Slide 1
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Colligative Properties
Practical uses of solutions
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Slide 2
Units of Concentration
Whatever units you use, the goal is the
same: specify the quantity of 1
component (the solutes) relative to the
quantity of another component (the
solvent).
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Slide 3
Common Units
% by mass
% by volume
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Mole %
ppm (parts per million)
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Molarity (M)
Molality (m)
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Slide 4
Common Units
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Slide 5
Solute as part of a solution
Note that, with the exception of molality,
all of the units of concentration are
expressed as some amount of solute
compared to some amount of solution.
 All the units of conversation are easily
convertible, although sometimes you may
need to know another piece of information
(molar mass, density, etc.)

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Slide 6
Molarity
Molarity is probably the most common unit
of concentration in chemistry.
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Why is that?
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Slide 7
Molarity
Molarity is probably the most common unit
of concentration in chemistry.
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Why is that?
UNITS! UNITS! UNITS!
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Slide 8
Molarity
Molarity (M) = moles solute
L solution
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This is both a chemically relevant unit and a practically
relevant unit.
L solution is very easy to measure in the lab!
MOLES! MOLES! MOLES!
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Reactions occur based on the relative number of moles.
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Slide 9
% by mass
This is one of the easiest units to use.
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Why?
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Slide 10
% by mass
This is one of the easiest units to use.
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Why?
UNITS! UNITS! UNITS!
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Mass is easy to measure in the lab!
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Slide 11
% by mass
% by mass = g solute
100 g solution
This is easy to use, but not as chemically relevant
as Molarity. After all, it’s all about the MOLES!
MOLES! MOLES!
Is this a problem?
Not at all, it is easy to convert these units, but you
need to know a few other things.
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Slide 12
Converting M to % by mass
% by mass = g solute
100 g solution
M = moles solute
L solution
What do we need to know to be able to convert
from one to the other?
UNITS! UNITS! UNITS! They shall lead the way.
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Slide 13
Converting M to % by mass
Compare the units:
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g solute
100 g solution
Moles solute
Liter solution
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What do you see?
Both numerators refer only to solute.
Both denominators refer to the whole solution.
PLEASE! PLEASE! PLEASE! Pay attention to the entire
unit – it’s not “grams” it’s “grams of something”
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Slide 14
Converting M to % by mass
g solute
100 g solution
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Moles solute
Liter solution
If I want to convert the numerators, what do I need to do?
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Slide 15
Converting M to % by mass
g solute
100 g solution
Moles solute
Liter solution
If I want to convert the numerators, what do I need to do?
(Do I need to say UNITS! UNITS! UNITS! again)?
g solute
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 moles solute
What’s the conversion factor?
g solute
Or
moles solute
moles solute
g solute
What do we call this?
Molar mass!
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Slide 16
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Converting M to % by mass
g solute
100 g solution
x moles solute = moles solute
g solute
100 g solution
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Or, if you’re going the other way:
moles solute
L solution
x g solute =
moles solute
g solute
L solution
What about the denominator?
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Slide 17
Converting M to % by mass
moles solute
100 g solution
g solute
Liter solution
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100 g solution  L solution
What’s the conversion factor?
g solution OR
L solution
L solution
g solution.
What’s this called?
Density! We need to know the density of the
solution and the molar mass of the solute to be
able to convert M to % by mass.
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Slide 18
Sample problem
You have a 1.2 M solution of NaCl in water.
What is its % by mass given that the
density of 1.2 M NaCl is 1.08 g/mL?
What do we do?
Simply convert the units!
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Slide 19
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Sample problem
You have a 1.2 M solution of NaCl in water.
What is its % by mass given that the
density of 1.2 M NaCl is 1.08 g/mL?

1.2 mol NaCl
L solution
g NaCl
100 g solution
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Slide 20
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Sample problem
You have a 1.2 M solution of NaCl in water.
What is its % by mass given that the
density of 1.2 M NaCl is 1.08 g/mL?
1.2 mol NaCl x 58.453 g NaCl
L solution
mol NaCl
=
70.143 g NaCl
L solution
70.143 g NaCl
L solution
= ? g NaCl
100 g solution
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We have density in g/mL, but we have volume in L, so…?
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Slide 21
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Sample problem
You have a 1.2 M solution of NaCl in water.
What is its % by mass given that the
density of 1.2 M NaCl is 1.08 g/mL?
1.2 mol NaCl x 58.453 g NaCl
L solution
mol NaCl
70.143 g NaCl
L solution
=
70.143 g NaCl
L solution
x 1 L solution
x 1 mL solution
= 0.06495 g NaCl
1000 mL solution
1.08 g solution
g solution
To get from g solution to 100 g solution???
Just multiply by 100.
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Slide 22
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Sample problem
You have a 1.2 M solution of NaCl in water.
What is its % by mass given that the
density of 1.2 M NaCl is 1.08 g/mL?
1.2 mol NaCl x 58.453 g NaCl
L solution
mol NaCl
70.143 g NaCl x 1 L solution x
L solution
1000 mL solution
=
1 mL solution
1.08 g solution
70.143 g NaCl
L solution
= 0.06495 g NaCl * 100
g solution
6.495 g NaCl/100 g solution OR 6.495% by mass
(6.5% by mass to the correct number of sig figs.)
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Slide 23
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Slightly more complicated problem
You have a 2.5 M solution of NaCl in water. What is its % by
mass given that the density of 1.0 M NaCl is 1.07 g/mL, the
density of 3.0 M NaCl is 1.10 g/mL and the density of 5.0 M
NaCl is 1.13 g/mL?
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This is a more common real-life example. There are tables of
solution densities, but they don’t have an infinite number of
values. So, what do we do…?
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Slide 24
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Slightly more complicated problem
You have a 2.5 M solution of NaCl in water. What is its % by
mass given that the density of 1.0 M NaCl is 1.07 g/mL, the
density of 3.0 M NaCl is 1.10 g/mL and the density of 5.0 M
NaCl is 1.13 g/mL?
2 options:
 Use the 3.0 M as being “close enough”.
 Do a linear interpolation to determine the approximate
value at 2.5 M.
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What is a “linear interpolation”?
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Slide 25
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Linear interpolation
Assume that the data should have a linear
trend, plot the line of the data you do have
and find the data you don’t have?
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Slide 26
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Linear interpolation
Density (g/mL)
Assume that the data should have a linear trend, plot the line of
the data you do have and find the data you don’t have?
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1.14
1.13
1.12
1.11
1.1
1.09
1.08
1.07
1.06
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0
1
2
3
4
5
6
Molarity
So, the density appears to be about 1.09 g/mL at 2.5 M.
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Slide 27
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The rest of the problem is the same.
You have a 2.5 M solution of NaCl in water. What is its % by
mass given that the density of 1.0 M NaCl is 1.06 g/mL, the
density of 3.0 M NaCl is 1.09 g/mL and the density of 5.0 M
NaCl is 1.13 g/mL?
2.5 mol NaCl x 58.453 g NaCl
L solution
mol NaCl
146.13 g NaCl
L solution
=
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146.13 g NaCl
L solution
x 1 L solution
x 1 mL solution
= 0.134 g NaCl * 100
1000 mL solution
1.09 g solution
g solution
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=13.4 g NaCl = 13.4% NaCl by mass
100 g solution
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Slide 28
Colligative Property
A colligative property is a property that
depends ONLY on the amount of the
substance present NOT on the identity of
the substance.
In other words, it doesn’t matter if it is salt,
sugar, gasoline, or tennis balls – it will
behave the same way!
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Slide 29
Solution

homogeneous mixtures


composition may vary from one sample to another
appears to be one substance, though really contains
multiple materials

most homogeneous materials we encounter
are actually solutions

nature has a tendency toward spontaneous
mixing


e.g., air and sea water
generally, uniform mixing is more energetically
favorable
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Slide 30
Solutions




solute is the dissolved substance
 seems to “disappear”
 “takes on the state” of the
solvent
solvent is the substance solute
dissolves in
 does not appear to change
state
when both solute and solvent
have the same state, the
solvent is the component
present in the highest
percentage
solutions in which the solvent is
water are called aqueous
solutions
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Slide 31
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..\chem152\dissolve NaCl.MOV
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Slide 32
Intermolecular Forces and the Solution Process
Enthalpy of Solution


energy changes in the formation of most solutions
also involve differences in attractive forces
between particles
must overcome solute-solute attractive forces



endothermic
at least some of the energy to do this comes from
making new solute-solvent attractions

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endothermic
must overcome some of the solvent-solvent
attractive forces

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exothermic
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Slide 33
Intermolecular Attractions
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Slide 34
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Relative Interactions
and Solution Formation
Solute-to-Solute
+
Solvent-toSolvent
Solute-to-Solvent
>
Solute-to-Solvent
Solute-to-Solute
+
=
Solvent-toSolvent
Solution Forms
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Solution Forms
when the solute-to-solvent
attractions are weaker
Solute-to-Solute
Solution
May
than the sum of the+solute-to-solute and
solventSolute-to-Solvent
<
May
Not if
to-solvent attractions,
the solution will or
only
form
Solvent-toForm
the energy difference
is
small
enough
to
be
Solvent
overcome by the entropy

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Slide 35
Solution Interactions
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Slide 36
Energetics of Solution Formation




overcome attractions between the solute particles
– endothermic
overcome some attractions between solvent
molecules – endothermic
for new attractions between solute particles and
solvent molecules – exothermic
the overall DH depends on the relative sizes of the
DH for these 3 processes
DHsol’n = Dhsolute+ DHsolvent + DHmix
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Slide 37
Solution Process
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1. add energy in to overcome solute-solute attractions
3. form new solute-solvent attractions, releasing energy
2. add energy in to overcome some solvent-solvent attractions
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Slide 38
Energetics of Solution Formation
if the total energy cost for
breaking attractions between
particles in the pure solute and
pure solvent is greater
less thanthan
the the
energy released in making the
new attractions between the
solute and solvent, the overall
process will be endothermic
exothermic
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Slide 39
Heats of Hydration

for aqueous ionic solutions, the energy added to
overcome the attractions between water
molecules and the energy released in forming
attractions between the water molecules and ions
is combined into a term called the heat of
hydration



attractive forces in water = H-bonds
attractive forces between ion and water = ion-dipole
DHhydration = heat released when 1 mole of gaseous ions
dissolves in water
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Slide 40
Heat of Hydration
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Slide 41
Some Examples:

Vapor Pressure Reduction

Freezing Point Depression
 Related to boiling point
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 Salt on the road
 Anti-freeze in your radiator

Boiling Point Elevation

Osmotic Pressure
 Anti-freeze in your radiator
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 Membrane diffusion
 The Great Sugar Fountain!
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Slide 42
Vapor Pressure Reduction
What is “vapor pressure”?
Vapor pressure is the amount (P is
proportional to n for ideal gases) of gas A
that is in equilibrium above the surface of
liquid A.
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Slide 43
Vapor Pressure
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At equilibrium, the rate of
evaporation (liquid to gas)
equals the rate of
condensation (gas to liquid).
The amount of gas is the
“vapor pressure”
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Surface of liquid
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Slide 44
What if you add a solute?
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At equilibrium, the rate of
evaporation still must equal
the rate of condensation. But
at any given temperature, the
# of solvent molecules at the
surface is decreased and,
therefore, so is the vapor
pressure
Surface of liquid
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Slide 45
Vapor Pressure of Solutions

the vapor pressure of a solvent above a
solution is lower than the vapor pressure
of the pure solvent

the solute particles replace some of the
solvent molecules at the surface
Eventually,
reAddition
of aequilibrium
nonvolatileissolute
The pure solvent
establishes
an
established,
smaller
number
reduces thebut
ratea of
vaporization,
liquidmolecules
 vapor equilibrium
ofdecreasing
vapor
– therefore
the
the amount
of vapor
vapor pressure will be lower
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Slide 46
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Thirsty Solutions
a concentrated solution will draw solvent
molecules toward it due to the natural
drive for materials in nature to mix
 similarly, a concentrated solution will draw
pure solvent vapor into it due to this
tendency to mix
 the result is reduction in vapor pressure

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Slide 47
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Thirsty Solutions
When equilibrium
Beakers
with equalis
liquid levels the
established,
of pure
liquid
solvent
level
in and
the solution
a solution
are place
beaker
is in
higher
a bellthan
jar.
Solvent
the
solution
molecules
level in the
evaporate
pure
solvent
from
beaker
each–
onethirsty
the
and fillsolution
the bell jar,
establishing
grabs
and holds
an solvent
equilibrium
vapor
more effectively
with the
liquids in the beakers.
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Slide 48
Raoult’s Law

the vapor pressure of a volatile solvent
above a solution is equal to its mole
fraction of its normal vapor pressure, P
Psolvent in solution = csolvent∙P

since the mole fraction is always less than 1,
the vapor pressure of the solvent in solution
will always be less than the vapor pressure of
the pure solvent
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Slide 49
Ex 12.5 – Calculate the vapor pressure of water in a
solution prepared by mixing 99.5 g of C12H22O11 with
300.0 mL of H2O
Given:
Find:
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99.5 g C12H22O11, 300.0 mL H2O
PH2O
mol A
mol C12H22O11 c  mol
Concept g C12H22O11
Plan:
mL H2O
P
H2O
g H2O
P  c  P
CH2O
PH2O
= 23.8 torr, 1 mol C12H22O11 = 342.30 g, 1 mol H2O =
Relationshi 18.02 g
ps:
Solve:
99.5cg C12 H 22O11 
116
mol
12 H H
22OO
11
.65Cmol
2
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total
mol H2O
 0.2907 mol
12 H
 0C.98
222
8 O11
H 22O11
0.2907 mol342.30
C12 H g22C
O12
11  16.65 mol H 2 O 
1.00 g 1 mol H 2O
300.0 mL H 2O 

 16.65 mol H 2O
PH 2O  c H 2O 1 mL
P  H 2O18.02
 0.g98
H228O23.8 torr 
PH 2O  23.4 torr
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Slide 50
Boiling Point Elevation
What is the “boiling point” of a liquid?
It is the temperature at which the vapor pressure
equals the atmospheric pressure. So…
…if you decrease the vapor pressure, you must
increase the boiling point – it will take a higher
temperature to get enough gas molecules (vapor
pressure) to equal the atmospheric pressure.
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Slide 51
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Boiling Pt. Elevation
Freezing Pt. Depression
Birds of a Feather:
ΔTb = T
boiling, solution
ΔTf = T
freezing, solution
–T
–T
boiling, pure solvent
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= Kb m
freezing, pure solvent
= - Kf m
m = molality of the solution
Kb = boiling constant
Kf = cryoscopic constant
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Slide 52
NOTE ABOUT THE SIGN
CONVENTION
ΔTb = T
boiling, solution
ΔTf = T
freezing, solution
–T
–T
boiling, pure solvent
= Kb m
freezing, pure solvent
= - Kf m
Δ = change
Δ = final state – initial state
Many texts do not put a minus sign before the cryscopic
constant and reverse the definition of ΔTf instead.
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Slide 53
K is a SOLVENT property
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Colligative properties don’t care what the
solute is – they only care how much solute
there is.
___________________________________
The boiling point constant and cryoscopic
constant are SOLVENT properties only.
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Slide 54
A simple problem
How many grams of sucrose (C11H22O11) are
needed to lower the freezing point of 100
g of water by 3 C?
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Slide 55
The Answer
We want to decrease the freezing point by 3 C.
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NOTE: Kf is the WATER cryoscopic
constant
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Slide 56
C11H22O11 - sucrose
Or, my preference is to use concentration as a conversion
factor:
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Slide 57
Kf
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Slide 58
Ex 12.8 – What is the freezing point of a 1.7 m aqueous
ethylene glycol solution, C2H6O2?
Given:
Find:
Concept
Plan:
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1.7 m C2H6O2(aq)
Tf, C
m
DT f  m  K f
DTf
DTf = m ∙Kf, Kf for H2O = 1.86 C/m, FPH2O =
Relationships 0.00 C
Solve:: DT  m  K
FPH 2O  FPsol'n  DT f
f
f ,H 2O

 1.7 m  1.86
C
m

DT f  3.2 C
0.00C  FPsol' n  3.2C
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FPsol'n  3.2C
Check: the unit is correct, the freezing point lower than
the normal freezing point makes sense
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Slide 59
Colligative Properties

colligative properties are properties whose value
depends only on the number of solute particles,
and not on what they are



Vapor Pressure Depression, Freezing Point Depression,
Boiling Point Elevation, Osmotic Pressure
the van’t Hoff factor, i, is the ratio of moles of
solute particles to moles of formula units dissolved
measured van’t Hoff factors are often lower than
you might expect due to ion pairing in solution
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Slide 60
Ionic compounds dissociate…
…but not always completely.
Ions of opposite charge are always a little
“sticky”. So depending on just how sticky,
an ionic compound may not completely
dissociate.
The degree of dissociation is given by the
“van’t Hoff factor” – abbreviated “i”
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Slide 61
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Slide 62
A few van’t Hoff factors
Solute
i Measured
FeCl3
3.4
K2SO4
2.6
MgCl2
2.7
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Notice that FeCl3 would seem to want to dissociate into
FOUR (4) pieces: 1 Fe3+ and 3 Cl-.
The van’t Hoff factor is less than 4: 3.4. This is because
the ions are a little sticky.
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Slide 63
An ionic compound in water

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Slide 64
Boiling Point Elevation
What is the “boiling point” of a liquid?
It is the temperature at which the vapor pressure
equals the atmospheric pressure. So…
…if you decrease the vapor pressure, you must
increase the boiling point – it will take a higher
temperature to get enough gas molecules (vapor
pressure) to equal the atmospheric pressure.
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Slide 65
What is the Microscopic View of
Freezing Point Depression?
These phase change properties arise from
the attractive interactions between all
molecules (Van der Waal’s forces).
When you make a solution, you break some
of the solvent-solvent interactions and
replace them with solvent-solute
interactions.
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Slide 66
Freezing is a dynamic equilibrium
between melting and freezing.
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Slide 67
Slight Variation on the problem
How many grams of NaCl are needed to
lower the freezing point of 100 g of water
by 3 C?
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Slide 68
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Slide 69
The Answer
We want to decrease the freezing point by 3 C.
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NOTE: Kf is the WATER cryoscopic
constant
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Slide 70
NaCl
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Slide 71
It’s all about the # of particles
NaCl is an electrolyte:
NaCl →Na+ + Cl-
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You get 2 moles of solute per mole NaCl (well, 1.9 because of
ion pairing)
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Slide 72
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Osmotic Pressure

Osmotic pressure is the “funky” colligative property, but it
is very important biologically
Osmotic pressure is the pressure required to overcome the
natural pressure exerted by a solution by virtue of having a
concentration.

Osmotic pressure looks just like the ideal gas law:

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ΠV = nRT
where Π is osmotic pressure
Π= (n/V) RT = M RT
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Slide 73
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Why is it important?
Consider two solutions separated by a semipermeable membrane (meaning water can
pass through but nothing else) at room
temperature (298 K).
On one side of the membrane is a 1 M NaCl
solution. On the other side is a 4 M NaCl
solution.
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Slide 74
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What Happens?
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1 M NaCl
4 M NaCl
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Slide 75
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The water moves from the 1 M side
to the 4 M side. Why?
1 M NaCl

4 M NaCl
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Slide 76
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OSMOTIC PRESSURE
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Π= iM RT
On the 1 M side, the pressure is:
Π= M RT = (1.9) 1 M (0.0821 Latm/mol K)(298 K)
Π=48 atm
On the 4 M side, the pressure is:
Π= iM RT =(1.9) 4 M (0.0821 Latm/mol K) (298 K)
Π= 190 atm
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Slide 77
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Both solutions “push” on the membrane
The bigger push wins!
48 atm
1 M NaCl


142 atm
4 M NaCl
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190 atm

Note the direction of the arrows. Osmotic pressure is pushing AGAINST
the solution.
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Slide 78
It’s all about sweating!

What happens when you sweat?
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Slide 79
It’s all about sweating!
What happens when you sweat?
Your body loses water and some
dissolved salts through your skin which
cools your body.
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What is the price you pay for this?
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Slide 80
The cost of sweating
Your cells have lost precious water and
salts.
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What do you do about it?
Drink water?
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Slide 81
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The problem with water
Your cells are slightly dehydrated from sweating.
The salt concentration is actually higher than it
should be.
Water has no salt concentration. It flows into the
cells rapidly (osmotic pressure) and fills the cells.
But the cells are now diluted.
Further, since the pure water will always have lower
osmotic pressure and flow into the cell until
either you run out of water or the cell bursts.
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Slide 82
Gatorade – Man’s answer to Osmotic
Pressure
Gatorade is flavored sweat!
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By adding salts, it matches its osmotic
pressure to what the cell should be. This
allows for smooth rehydrating of the cell
without dilution or lysing.
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Slide 83
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continued…
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Slide 84
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Slide 85
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continued…
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Slide 86
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Slide 87
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continued…
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Slide 88
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Slide 89
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continued…
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Slide 90
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Slide 91
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