Classification of Matter
and Solutions Notes
Part 1
I. Classification of Matter
Yes
Matter
Can it be physically separated?
Mixtures
Is the composition uniform?
Yes
Homogeneous
Mixtures
(Solutions)
(air, sugar water,
salt water)
No
Heterogeneous
Mixtures
(Suspensions
or Colliods)
(granite, wood,
muddy water)
No
Pure Substances
Can it be decomposed by an
ordinary chemical reaction?
Yes
No
Compounds
Elements
(water, sodium
chloride, sucrose)
(gold, oxygen,
carbon)
Mixtures: matter that can be physically
separated into component parts.
 a. homogeneous mixture –has uniform
composition; also called a solution
 b. heterogeneous mixture – does not
have a uniform composition; suspensions
or colloids

Pure Substances: when component parts of a
mixture can no longer be physically separated
into simpler substances. Pure substances are
either compounds or elements.
 a. Compounds – can be decomposed by a
chemical change.
 b. Elements – cannot be decomposed by a
chemical change.

II. Types of Mixtures (Solutions,
Suspensions, and Colloids) Table 13-3 page
398
1. Solution (homogeneous mixture)- any
substance (solid, liquid, gas) that is evenly
dispersed throughout another substance. page
398 (Not the same as a chemical reaction!!)
 Ex: sugar water, salt water (do not scatter
light)
Components of a Solution
 1. Solute – substance dissolved
 2. Solvent – substance that does the dissolving

(water is the universal solvent)
2. Suspensions (heterogeneous
mixtures) – particles in a solvent are so
large that they settle out unless the
mixture is constantly stirred Ex: muddy
water, vegetable soup, page 398 (may
scatter light, but are transparent)
3. Colloids (heterogeneous mixtures)
– particles are intermediate in size
between those in solutions and
suspensions. Example: After large soil
particles settle out of muddy water the
water is often still cloudy because colloidal
particles remain dispersed in the water.
Ex: milk, mayonnaise , page 398 (do
scatter light – Tyndall Effect)
III. The Solution Process (Solvation)
Solvation is the process by which a solute
dissolves in a solvent.
Miscible: when solutes and solvents are
soluble in each other (solvation occurs)
Immiscible: when solutes and solvents
are not soluble in each other (solvation
does not occur)
Aqueous solutions – solvent is water.
What happens when:
Ionic compound as the solute
and water is the solvent?
Dissociation of ionic compound occurs (ions
separate). Water is then attracted to the
positive and negative ions. When all
molecules have been “surrounded” the
molecule is called hydrated.
Miscible
What happens when:
Polar covalent molecules are the solute
and water is the solvent? Dissociation
does NOT occur. Water is polar and its
“oppositely charged poles” will be
attracted to other polar molecules'
“oppositely charged poles.” When a
solution is made between two polar
molecules it is called molecular solvation.
Miscible
What happens when:
Nonpolar covalent molecules are the
soluteand water is the solvent? A solution
will NOT occur. Water and any nonpolar
molecule will not mix! Think of putting water
and oil together. Water is polar and oil is
nonpolar. The polar water is not attracted to the
oil, because the nonpolar oil does not have any
oppositely charged poles!
Immiscible
IV. Like Dissolves Like
We don’t always use water as the solvent!
Solutions can be made from various
substances – a rule of thumb to follow
when trying to determine if two
substances will form a solution is “like
dissolves like.”
Polar Molecules + Polar Molecules
Water and Alcohol : Miscible
Nonpolar Molec. + Nonpolar Molec.
Oil and Hexane : Miscible
Ionic and Ionic
Chemical Reaction
Ionic + Polar Molecules
Salt and Water : Miscible
Polar Molecules + Nonpolar Molecules
Water and Oil : Immiscible
Ionic + Nonpolar Molecules
Salt and Oil:
Immiscible
V. Solubility - Now that we can figure
out what we can mix to make a
solution, how do we know how much
solvent and how much solute to use?
Solubility: the maximum amount of a
substance that will dissolve in a solvent (at
a specific temperature)
According to solubility, solutions can be
either:
unsaturated – a solution that is able to
dissolve more solute (not enough)
saturated – a solution that cannot dissolve
any more solute (just enough)
supersaturated – a solution that contains
more solute than can be dissolved (too
much!!)
The solubility of substances varies widely.
For example 0.189 grams of Ca(OH)2
dissolves in 100 grams of water at 0C.
122 grams of AgNO3 dissolves in 100
grams of water at 0C. (page 404 in your
book)
VI. Factors Effecting Rate and
Solubility
A.
Factors Effecting Rate:
1. Agitation – stirring or mixing the solution will
increase the rate or how fast the solute
dissolves, but it will not change how much
solute can be dissolved. If you add 35.9 grams
of salt to water (at 20C) it will all eventually
dissolve, but if you stir the solution it will
dissolve much quicker. (As you stir the particles
are constantly being moved, allowing for
interactions between solute and solvent to occur
more quickly.)
2.
Surface Area – increasing the surface
area of the solute will increase the rate or
how fast the solute dissolves, but it will
not change how much solute can be
dissolved.
3. Temperature – increasing temperature
will increase the rate or how fast the
solute dissolves in the solvent. (As
temperature increases the particles begin
to move faster and faster and collide with
more particles quicker, which means the
solute and solvent particles have an
increased chance of coming into contact
with each other.)
B. Factors Effecting Solubility:
1. Increasing Temperature - solubility of
a solid solute in a liquid solvent generally
increases with an increase in temperature.
At 20C 35.9 grams of salt will dissolve in
100 grams of water, but at 100 C 39.2
grams of salt will dissolve in 100 grams of
water!
2. Decreasing Temperatureincreases the solubility of a
gaseous solute in a liquid
solvent. What would you
rather drink, a hot coke or a
cold coke?
3. Pressure – The solubility of a gas
increases as the pressure of the gas
above the liquid increases.
Carbonated drinks have CO2
dissolved in them. They are also
bottled under a high pressure of CO2,
which forces the CO2 into solution.
When the bottle is opened, the
pressure above the solution
decreases, and bubbles of CO2 form
in the liquid, then escape. Eventually,
most of the CO2 escapes and the
drink becomes “flat.”
Henry’s Law- “At a given
temperature, the solubility, S,
of a gas in a liquid is directly
proportional to the pressure,
P, of the gas above the
liquid.”
S1 = S2
P1
P2
VII. Electrolyte VS
Nonelectrolyte
1. Electrolyte – compounds that conduct
an electric current in an aqueous solution
OR in the molten state. An electrolyte
solution contains charged particles (ions),
which can move. Any salt dissolved in
water is an electrolyte: NaCl, KI, etc.
Some polar molecules also conduct
electricity (most acids are electrolytes
because H is the only nonmetal that has a
+ charge).
Types of Electrolytes
1. Strong electrolytes – a large portion
of the solute exists as ions:
a. aqueous solutions of all ionic compounds
b. strong acids: have at least 2 oxygens per
hydrogen (H2SO4, HNO3)
c. strong bases – these are hydroxides from
Group I and II, except Be. (NaOH, CsOH,
etc)
2. Weak electrolytes – these are solutions in
which only a small portion of the solute exists
as ions
a. weak acids:
-all binary acids – HF, H2S, etc
-weak acids that have less than 2 oxygen's per
hydrogen
b. weak bases – hydroxides of everything else not
in Group I or II, including Be(OH)2
2. Non-Electrolytes- compounds that do
NOT conduct electricity in either aqueous
solution or when melted:
 distilled water
 gases
 molecular compounds (2 nonmetals)
 organic compounds – alcohols, sugars,
etc. anything containing a Carbon
Practice Problems: Tell whether each of the
following aqueous solutions would be a
STRONG, WEAK, or NON electrolyte.
1. NaCl
2. CH3Br (l) 3. HMnO4
4. HC2H3O2
5. LiOH
6. HC6H7O6
7. CO2 (l)
8. HF
VIII. Concentration: the concentration
of a solution is a measure of the amount
of solute in a given amount of solvent or
solution.
1. Molarity: the number of moles of
solute in one liter of solution.
M = amount of solute (moles)
volume of solution (liters)
OR
M = grams of solute/molar mass of solute
L of solution
A bottle labeled 6M HCl is pronounced 6
molar HCl and it was prepared by mixing 6
moles of HCl with enough water to make 1
liter of solution. Molarity is always
moles/liter. Ex: 6M is 6 moles/ liter
6M HCl
Molarity is probably the MOST
IMPORTANT unit of concentration that
we work with in chemistry. Knowing the
correct technique for preparing molar
solutions is extremely important. A
volumetric flask MUST be used. Here is
the technique: (Know it)
Determine the correct mass of solute
needed, add to a volumetric flask and fill
flask with distilled water until it reaches
the line.
Example Molarity Problems:
1. Calculate the molarity, M, of a solution prepared by
dissolving 11.85 g of potassium permanganate in
enough water to make 750. mL of solution.
2. Calculate the mass of NaCl needed to prepare 175 ml of
0.500 M saline solution.
3. Calculate the volume (in mL) needed to prepare a 2.48
M sodium hydroxide solution containing 31.52 g of the
dissolved solid.
4. How many grams of calcium chloride must be dissolved
in water to make 350. mL of a l.75 M solution?
5. What is the molar mass of 55.0 grams of a solute that
has been dissolved in enough solvent to make 500. mL
of a l.5 M solution?
2. Molarity of Ions in solution – Most
ionic solids when dissolved in water, ionize
Ex: CaCl2 (aq)  Ca2+ + 2 ClNa3PO4  3 Na+ + PO43Mg(OH)2  Mg2+ + 2 OH-
Example Molarity of Ions Problems:
Calculate the molarity of the ions in the
following solutions:
1. 0.25 M calcium phosphide
2. 2.0 M Chromium (III) chloride
3. 0.25 M barium hydroxide
4. 0.55 M aluminum nitride
3. Dilutions – you need to make 800.0 mL of a 0.25 M
solution of HCl. The only available HCl is concentrated
(12 M). How would you do this? Being able to prepare
dilutions is a very common application of chemistry. Our
department buys concentrated acids, but normally uses
more dilute solutions of these acids in our labs.
Therefore, it is important to know how to correctly
dilute.
M1V1 = M2V2
where M1 = concentrated solution
V1 = volume of concentrated solution
M2 = dilute solution
V2 = volume of dilute solution
In the above problem,
M1 = 12.0 M
M2 = 0.25 M
Vl = ?
V2 = 800.0 ml therefore
(12.0 M) (V1) = (0.25 M)(800 mL)
and V1 = 16.67 ml.
This 16.67 mL is the amount of concentrated acid
we will take out, but we still have to make 800
mL of 0.25 M. The other 783.3 mL of solution
must be water (800 – 16.67). Therefore we
would place 16.67 mL of HCl into 783.3 mL of
water and we would have our diluted solution
Practice Dilutions Problems:
l. How would you prepare 485 mL of 0.39 M
solution of NaCl when a l.0 M solution of NaCl is
all you can find?
2. If 300.0 mL of a 2.5 M solution of nitric acid is
added to 500.0 mL of water, what is the molarity
of the dilute solution?
3. Prepare 500. mL of a dilute solution (0.50 M )
of nitric acid from the l5.0 M stock solution. You
will need 600. mL of the dilute solution.
4. How would you prepare 500. mL of a 0.250 M
solution of NaCl from a 3.00 M stock solution?
5. Tell how you would prepare enough 0.75 M
NaCl solution so that 78 students working in
groups of 2 will have 12 mL of solution for a lab.
The stock solution is 3.5 M NaCl.
4. Percent Solutions (2 types):
1. Percent mass or % (m/m)– used
when a solid solute is dissolved in liquid,
usually water.
% (m/m) = grams of solute
grams of solution
Example 1: Prepare a 10.00 % NaCl
solution using 50.0 g water and solid salt:
0.l0 = x / 50 + x
and x = 5.67 g NaCl
in 50 grams of water
Example 2: How many grams of water must
be added to 25.0 g salt in order to have a
2.00 % (by mass) salt solution?
Example 3. Prepare 600.0 g of a 3.00 %
saline solution (NaCl solution).
2. Percent volume or %(v/v) – used
when a liquid solute is mixed with a liquid
solvent. The units are mL or L, but are
worked the same.
% (v/v) = mL solute
mL solution
Example 1: Prepare a 20.00 % alcohol
solution using 400.0 mL of water:
0.20 = x / 400 + x
and
x = 100 mL alcohol + 400 mL water
Example 2: What is the percent (v/v) of
ethanol in the final solution when 90.0 mL
of it are diluted to a volume of 300. mL
with water?
5. Molality: is the concentration of a solution expressed in moles of
solute per kilogram of solvent. Molality is represented by a lower
case m.
molality = amount of solute (moles)
mass of solvent (kg)
OR
molality = grams of solute/molar mass
kg of solvent
5m NaOH is pronounced 5 molal NaOH solution.
5m = 5 moles of NaOH
1 kg of water
5m or 5moles/kg
Example Molality Problems:
1. Calculate the molality of a solution prepared by
dissolving 5.0l g sodium sulfate in 700.0 g water.
2. Prepare a solution that is 2.50 molal barium nitrate in
1500. grams of water.
3. A solution is prepared by dissolving 3.00 g of potassium
chromate in 58.5 g of water. Calculate the molality of
the solution.
4. How many kg of water must be added to 8.3 g of oxalic
acid, H2C204, to prepare a 0.050 m solution?
IX. Problems involving percent
solution and density
density – ratio of mass to volume
D=m/v
Example: Calculate the mass of sodium
hydroxide in 300.0 ml of solution that is
8.00 % NaOH. The density of the
solution is 1.09 g/ml.
(1) find the number of grams of solution by
using the density of the solution
300 mL
1.09 g
= 327 g solution
mL
(2) find the grams of NaOH by using the
percent solution:
0.08 = x / 327 g x = 26.2 g
1.