
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Boiling Point Elevation and
Freezing Point Depression
Nonvolatile solutesolvent interactions
also cause solutions
to have higher boiling
points and lower
freezing points than
the pure solvent.
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Boiling Point Elevation
 # particles of nonvolatile solute   VP
 boiling point
- Since bp = VP at atm P, if the VP is lower,
more energy must be used to overcome atm P
= higher T at boiling
- It takes more energy to break apart solvent/solute bonds
to allow the solvent to vaporize.
-
For H2O: 1 mole particles = + 0.512oC (molal bp elevation constant)
1 kg H2O
Note: 1 m solution of NaCl is 2 m in total solute particles
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Boiling Point Elevation
• The change in boiling
point is proportional to
the molality of the
solution:
Tb = Kb  m
Tb is added to the normal
boiling point of the solvent.
where Kb is the molal
boiling point elevation
constant, a property of
the solvent.
m = molality
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Boiling Point Elevation
• The change in freezing
point can be found
similarly:
Tf = Kf  m
• Here Kf is the molal
freezing point
depression constant of
the solvent.
Tf is subtracted from the normal
boiling point of the solvent.
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Freezing Point Depression
 # particles of nonvolatile solute   freezing point
- When a solution freezes, crystals of almost pure solvent form
first
- Solute molecules are usually not soluble in the solid phase of
the solvent, so become more concentrated in the liquid phase
  VP
 triple point occurs at a lower T because of the lower VP for
the solution.
-
-
The melting-point (freezing-point) curve is a vertical line from the
triple point
 the solution freezes at a lower temperature than the pure
solvent
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Boiling Point Elevation and
Freezing Point Depression
Note that in both
equations, T does
not depend on what
the solute is, but
only on how many
particles are
dissolved.
Tb = Kb  m
Tf = Kf  m
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Boiling Point Elevation and
Freezing Point Depression
e.g. What is the boiling point of a solution that
contains 1.25 mol CaCl2 in 1400. g of water?
mol CaCl2 = 1.25 mol CaCl2 x 1000. g = 0.893 m x 3 = 2.68 m
1000. g H2O 1400. g H2O
1 kg
= # particles when CaCl2
dissociates in H2O
Tb = Kb  m
= 0.512oC x 2.68 m = 1.37oC + 100.oC = 101.37oC
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Sample Exercise 13.9 (p. 550)
Automotive antifreeze consists of
ethylene glycol (C2H6O2), a nonvolatile
nonelectrolyte. Calculate the boiling
point and freezing point of a 25.0 mass
% solution of ethylene glycol in water.
(102.7oC; -10.0oC)
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Practice Exercise 13.9
Calculate the freezing point of a solution
containing 0.600 kg of chloroform
(CHCl3) and 42.0 g of eucalyptol
(C10H18O), a fragrant substance found
in the leaves of eucalyptus trees.
(See Table 13.4)
(-65.6oC)
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Sample Exercise 13.10 (p. 551)
List the following aqueous solutions in
order of their expected freezing points:
0.050 m CaCl2; 0.15 m HCl; 0.050 m
HC2H3O2; 0.10 m C12H22O11.
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Practice Exercise 13.10
Which of the following solutes will
produce the largest increase in boiling
point upon addition of 1 kg of water:
1 mol Co(NO3)2, 2 mol of KCl, 3 mol of
ethylene glycol (C2H6O2)?
(2 mol KCl)
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Check in
End of last semester
colligative properties, esp. freezing point
depression and boiling point elevation
Calculations: predict new bp or fp
Next: Colligative props of electrolytes
Osmotic pressure
How to determine molar mass using colligative
properties
- freezing point depression lab
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Colligative Properties of
Electrolytes
Since these properties depend on the number of
particles dissolved, solutions of electrolytes (which
dissociate in solution) should show greater changes
than those of nonelectrolytes.
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Colligative Properties of
Electrolytes
However, a 1M solution of NaCl does not show
twice the change in freezing point that a 1M
solution of methanol does.
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van’t Hoff Factor
One mole of NaCl in
water does not
really give rise to
two moles of ions.
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van’t Hoff Factor
Some Na+ and Clreassociate for a short
time, so the true
concentration of
particles is somewhat
less than two times the
concentration of NaCl.
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van’t Hoff Factor
• Reassociation is
more likely at higher
concentration.
• Therefore, the
number of particles
present is
concentrationdependent.
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van’t Hoff Factor
• We modify the
previous equations
by multiplying by the
van’t Hoff factor, i.
Tf = Kf  m  i
(is is used
incorrectly in your
handout on p. 18)
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Osmosis
• Some substances form semipermeable
membranes, allowing some smaller
particles to pass through, but blocking
other larger particles.
• In biological systems, most
semipermeable membranes allow water
to pass through, but solutes are not free
to do so.
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Osmosis
In osmosis, there is net movement of solvent
from the area of higher solvent
concentration (lower solute concentration) to
the are of lower solvent concentration
(higher solute concentration).
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Osmotic Pressure
The pressure required to stop osmosis,
known as osmotic pressure, , is
=(
n
)
RT = MRT
V
where M is the molarity of the solution.
If the osmotic pressure is the same on both sides
of a membrane (i.e., the concentrations are the
same), the solutions are isotonic.
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Osmosis in Blood Cells
• If the solute
concentration outside
the cell is greater than
that inside the cell, the
solution is hypertonic.
• Water will flow out of
the cell, and crenation
results.
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Osmosis in Cells
• If the solute
concentration outside
the cell is less than
that inside the cell, the
solution is hypotonic.
• Water will flow into the
cell, and hemolysis
results.
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Sample Exercise 13.11 (p. 553)
(on your own)
The average osmotic pressure of blood
is 7.7 atm at 25oC. What concentration
of glucose (C6H12O6) will be isotonic
with blood?
(0.31 M)
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Practice Exercise 13.11
(on your own)
What is the osmotic pressure at 20oC of
a 0.0020 M sucrose (C12H22O11)
solution?
(0.048 atm or 37 torr)
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Using Colligative Properties to
Determine Molar Mass
• Any of the four colligative properties
may be used to determine molar mass
• Thursday’s lab
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Sample Exercise 13.12 (p. 555)
A solution of an unknown nonvolatile
nonelectrolyte was prepared by dissolving
0.250 g of the substance in 40.0 g of CCl4.
The boiling point of the resultant solution was
0.357oC higher than that of the pure solvent.
Calculate the molar mass of the solute.
(88.0 g/mol)
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Practice Exercise 13.12
Camphor (C10H16O) melts at 179.8oC, and it
has a particularly large freezing-pointdepression constant, Kf = 40.0oC/m. When
0.186 g of an organic substance of unknown
molar mass is dissolved in 22.01 g of liquid
camphor, the freezing point of the mixture is
found to be 176.7oC. What is the molar mass
of the solute?
(110 g/mol)
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Sample Exercise 13.13 (p. 556)
The osmotic pressure of an aqueous solution
of a certain protein was measured in order to
determine its molar mass. The solution
contained 3.50 mg of protein dissolved in
sufficient water to form 5.00 mL of solution.
The osmotic pressure of the solution at 25oC
was found to be 1.54 torr. Calculate the
molar mass of the protein.
(8.45 x 103 g/mol)
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Practice Exercise 13.13
A sample of 2.05 g of the plastic
polystyrene was dissolved in enough
toluene to form 0.100 L of solution. The
osmotic pressure of this solution was
found to be 1.21 kPa at 25oC.
Calculate the molar mass of the
polystyrene.
(4.20 x 104 g/mol)
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Colloids
(review on your own)
Suspensions of particles larger than
individual ions or molecules, but too small to
be settled out by gravity are called colloids.
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Tyndall Effect
• Colloidal suspensions
can scatter rays of light.
• This phenomenon is
known as the Tyndall
effect.
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Colloids in Biological Systems
Some molecules have
a polar, hydrophilic
(water-loving) end and
a non-polar,
hydrophobic (waterhating) end.
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Colloids in Biological Systems
Sodium stearate
is one example
of such a
molecule.
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Colloids in Biological Systems
These molecules
can aid in the
emulsification of fats
and oils in aqueous
solutions.
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Sample Integrative Exercise 13
(p. 560)
A 0.100-L solution is made by dissolving 0.441 g of
CaCl2(s) in water.
a) Calculate the osmotic pressure of this solution at
27oC, assuming that it is completely dissociated into
its component ions.
(2.93 atm)
b) The measured osmotic pressure of this solution is
2.56 atm at 27oC. Explain why it is less than the
value calculated in (a), and calculate the van’t Hoff
factor, i, for the solute in this solution.
(2.62)
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Sample Integrative Exercise 13
(p. 560)
c) The enthalpy of solution for CaCl2 is H = 81.3 kJ/mol. If the final temperature of the
solution was 27.0oC, what was its initial
temperature? (Assume that the density of the
solution is 1.00 g/mL, that its specific heat is
4.18 J/g-K, and that the solution loses no heat
to its surroundings.)
(26.2oC)
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