Transcript Everyone has Problems, but Chemists have Solutions

Everyone has Problems,
but Chemists have
Solutions
Solutions 2009-2010
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A solution is defined as a homogeneous
mixture of two or more substances.
What you are
dissolving
What you are
dissolving into
Types of Solutions
Solute
Solvent
Example
Gas
Gas
Air
Gas
Liquid
Carbonated soda
Liquid
Liquid
3% Hydrogen peroxide
Solid
Liquid
Salt water
Solid
Solid
Brass (Cu/Zn)
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Solubility is a measure of how much solute will
dissolve in a solvent at a specific
temperature.
The saying “like dissolves like” is helpful in
predicting the solubility of a substance in a
given solvent. What this expression means is
that two substances with intermolecular
forces of similar type and magnitude are
likely to be soluble in each other.
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“Like Dissolves Like”
Will bromine be
more soluble in
water or in
carbon
tetrachloride?
H2O
Bromine (Br2) is
nonpolar, therefore it
will be more soluble in
the nonpolar carbon
tetrachloride
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CCl4
4
FYI-When eating hot peppers or hot spicy foods has
left you "breathing fire“, drinking water will not
relieve you, because the water cannot dissolve the
oils which give the spicy taste. A chemist might
suggest that the ideal would be to drink a chaser of a
non-polar liquid. But most such liquids are toxic, and
even gargling with benzene or toluene doesn’t sound
very pleasant. A better solution is to eat greasy foods
like meats which will dissolve the oils. Alternatively,
food like pasta or bread, that can absorb the oils will
help "put out the fire." The smell and taste of onions
and garlic are also due to oils that are not easily
washed away by water and are not absorbed well.
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Two liquids are said to be miscible if they
are completely soluble in each other in
all proportions.
•
Oil and water are immiscible.
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Chemists characterize solutions by their
capacity to dissolve a solute.
• An unsaturated solution contains less solute
than it has the capacity to dissolve.
• A saturated solution contains the maximum
amount of a solute that will dissolve in a
given solvent at a specific temperature
• A supersaturated solution contains more
solute than is present in a saturated
solution. The supersaturated solution is not
very stable. In time, some of the solute will
come out of solution and form crystals. The
remaining solution will then be _________.
saturated
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Concentration Units
Chemists use several different
concentration units, each of which has
advantages as well as limitations. The
four most common units of
concentration: percent by mass, mole
fraction, molarity and molality.
The choice of concentration unit is based
on the purpose of the measurement.
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Mole Fraction (X)
XA =
Moles of A
Moles of A + Moles of B + Moles of C …
•mole fraction is appropriate for
calculating partial pressures of gases and
for dealing with vapor pressures of
solutions
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Calculate the mol fraction ethanol (C2H5OH) if
25.0 g ethanol is mixed with 50.0 grams water.
XA =
Moles of A
Moles of A + Moles of B + Moles of C …
25.0 g C2H5OH x 1 mol C2H5OH = .543 mol C2H5OH
46.07 g C2H5OH
50.0 g H2O x 1 mol H2O = 2.77 mol H2O
18.02 g H2O
XEthanol =
.543
.543 + 2.77
= .164
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Molarity (M)
Molarity =
moles of solute
liters of solution
•the advantage of molarity is that it is
generally easier to measure the volume
of a solution using precisely calibrated
volumetric flasks, than to mass the
solvent
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Calculate the molarity of an aqueous solution
containing 34.2 grams of potassium chlorate in
225 mL of solution.
Molarity =
moles of solute
liters of solution
34.2 g KClO3 x 1 mol KClO3 = .279 mol KClO3
122.55 g KClO3
.279 mol KClO3
.225 L
= 1.24 M
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Molality (m)
molality =
moles of solute
mass of solvent in kg
•molality is independent of temperature
(the volume of a solution typically
increases with increasing temperature,
so that a solution that is 1.0 M at 25 oC
may become 0.97 M at 45 oC because of
the increase in volume)
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What is the molality of a solution containing
14.5 g of ferric nitrate in 285 mL of
water?
14.5 g Fe(NO3)3 x 1 mol Fe(NO3)3 = .0599 mol Fe(NO3)3
241.88 g Fe(NO3)3
.0599 mol Fe(NO3)3
.285 kg water
= .210 m
Dwater
1 g = 1 mL = 1 cm3
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Percent by Mass (also called percent by
weight or weight percent)
Mass of solute
Mass of solution X 100%
•like molality, percent by mass is
independent of temperature
•we do not need to know the molar mass
of the solute in order to calculate the
percent by mass
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What is the mass percent table salt in a
solution containing 25.0 grams of table salt
dissolved in 100.0 mL of water?
25
X 100% = 20.0 %
125
Dwater
1 g = 1 mL = 1 cm3
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The effect of temperature on
solubility
• The solubility of gases in water usually
decreases with increasing temperature.
• In most, but not all cases, the solubility
of a solid substance increases with
temperature.
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Solubility curves,
which are used to
illustrate how the
solubility of a
solute is affected
by temperature,
are determined
experimentally
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Because the solubility of solids generally
decrease with decreasing temperatures
fractional crystallization, the process
of separating a mixture of substances
into pure components on the basis of
their differing solubilities, can be used
to purify substances.
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Crystallization is
the process in
which dissolved
solute comes out
of solution and
forms crystals.
This occurs because supersaturated
solutions are not very stable. In time,
some of the solute will come out of a
supersaturated solution as crystals.
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A 90.0 g sample of KNO3 is contaminated with a small amount (less
than 5.0 g) of NaCl. Explain how you would purify the KNO3.
 The goal is not to necessarily get all 90.0 g of KNO3, but to get as
much pure KNO3 as possible.
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Add water to the crystals
(100.0 g at least).
Heat the solution until all of
the crystals dissolve
(excess of 50
__ oC).
Slowly cool the solution until
it reaches 0 oC. At that
temperature, about _____
13 g of
KNO3 is still soluble and
____
5 g of the NaCl is still
soluble.
That means that ____
77 g of KNO3 will have formed a pure
crystal free from NaCl.
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Mixtures of ions can be separated by
precipitation. Precipitation is when an
insoluble solid forms and separates
from a solution.
*Note that both precipitation and
crystallization describe the separation
of excess solid substance from a
supersaturated solution.
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The Effect of Pressure on the
Solubility of Gases
For all practical purposes, external
pressure has no influence on the
solubilities of liquids and solids, but it
does greatly affect the solubility of
gases. The quantitative relationship
between gas solubility and pressure is
given by Henry’s Law, which states the
the solubility of gas in a liquid is
proportional to the pressure of the gas
over the solution.
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c = kP
Here c is the molar concentration (mol/L) of
the dissolved gas; P is the pressure (in atm)
of the gas over the solution (if several
gases are present , P is the partial pressure
of the specific gas of interest); and k is a
constant (for that specific gas) that
depends only on temperature. (For example,
at a given temperature, CO2 has the same k
value, AS LONG AS THE TEMP STAYS
CONSTANT.) The constant k has the units
mol/L . atm.
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The solubility of pure nitrogen gas at 25 oC and 1 atm is
6.8 x 10-4 mol/L. What is the concentration of
nitrogen dissolved in water under atmospheric
conditions? The partial pressure of nitrogen gas in
the atmosphere is 0.78 atm.
Step 1: Solve for k (which remains constant for this gas at this temp)
c = kP
Now you can use this same
-4
.
6.8 x 10 mol/L = k (1 atm)
k value at a different
K = 6.8 x
10-4
mol/L
. atm
pressure, as long as the
temp stays the same.
Step 2: Solve for the solubility of nitrogen gas at 0.78 atm
c = kP
c = (6.8 x 10-4 mol/L . atm)(0.78 atm)
c = 5.3 x 10-4 M
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Explain how the effervescence of a soft
drink when the cap is removed is a
demonstration of Henry’s Law.
Before the bottle is sealed, it is pressurized
with a mixture of air and CO2 and water
vapor. Because of the high partial pressure
of CO2, the amount dissolved in the soft
drink is many times the amount that would
dissolve under normal atmospheric conditions.
When the cap is removed, the pressurized
gases escape, and the excess CO2 comes out
of solution causing the effervescence.
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Most gases obey Henry’s law, but there are
some important exceptions…
• Dissolved gases that react with water have
higher solubilities
NH3 + H2O  NH4+ + OHCO2 + H2O  H2CO3
• Normally, oxygen gas is only sparingly soluble
in water, however, its solubility in blood is
dramatically greater because hemoglobin can
bind up to four oxygen molecules
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Colligative Properties
Colligative properties (or collective
properties) are properties that depend
only on the number of solute particles in
solution and NOT on the nature of the
solute particles. The colligative
properties are vapor-pressure lowering,
boiling point elevation, freezing point
depression, and osmotic pressure.
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Electrolytes and
Nonelectrolytes
All solutes that dissolve in water fit into
one of two categories: electrolytes and
nonelectrolytes.
• An electrolyte is a substance that, when
dissolved in water, results in a solution
that can conduct electricity.
• A nonelectrolyte does not conduct
electricity when dissolved in water
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A solution’s ability to conduct electricity depends on the
number of ions it contains.
(a) A nonelectrolyte solution does not contain ions, and the light bulb is not lit.
(b) A weak electrolyte solution contains a small number of ions, and the light
bulb is dimly lit.
(c) A strong electrolyte solution contains a large number of ions, and the light
bulb is brightly lit.
The molar amounts of the dissolved solutes are equal in all three cases.
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Vapor Pressure
When a liquid evaporates, its gaseous
molecules exert a vapor pressure. Vapor
pressure is defined as the pressure of
the vapor of a substance in contact with
its liquid or solid phase. Vapor pressure
changes with temperature; the higher
the temperature, the higher the vapor
pressure.
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At the boiling point,
bubbles form within a
liquid. The pressure
inside the bubble is
due solely to the
vapor pressure of the
liquid. The pressure
exerted on the bubble
is largely atmospheric.
When the vapor
pressure (internal)
equals the external
(atmospheric)
pressure, the bubble
rises to the surface
of the liquid and
bursts.
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Phase Diagram
• A phase diagram shows the
conditions at which a
substance exists as a solid,
liquid or gas. This is a
phase diagram for water.
• At 1 atm (760 mm Hg) the
vapor pressure of water
(inside the bubble) is equal
to the external
atmospheric pressure at
100 oC, the point at which
water boils. (See your
Water Vapor Reference
Sheet to obtain water’s
vapor pressure at
different temperatures.)
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If you live at higher
altitudes, the
atmospheric pressure
is lower, therefore, the
liquid boils at a lower
temperature. higher/lower
What would be
true about
water’s vapor
pressure at the
lower temp?
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Do you need to boil your eggs for less
time or more time if you live in Denver,
Colorado? Explain.
It takes a certain amount of heat (total
kinetic energy) to cook a hard-boiled egg.
Because water boils at a lower temperature
in Denver, (approx 95 oC) it is supplying less
heat to the egg, therefore the egg needs
to be boiled for longer.
*Remember, liquid water in Denver will not
get hotter than 95 oC if that is its boiling
point!!!
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Vapor-pressure Lowering
If a solute is nonvolatile (therefore the solute
does not have a measurable vapor pressure), the
vapor pressure of the solution will always be less
than that of the pure solvent. Thus, the
relationship between solution vapor pressure
and solvent vapor pressure depends on the
concentration of the solute in the solution. This
relationship is expressed by Raoult’s law
P1 = X1 . P01 , where
P1 = partial pressure of the solvent over a solution
P01 = vapor pressure of the pure solvent (torr, mm Hg, atm or kPa)
X1 = mole fraction of the solvent in the solution
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Calculate the new vapor pressure when 10.0 mL glycerol
(C3H8O3), a nonvolatile solute, is added to 500.0 mL
water at 50 oC. At this temperature, the vapor
pressure of pure water is 92.5 torr and its density is
0.988 g/mL. The density of glycerol is 1.26 g/mL.
We must first must calculate the mole fraction of the solvent (water).
500.0 mL x .988 g
x 1 mol H2O = 27.4 mol H2O
1 mL
10.0 mL x
1.26 g
18.02 g
x 1 mol C3H8O3 = 0.137 mol C3H8O3
1 mL
Xwater =
92.09 g
27.4
= .995
Notice, the
vapor pressure
has been
lowered, just
as we
expected!
27.4 + 0.137
Now we can use Raoult’s law to calculate the new vapor pressure
Psolution = Xwater . P0water
Psolution = (.995) (92.5 torr) = 92.0 torr
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In a solution containing only one solute, X1 = 1 –
X2, where X1 is the mole fraction of the
solvent, and X2 is the mole fraction of the
solute. The previous equation can therefore
be rewritten substituting 1 – X2 for X1
P1 = (1 – X2) . P01
P1 = P01 – X2P01
-P01 + P1 = - X2P01
P01 - P1 = X2P01
Using this formula will
give you the CHANGE in
vapor pressure.
DP = X2P01
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Let’s solve the previous problem again, this time solving for change
(decrease) in vapor pressure using this derived formula.
Calculate the vapor pressure lowering when 10.0 mL glycerol (0.137
mol) is added to 500.0 mL (27.4 mol) water at 50 oC. At this
temperature, the vapor pressure of pure water is 92.5 torr
We must first must calculate the mole fraction of the solute (glycerol).
Xglycerol =
0.137
= .00498
27.4 + 0.137
Now we can use Raoult’s law to calculate how much the vapor pressure was
lowered by the addition of the glycerol
This shows that the vapor pressure
would be lowered by .461 torr
DP = X2P01
(92.5 torr – 0.461 torr = 92.0 torr),
DP = Xglycerol . P0water
the same answer we obtained two
DP = (.00498) (92.5 torr) = 0.461 torr slides ago!
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Why is the vapor pressure of a solution
less than that of the pure solvent?
One of the driving forces in physical and chemical
processes is an increase in disorder – the greater the
disorder, the more favorable the process.
Vaporization increases the disorder of a system
because molecules in a vapor are not as closely packed
and therefore have less order than those in a liquid.
Because a solution is more disordered that a pure
solvent, the difference in disorder between a solution
and a vapor is less than that between a pure solvent
and a vapor. Thus solvent molecules in a solution have
less of a tendency to leave the solution than to leave
the pure solvent to become vapor, and the vapor
pressure of a solution is less than that of the pure
solvent.
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Nature tends towards disorder (entropy)
Pure
solvent
The pure
solvent is
highly ordered,
therefore many
of the particles
will leave the
solvent
because it
GREATLY
increases
disorder, this
results in high
vapor
pressure.
Nonvolatile
solute
This solution is already somewhat
disordered, therefore less of the
solvent particles will leave the
solvent, this results in lower vapor
pressure.
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If both components of a solution are
volatile (that means they both have
measurable vapor pressure), the vapor
pressure of the solution is the sum of
the individual partial pressures.
Raoult’s law holds equally well in this case.
PT = XAP0A + XBP0B
Where PT is equal to the total pressure
(PA + PB)
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Fractional Distillation
Solution vapor pressure has a direct
bearing on fractional distillation, a
procedure for separating liquid
components of a mixture based on their
different boiling points.
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Fractional distillation is somewhat analogous to
fractional crystallization.
When you boil a mixture containing two
substances with appreciably different boiling
points (80.1 oC and 110.6 oC), the vapor formed
will be somewhat richer in the more volatile
compound (the one that vaporizes/boils at the
lower temp). If the vapor is condensed in a
separate container and that liquid is boiled
again, a still higher concentration of the more
volatile substance will be obtained in the
vapor phase. By repeating this process many
times, it is possible to obtain a pure sample of
the more volatile substance.
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Boiling-point elevation
Because the presence of a nonvolatile
solute lowers the vapor pressure of a
solution, it must also affect the boiling
point of the solution.
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Phase Diagram of a pure substance.
“Normal” boiling and freezing points
are indicated.
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Phase diagram of a pure substance and curves
representing the effect of adding a solute.
For each temperature the solute decreases the vapor pressure
of the solution.
Notice that the solution boils at a higher temperature than the
pure substance (boiling occurs when the vapor pressure is equal
to atmospheric pressure – in this case 1 atm - so now the solution
needs to be hotter to reach a vapor pressure of 1 atm.)
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Boiling-point elevation (DTb) is defined as the
CHANGE in the boiling point (boiling point of
the solution minus the boiling point of the
pure solvent).
The value of DTb is proportional to the concentration
(molality) of the solution. That is,
DTb = kbm
where kb is the molal boiling point elevation constant.
The units of kb are oC/m.
(You can find a table of molal boiling-point elevation and freezing-point
depression constants of several common liquids on your reference
sheet.)
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What is the new boiling point when 175 grams of
sucrose, C12H22O11, are added to 285 mL of water?
DTb = kbm
175 g C12H22O11 x 1 mol C12H22O11 = .511 mol C12H22O11
342.30 g C12H22O11
.511 mol C12H22O11
.285 kg water
= 1.79 m
DTb = (.52 oC/m)(1.79 m) = .93 oC
.93 oC + 100.00 = 100.93 oC
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Freezing-point depression
Freezing involves a transition from the disordered state to
the ordered state. For this to happen, energy must be
removed from the system. Because a solution has greater
disorder than the solvent, more energy needs to be
removed from it to create order than in the case of the
pure solvent. Therefore, it needs to get colder to freeze a
solution, giving the solution a lower freezing point than the
solvent.
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Freezing-point depression (DTf) is defined as the
CHANGE in the freezing point (freezing point of the
pure solvent minus the freezing point of the solution).
Note** DTf is ALWAYS positive.
The value of DTf is proportional to the
concentration (molality) of the
solution. That is,
DTf = kfm
where kf is the molal freezing point
depression constant.
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An aqueous solution of glucose freezes at –2.56 oC.
How many grams of glucose must have been added
to 150.0 mL of water to form this solution?
DTf = (1.86 oC/m)(m) = 2.56 oC
m = 1.38
1.38 mol C6H12O6
1 kg water
=
X mol C6H12O6
.150 kg water
X = .207 mol
.207 mol C6H12O6 x 180.15 g C6H12O6 = 37.3 g C6H12O6
1 mol C6H12O6
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Explain why salt is placed on frozen roads
and sidewalks.
Salt depresses the freezing point of
water, which means that the ice on the
roads and sidewalks will melt into water,
even at temperatures below 0 oC.
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Osmotic Pressure
The net movement of solvent molecules
through a semipermeable membrane
from a pure solvent of a dilute solution
to a more concentrated solution is
called osmosis. The osmotic pressure
(p) of a solution is the pressure required
to stop osmosis. This pressure can be
measured directly from the difference
in the final fluid levels.
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Like boiling point elevation and freezing point
depression, osmotic pressure is directly
proportional to the concentration of solution.
If two solutions are of equal concentration
(hence, equal osmotic pressure), they are said
to be isotonic. If two solutions are of unequal
osmotic pressures, the more concentrated
solution is said to be hypertonic and the more
dilute solution is described as hypotonic.
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Using colligative properties to
determine molar mass
The colligative properties of
nonelectrolyte solutions provide a means
of determining the molar mass of a
solute. Theoretically, any of the four
colligative properties is suitable for this
purpose. In practice, however, only
freezing-point depression and osmotic
pressure are used because they show
the most pronounced changes.
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A 7.85 g sample of a compound is dissolved in
301 g of benzene. The freezing-point of the
solution is 1.05 oC below that of pure benzene.
What is the molar mass of this compound?
•Our first step is to calculate the molality of the solution.
= 0.205 m
Molality = DTf = 1.05 oC
kf
5.12 oC/m
•Since there is 0.205 mole of the solute in 1 kg
of solvent, (0.250 m) , the number of moles of
solute in 301 g, or 0.301 kg of solvent is
0.205 mole =
1 kg solvent
.0617 mol
.301 kg solvent
•Finally, because we know that 7.85 g of the substance is equal to .0617
mol, we can calculate the molar mass of the solute as follows
7.85 g
= 127 g/mol
0.0617 mol
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The colligative properties of electrolytes
require a slightly different approach than the
one used for the colligative properties of
nonelectrolytes. The reason is that
electrolytes dissociate into ions in solution,
and so one unit of an electrolyte compound
separates into two or more particles when it
dissolves. (Remember, it is the number of
solute particles that determines the
colligative properties of a solution.)
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For example, each unit of NaCl dissociates into two ions –
Na+ and Cl-. Thus the colligative properties of a 0.1 m
solution of NaCl should be twice as great as those of a
0.1 m solution containing a nonelectrolyte, such as
sucrose. Similarly, we would expect a 0.1 m Ba(NO3)2
solution to depress the freezing point by three times
as much as a 0.1 m sucrose solution because Ba(NO3)2
produces three ions. To account for this effect we
must modify the equations for colligative properties as
follows:
DTb = ikbm
DTf = ikfm
The variable i is the van’t Hoff factor which is equal to
the number of particles in solution after dissociation.
(Remember, only electrolytes dissociate!)
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What is the new freezing point when 25 grams of
ammonium nitrate are added to 285 mL of water?
DTb = ikbm
25 g NH4NO3 x 1 mol NH4NO3 = .312 mol NH4NO3
80.05 g NH4NO3
.312 mol NH4NO3
.285 kg water
= 1.10 m
DTb = (2)(1.86 oC/m)(1.10 m) = 4.09 oC
New freezing point = -4.09 oC
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Colloids
The solutions we discussed so far are true
homogeneous mixtures. Now consider what
happens if we add fine sand to a beaker of
water and stir. The sand particles are
suspended at first but then gradually settle
to the bottom. This is an example of a
heterogeneous mixture. Between these two
extremes is an intermediate state called a
colloidal suspension, or simply, a colloid. A
colloid is a dispersion of particles of one
substance (the dispersed phase) throughout a
dispersing medium made of another
substance.
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Types of Colloids
Dispersing
Medium
Dispersed
phase
Name
Example
Gas
Liquid
Aerosol
Fog, mist
Gas
Solid
Aerosol
Smoke
Liquid
Gas
Foam
Whipped
cream
Liquid
Liquid
Emulsion
Mayonnaise
Liquid
Solid
Sol
Milk of
magnesia
Solid
Gas
Foam
Plastic foams
Solid
Liquid
Gel
Jelly, butter
Solid
Solid
Solid gel
Certain alloys
(steel),
gemstones
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One way to distinguish a solution from a colloid is by the
Tyndall effect. When a beam of light passes through
a colloid, it is scattered by the dispersed medium. No
such scattering is observed with ordinary solutions
because the solute molecules are too small to interact
with visible light.
Another demonstration of the Tyndall effect is the scattering of
sunlight by dust or smoke in the air.
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The End
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