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mixture = a blend of two or more
kinds of matter, each of which retains
its own identity and properties
1)
•
can be physically separated
(filtration, evaporation, decanting,
magnetism, etc)
a.) homogeneous mixture = a
mixture that is uniform in
composition throughout
•Ex:
Food coloring and water
b.) heterogeneous mixture = a
mixture that is NOT uniform in
composition throughout
•Ex:
Oil and water
1) solution = a homogeneous mixture
2) suspension = a mixture in which the
particles are so large that they settle out
unless the mixture is constantly stirred or
agitated
• Heterogeneous
• Ex:
mixture
Sand and water
3) colloid = a mixture consisting of particles
that are intermediate in size between those
in solutions and those in suspensions
• Heterogeneous
• Ex:
Mixture
Milk, mayonnaise, smog, butter,
whipped cream
What property of a colloid helps to
prevent colloid particles from settling
out of a mixture?
a)
•
Brownian motion = the random
continuous motions of colloidal
particles
b) Tyndall effect = visible
pattern caused by the
reflection of light from
suspended particles in a
colloid (or from suspended
particles in a suspension if
the particles have not
settled out)

Ex: visibility of a headlight beam
on a foggy night
Solutions
Colloids
Suspensions
Homogeneous Heterogeneous Heterogeneous
Particle size: 1- Particle size:
Particle size:
1000 nm,
over 1000 nm,
0.01-1 nm; can
dispersed; can suspended; can
be atoms,
be aggregates
be large
ions,
or large
particles or
molecules
molecules
aggregates
Solutions
Colloids
Do not separate Do not separate
on standing
on standing
Cannot be
separated by
filtration
Do not scatter
light
Suspensions
Particles settle
out
Cannot be
separated by
filtration
Can be separated
by filtration
Scatter light
(Tyndall effect)
May scatter
light, but are not
transparent
1.) If particles settle or can be filtered out =
suspension
2.) If particles DO NOT settle or filter out shine a
beam of light (Tyndall effect) through the
mixture
• If
• If
the Tyndall effect is observed = colloid
the Tyndall effect is NOT observed =
solution
1) Solvent = the substance that
does the dissolving in a solution
a)
Typically present in the
greatest amount
b)
Typically a liquid
c)
Water is the most common
or “universal” solvent
•B/c
•
•
•
water molecules are polar
The hydrogen side of each water (H2O) molecule
carries a slight positive electric charge, while the
oxygen side carries a slight negative electric charge.
water can dissociate ionic compounds into their
positive and negative ions.
The positive part of an ionic compound is attracted
to the oxygen side of water while the negative
portion of the compound is attracted to the
hydrogen side of water.
• Water
won't dissolve or won't dissolve
well. If the attraction is high between the
opposite-charged ions in a compound,
then the solubility will be low.
•Ex:
hydroxides exhibit low solubility in
water.
•Ex:
nonpolar molecules don't dissolve
very well in water (fats and waxes)
2) Solute = substance being dissolved in a
solution
a)
Typically present in the least amount
b)
Typically a solid
Solvent Solute
Gas
Gas
Gas
Liquid
Gas
Solid
Liquid
Gas
Liquid
Liquid
Liquid
Solid
Solid
Gas
Solid
Solid
Liquid
Solid
Common Example
Diver’s tank
oNOT all solutions
Humidity
are liquids/solids!
Moth ball
Carbonation
Vinegar
Seawater
oSolutions are
formed in ALL 3
states!
Gas stove lighter
Dental fillings
Sterling Silver (Ag + Cu)
1.) Solvation = the process of dissolving
a.) First- solute particles are surrounded
by solvent particles
b.) Then- solute particles are separated
and pulled into solution
C. Johannesson
2.) Dissociation = separation of an ionic solid
into aqueous ions
• Ex:
NaCl + H2O – the Na ion and Cl ion
become hydrated and gradually move away
from the crystal into solution.
• Each
ion in the solution acts as though it
were present alone: So there is only a
solution containing Na+ and Cl- ions
uniformly mixed with H2O particles
• NaCl(s)
 Na+(aq) + Cl–(aq)
Animation of Salt Dissolving:
http://www.northland.cc.mn.us/biology/Biology1111/animations/dissolve.html
3.) Ionization = breaking apart of some
polar molecules into aqueous ions
•Ex:
HNO3(aq) + H2O(l)  H3O+(aq) + NO3–(aq)
4.) Molecular Solvation =molecules stay
intact
•Ex:
C6H12O6(s)  C6H12O6(aq)
1)
Grinding: increases surface area
exposed to solvent
2)
Stirring: allows solvent
continual contact with solute
3)
Heating: increases kinetic
energy; increases mixing
1) electrolyte = a substance that dissolves
in water to give a solution that conducts
electric current
2) nonelectrolyte = a substance that
dissolves in water to give a solution that
does NOT conduct an electric current
3) Solutions of electrolytes can conduct
electric current:
a) The positive ions and the negative ions
disassociate (separate) in solution. The
mobile ions can move a charge from one
point in the solution to another point
4) Solutions of nonelectrolytes CANNOT
conduct electric current:
a)
When a nonelectrolyte dissolves in water
there are NO charged particles in
solution.
b)
Ex: Solute exists as molecules
5) Weak Electrolytes
a) Only a portion of dissolved molecules
ionize
6) Solid ionic compounds CANNOT conduct
electric current:
a) Ions are present but they are NOT
mobile.
-
-
+
sugar
-
+
acetic acid
+
salt
NonElectrolyte
Weak
Electrolyte
Strong
Electrolyte
solute exists as
molecules
only
solute exists as
ions and
molecules
solute exists as
ions only
1) Solubility = quantity of solute that will
dissolve in specific amount of solvent at a
certain temperature. (pressure must also be
specified for gases).
a)
Ex: 204 g of sugar will dissolve in 100 g
of water at 20C
b)
soluble and insoluble are relative terms
c)
solubility should NOT be confused with
the rate at which a substance dissolves
2) saturated solution = a stable solution in
which the maximum amount of solute has
been dissolved.
a)
Visual evidence: a quantity of
undissolved solute remains in contact
with the solution
3) solubility equilibrium = state where the
solute is dissolving at the same rate that the
solute is coming out of solution (crystallizing).
a)
Opposing processes of the dissolving and
crystallizing of a solute occur at equal
rates.
b)
solute + solvent
solution
4) unsaturated solution = a solution that
contains less solute than a saturated solution
under existing conditions
5) supersaturated solution = a solution that
temporarily contains more than the saturation
amount of solute than the solvent can hold
• Unstable
– if disturbed, the excess solute
will crystallize out of solution
UNSATURATED
SOLUTION
more solute
dissolves
SATURATED
SOLUTION
no more solute
dissolves
concentration
SUPERSATURATED
SOLUTION
becomes unstable,
crystals form
1) Nature of solute and solvent
a) “Like dissolves like” = rule of thumb for predicting
whether or not one substance dissolves in
another “Alikeness” depends on:
•Intermolecular
•Type
of bonding
•Polarity
•ionic
forces
or nonpolarity of molecules:
solutes tend to dissolve in
polar solvents but not in nonpolar
solvents
b.) Solvent-Solute Combinations:
Solvent Type
Solute Type
Polar
Polar
Polar
Nonpolar
Nonpolar
Polar
Nonpolar
Nonpolar
Is Solution
Likely?
YES
NO
NO
YES
“Like Dissolves
Like”
2) Pressure:
a)
Pressure has little effect on the
solubility of liquids or solids in liquid
solvents.
b)
The solubility of a gas in a liquid
solvent INCREASES when pressure
increases. It is a direct relationship.
3) Temperature:
a)
The solubility of a gas in a liquid solvent
DECREASES with an increase in
temperature.
b)
The solubility of a solid in a liquid solvent
MOST OFTEN increases with an increase
in temperature. However, solubility
changes vary widely with temperature
changes sometimes decreasing with
temperature increases.
1)
Saturated = any point on the line or
ABOVE the line
2) Unsaturated = any point BELOW the line
1. What is the
solubility of the
following solutes in
water?
a) NaCl at 60ºC= 38g
b) KCl at 40ºC=
c) KNO3 at 20ºC=
39g
31g
2. Are the following solutions
saturated or unsaturated?
Each solution contains 100 g
of H20.
a) 31.2 g of KCl at 30ºC =
Unsaturated
b) 106g KNO3 at 60ºC=
Saturated
(on the line)
2. Are the following solutions
saturated or unsaturated? Each
solution contains 100 g of H20.
c)40 g NaCl at 10ºC=
Saturated
d)150g KNO3 at 90ºC =
Unsaturated
3. For each of the following
solutions, explain how much of the
solute will dissolve and how much
will remain undissolved at the
bottom of the test tube?
a) 180 g of KNO3 in
100 g of water at
80ºC
169g dissolved
11g undissolved
3. For each of the following
solutions, explain how much of the
solute will dissolve and how much
will remain undissolved at the
bottom of the test tube?
b) 180 g of KNO3 in
100 g of water at
20ºC
31g dissolved
149g undissolved
3. For each of the following
solutions, explain how much of the
solute will dissolve and how much
will remain undissolved at the
bottom of the test tube?
C) 60 g of NaCl in
100 g of water at
60ºC
39g dissolved
21g undissolved
4. A
saturated solution of
KNO3 is formed from one
hundred grams of water. If
the saturated solution is
cooled from 90°C to 30°C,
how many grams of
precipitate are formed?
(200g-45g) = 155g
5. A
saturated solution of KCl is
formed from one hundred
grams of water. If the saturated
solution is cooled from 90°C to
40°C, how many grams of
precipitate are formed?
(53g-39g) = 14g
1.) Concentration = The amount of solute in a
solution.
a.) Describing Concentration:
•%
by mass - medicated creams
•%
by volume - rubbing alcohol
• ppm,
ppb - water contaminants
• molarity
- used by chemists
• molality
- used by chemists
1.) Molarity = unit of concentration of a
solution.
substance being dissolved
•Formula:
moles of solute
Molarity(M) 
liters of solution
total combined volume
2M HCl What does this mean?
mol
M
L
2 mol HCl
2M HCl 
1L
To solve for Molarity when given grams of solute,
you will use gram/mole conversion and the
equation for Molarity. Use the following steps:
1)
a)
Change grams of solute to moles. To do this you
need to use the following conversion:
Grams of solute 1 mole of solute
=moles of solute
Molar mass of solute*
*make sure formulas are correct; Use ion list!
b) Change the volume of solution to liters (L)
by moving the decimal 3 places to the LEFT.
• Remember
that 1000 mL = 1 L. Water will
most often be the solvent in a solution.
c) Substitute the information into the molarity
equation and solve.
moles of solute
Molarity(M) 
liters of solution
2) Ex: What is the molarity of a solution composed of
22.4g of sodium chloride dissolved in enough water to
make 500 mL of solution?
22.4 g of NaCl 1 mole NaCl =0.383 moles NaCl
58.443 g NaCl
0.383 mole NaCl = 0.766 M NaCl
0.500 L soln
Note: You need to
change mL of solution
to L. So move decimal
3 places to the left!
Use the Molarity equation to solve for grams
of solute. You will need to solve the Molarity
equation for moles of solute. Then convert
moles of solute to grams of solute.
mol
M
L
a) Substitute the information and solve for
moles by cross multiplying or multiplying by
the reciprocal:
“X” moles LiBr___
mol
M
L
= 0.4M LiBr
0.300 L solution
X = 0.12 moles of LiBr are needed
b) Convert moles to grams using the following conversion:
Moles of solute molar mass of solute
= grams of solute
1 mole of solute
•
The substitution would look like:
0.12 moles LiBr
86.845 g LiBr
1 mol LiBr
= 10. g LiBr
1.) Sometimes solutions of lower concentrations are
made from existing solutions.
2.) Moles of solute remain the same.
3.) Formula:
M1V1 = M2V2
• M1 is the original molarity concentration
• V1 is the volume of the original solution (in L)
• M2 is the new concentration
• V2 is the amount of the new solution needed
(in L)
Ex 1: What volume of 15.8M HNO3 is required to
make 250 mL of a 6.0M solution?
M 1V1  M 2V2
GIVEN:
M1 = 15.8M
V1 = ?
M2 = 6.0M
V2 = 250 mL
WORK:
M1 V1 = M2 V2
(15.8M) V1 = (6.0M)(250mL)
V1 = 95 mL of 15.8M HNO3
1.) Molality (m) = a unit of
concentration of a solution expressed in
moles of solute per kilogram of solvent.
a) Ex: A solution that contains 1 mol of
solute, ammonia (NH3), dissolved in exactly
1 kg solvent, is a “one molal” solution.
b) Formula:
moles of solute
molality(m) 
kg of solvent
mass of solvent only
c) Note: 1 kg water = 1 L water
1) If needed convert:
•
grams of solute
•
grams of solvent
kg of solvent
(move decimal 3 places to the left!)
2) Solve for molality
moles of solute
m = moles of solute
kg of solvent
3) Ex: Find the molality of a solution containing
75 g of MgCl2 in 250 mL of water.
75 g MgCl2 1 mol MgCl2
95.211 g MgCl2
0.79 mol MgCl2
0.250 kg water
mol
m
kg
= 0.79 mol
MgCl2
= 3.2m MgCl2
1) Use Molality equation and solve for
moles
2) Then convert moles of solute to grams of
solute
mol
m
kg
Ex: How many grams of NaCl are required to make
a 1.54m solution using 0.500 kg of water?
“X” moles NaCl___
0.500 kg H2O
= 1.54m NaCl
X = 0.770 moles NaCl
0.770 mol NaCl 58.443 g NaCl = 45.0 g NaCl
1 mol NaCl
1.) Colligative property = A property that
depends on the NUMBER of solute
particles rather than the type of particle.
C. Johannesson
1.) Vapor Pressure LOWERING of a
solution
a) Ex: We have two beakers. One contains
H2O, the other contains a salt solution
(NaCl) in H2O. Both beakers are placed
into a sealed chamber:
DAY 1
DAY 3: If we leave
it for a couple of
days and then come
back and take a
look, this is what it
might look like.
b) WHAT HAPPENED?:
•Na+
and Cl- ions are non-volatile ions. In other words, they will
not leave and go into the vapor phase.
•Those
H2O molecules with enough kinetic energy will leave the
surface of the solutions and enter into the vapor phase.
•The
fact that Na+ and Cl- ions are dissolved in H2O indicates
that there is an attractive interaction between the solutes and
the H2O.
DAY 1
DAY 3: If we leave it
for a couple of days
and then come back
and take a look, this
is what it might look
like.
•The
interaction of the Na+ and Cl- ions for H2O
will act to hinder the ability of the solvated H2O
molecules to leave and go into the vapor phase
(Na+ and Cl- ions are non-volatile).
•This
hindrance of the H2O molecules to enter
the vapor phase will reduce the vapor pressure
of the H2O in the salt-containing solution.
DAY 1
DAY 3: If we leave it
for a couple of days
and then come back
and take a look, this
is what it might look
like.
•The
two beakers are in the same sealed container,
thus, the vapor pressure above the solutions is
identical.
•The
rate of H2O entering the solutions by collisions
from the vapor state will be identical
•The
rate of H2O leaving the liquid phase and
entering the vapor phase is slower for the NaClcontaining solution
DAY 1
DAY 3: If we leave it
for a couple of days
and then come back
and take a look, this
is what it might look
like.
2.) Osmotic Pressure = If two solutions, with
different solute concentrations, are separated
by a semi-permeable membrane, there can be
a net flow of solvent across the membrane.
3.) Freezing Point depression= The solution
will begin to freeze at a temperature BELOW
that of the pure solvent.
4.) Boiling Point elevation = the solution will
begin to boil at a temperature ABOVE that of
the pure solvent.
1.) One of the ways in which they are different,
is that when you add a solute to a liquid both
the freezing point and boiling point of the
solution CHANGE.
a) Freezing point – In a solution, the solute
particles interfere with the attractive forces
among the solvent particles.
•This
prevents the solvent from entering
the solid state at its normal freezing
point
b) Boiling point- When the temperature of a solution
containing a nonvolatile solute is raised to the boiling point of
the pure solvent, the resulting vapor pressure is still less than
the atmospheric pressure and the solution will not boil.
• Thus,
the solution must be heated to a higher
temperature to supply the additional KE needed to raise
the vapor pressure to atmospheric pressure.
2) Water is the liquid we will be dealing with most
often
a)
The freezing point of pure water is 0°C.
b)
The normal boiling point of water is
100°C.
c)
But if you make a solution using water as
the solvent, the freezing point of that
solution will NOT be 0°C nor will the
boiling point be 100°C!
1) Freezing point depression (TFP) –
solutions will freeze at LOWER
temperature than the pure solvent
a) The more solute dissolved, the
greater the effect.
b) Ex: ethylene glycol (antifreeze) protects
against freezing of the water in the cooling
system by lowering the freezing point to
about -40°C
c) Ex: making of homemade ice
cream- The ice cream mix is
put into a metal container
which is surrounded by
crushed ice. Then salt is put on
the ice to lower its melting
point. This gives a temperature
gradient across the metal
container into the saltwater-ice
solution which is lower than
0°C. The heat transfer out of
the ice cream mix allows it to
freeze.
2.) Boiling point elevation (TBP) – solutions
will boil at HIGHER temperatures than the pure
solvent
a) The boiling point of pure water is
100°C, but that boiling point can be
elevated by the addition of a solute such
as a salt.
b) The more solute dissolved, the greater
the effect.
1) Solution concentrations are given in
molality (m)
2) Colligative properties are directly
proportional to the molal concentration of a
solute
3)Account for particle molality for
ELECTROLYTES
• Ex:
NaCl = 2 ions (Na+ & Cl-)
• Ex:
MgCl2 = 3 ions (Mg+2 and 2 Cl-)
4) A change in the concentration (m),
changes the freezing point and boiling
point of a solution
a) You will need to figure out the change in
temperature for the normal freezing point (FP) or
boiling point (BP) and adjust them with the
following steps:
∆T = mK
∆T = change in temperature
m = moles solute/kg solvent (MOLALITY)
K = constant
•You
will mostly use the constants for
water: Kfp = 1.853 ˚C/m
Kbp = 0.515˚C/m
•
Other values can be found below:
FREEZING Point Depression Constants
Freezing
oC/m)
Compound
K
(
fp
Point (oC)
water
0
1.853
acetic acid
16.66
3.90
benzene
5.53
5.12
p-xylene
13.26
4.3
naphthalene
80.29
6.94
cyclohexane
6.54
20.0
BOILING Point Elevation Constants
Boiling
Point (oC)
Kbp(oC/m)
100
34.55
0.515
1.824
carbon
disulfide
46.23
2.35
benzene
80.10
2.53
carbon
tetrachloride
76.75
4.48
camphor
207.42
5.611
Compound
water
ethyl ether
4.) Steps solving for ∆T:
a) Solve for molality (see steps from
previous problems)
m = moles of solute
kg of solvent
mol
m
kg
b) Find your K constant from the chart
listed in your notes.
•
Be sure to find the correct K constant
for what you are solving for-- either
FP or BP!!
c) Solve for ∆T
∆T = mK
d) If solving for:
• Freezing
point: Take the normal freezing
point of the solvent and SUBTRACT the ∆T
value
• Round
final answer to 3 significant
figures!
• Boiling
point: Take the normal boiling point
of the solvent and ADD the ∆T value
• Round
final answer to 6 significant
figures!
5.) At what temperature will a solution that is composed of 32.8 g of glucose (C6H12O6) in 225 g of water boil?
32.8g C6H12O6 1 mol C6H12O6
= 0.182 mol
180.155g C6H12O6 C6H12O6
m=
0.182 mol C6H12O6
0.225 kg water
= 0.809 m C6H12O6
TBP = mKbp
TBP = (0.809 m) (0.515 C/m)
BP = 100C + 0.417 C
= 0.417C
= 100.417C
6.) Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 115 g water.
28.0g NaCl 1 mol NaCl
= 0.479 mol NaCl
58.443g NaCl
m=
0.479 mol NaCl
0.115 kg water
= 4.17m NaCl
Particle molality = 4.17m X 2 = 8.34m
TFP = mKfp
TFP = (8.34 m) (1.853 C/m) = 15.5C
FP = 0C – 15.5 C
= -15.5C

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