Solution Chemistry
Characteristics of solutions:

solute - the material that is being dissolved into something else.
Example : when we dissolve sugar into a cup of coffee, the sugar is the solute.

solvent - the material into which a solute is being dissolved. In the above example, the coffee
is a solvent. Solvents can be almost any material or in any state but for the most part, we will
be using water as a solvent.

solution - a combination of a solute and a solvent. We can not distinguish the identify a
solute and a solvent- it is a HOMOGENEOUS MIXTURE!

A solution can be in any combination of states:
- a solid/liquid combination (dissolving a solid in water such as salt dissolved in
water).
- a liquid/liquid combination (table vinegar is a mixture of water and acetic acid).
- solid/solid combination (amalgam or any type of alloy).
- Air is an example of gas solution: nitrogen
oxygen
argon
Carbon dioxide
78.08%
20.955
0.93%
0.03%
-solutions having water as the solvent is referred to as AQUEOUS solution (blood,
saliva, etc).

There comes a point where no more solute can dissolve into the solution. At this point in time
the solution is said to be saturated.
- When solute can still dissolve in a solution, it is said to be unsaturated.
- To increase the solubility of the chemical, we can usually heat the solution up.
- The result will be a solution that contains more solute than what would normally
be dissolved in the solvent. This is called a supersaturated solution.

Solubility of a solute is the maximum amount of solute that can dissolve in a given amount
of solution at a given temperature.
-
Solubility is temperature dependent. Usually an increase in temperature
increases the solubility and increases how fast a solid dissolves in water.
-
Solubility is measured in terms of g/L, g/mL, or mol/L (at a given temp.)
-

Solubility of solids in liquids vary widely:
e.g.
Solubility of NaCl in water at 25oC: ~6 mol/L
Solubility of NaCl in ethanol at 25oC: ~ 0.0009 mol/L
e.g.
Solubility of AgNO3 in water at 25oC: ~1 mol/L
Solubility of AgCl in water at 25oC: ~0.00001mol/L
There is an upper limit to the solubility of a solid. It important to note that even
the least soluble solid will have a few particles dissolve per litre of solution.
So, we will say that a substance is soluble if 0.1 mol of the substance dissolves
per litre.
When there is no apparent limit to the solubility of one substance in another, the components
are said to be MISCIBLE.
Electrical Conductivity:

Metals (solid and liquid phase):
conduct electricity

Non-metals (solid and liquid phase):
non-conductor of electricity

Ionic compound in solid phase:
e.g. NaCl(s),
non-conductor

Ionic compound in liquid and aqueous phase:
conductor

Covalent compounds in solid and liquid phase:
non-conductor (usually!)

Acid and Base:
conductor
Examples: Which of the following would you expect to form conducting (ionic) solutions?
a) KI
b) ICl
c) HBr
f) CH3OCH3
g) N2O
h) HCl
d) CH4
e) LiOH
Examples: Which of the following would you expect to conduct electricity?
a) NaCl(s)
b) CH3OH(l)
c) Cu(s)
d) Fe(l)
e) K2CrO4(aq)
f) C14H10(s)
g) CO2(s)
h) KBr(aq)
i) KBr(s)
j) Ag(s)
k) Hg(l)
l) Na(s)
Polarity of Molecules:
A. Van der Waal’s Forces (intermolecular interaction):
1. Dipole-dipole interactions:

molecules that are dipolar are attracted
electrostatically with one another.

Polar molecules have a partial positive
and a partial negative portion of the
molecule.

They are made of atoms having
differences in ELECTRONEGATIVITY.

Polar molecule must also be
ASYMETRICAL.
2. London Forces:

are weak attractive forces which arise as
the result of temporary dipolar attraction
between neighbouring atoms.

London Forces are always present, even
in species that have dipolar attraction.

This is a weaker attraction than dipoledipole forces.
B. Hydrogen Bonding (intermolecular interaction):

Hydrogen bonding occurs when hydrogen
is bonded covalently to N, O, or F (high
electronegativity). Highly polar.

Hydrogen bonding is stronger than the
Van der Waal’s Forces.

The presence of H-O- , H-N-, H-F indicate
a highly polar section of the molecule.

Asymetrical molecules are POLAR
molecules.
Polar and Non-Polar Solvents:
Polar Molecules:
- Dipole is present
- Molecule is ASYMETRICAL (not symmetrical).
eg. HCl
Chloromethane
eg. formaldehyde
Polar Solvent:
Non-polar Solvent:
acetylene
Water
HBr
Ammonia
propanol
Methanol
benzene
dimethyl ether
ethanol
chloroform
“Like dissolves like” :


polar solvents dissolve polar solutes and ionic compounds
Non-polar solvents dissolve non-polar solvents
A. Polar and ionic molecules are soluble in
polar solvents.
For example:
For example:
Salt (ionic) dissolves in water (polar solvent).
I2(s) is more soluble in water than gasoline.
non-polar molecules are soluble in non-polar solvents.
B. Polar and ionic compounds are generally
held together quite tightly.
Polar solvents are strongly attracted to the ions
or the polar compounds causing separation of
the solute out of its crystal structure and into
solution.
Non-polar solvents do not have this attraction,
therefore, polar solutes are unable to separate
into a non-polar solution.
C. Non-polar solutes are held together by
London Forces.
Solvents with enough London Forces to attract
the solutes and break off the solid crystal
structure are capable of dissolving the solid.
Generally speaking, non-polar solvents have
high London Forces, and dissolve other
non-polar compounds.

Hydrogen bonding holds water molecules together. For example, water
molecules undergo hydrogen bonding. When water freezes, the six sided
structure of water crystals is due to hydrogen bonding.
The nature of Solutions of ions:

The formation of solution depends on the ability of the solute to dissolve in the solvent.

Ionic solid are crystal structure made up of ions.

Covalent solids are crystal structure made up of covalent molecules.

Crystal lattice is an orderly arrangement of particles that exists within a crystal.

Water: “Universal Solvent”, polar, V-shaped molecule. Its positive end attracts the
negative ions and the negative end attracts the positive ions.

Dissociation: reaction that involves the separation of ions in an ionic compound.
KBr(s) ------- > K+(aq) + Br-(aq)

Ionization: reaction that involves the breaking up of a neutral molecule into ions.
CH3COOH(l) ----------- > CH3COO - (aq) + H +(aq)
Assignment: read p. 133-148, Zumdahl. Exercises:p. 180-181 # 11, 13, 15, 17, 25
Calculating The concentration of Ions in Solution:
concentration - this is the amount of solute that is in a certain amount of solution. Because
'amount of solute' can be interpreted in a number of different ways (i.e. grams, moles, molecules
etc.) we will define a value for concentration :
moles solute (moles
concentration =
------------------------------volume solution (Litre)
This is also defined as MOLARITY.
Molarity =
moles of solute
-----------------------------------volume (in litres) solution
The units for molarity are moles/litre, mol/L or M for short.
Examples :
1) What is the concentration of a solution containing 0.055 moles of HCl dissolved in water to
make 250.0 mL of solution ?
Solution:
moles
0.055 mol
M=
-------- -=
--------- --= 0.22 M HCl
volume
0.2500 L
2) What is the concentration of a solution containing 0.055 grams of HCl dissolved in water to
make 250.0 mL of solution ?
Solution :
0.055 g x
moles HCl =
molarity =
0.0015 mol
----------0.250 L
1 mol
---------36.5 g
=
0.0015 mol HCl
=
0.0060 M HCl
3) How many molecules of NaOH are dissolved in 125.3 mL of a 3.00 M NaOH solution ?
Solution :
moles NaOH = Molarity X volume
= (3.00 M)(.1253 L)
= 0.376 mol NaOH
molecules NaOH = moles X Avogadro's number
= (0.376 mol)(6.02 x 1023)
= 2.26 x 1023 molecules NaOH
How to make a solution of a certain concentration:
In order to make a solution of certain concentration, we do NOT dissolve the amount of
solute needed in the volume of solvent. If we did that, the volume would increase somewhat and
we would not have the precise concentration we require. In order to accomplish this, we simply
determine the amount of solute needed, dissolve this in about 1/2 the amount of solvent and then
fill up to the proper amount of solution. It is best to use a volumetric flask for this purpose.
Example :
Describe how you would make 125 mL of a 0.250 M solution of sodium hydroxide.
Solution :
mol NaOH = Molarity X volume
= (0.250 M)(0.125 L)
= 0.03125 mol NaOH
mass NaOH = moles X molar mass
= (0.03125 mol)(40.0 g/mol)
= 1.25 g NaOH
Dissolve 1.25 g of NaOH into about 60 mL distilled
water. Add more distilled water to raise volume up
to 125 mL.
- Assignment :
- Solution Worksheet #1 - Molarity
Ionic concentrations
When a solid dissolves in a solvent it can act one of three ways - it can either dissociate
into its ions (only if it is an ionic compound);it can remain as a molecule dissolved in water (if it is
not an ionic compound);or it can react with the water (we will not consider this at the present
time). For the sake of Chemistry 11, we will discuss only what happens when an ionic substance
dissolves in a solvent.
First of all, why do some ionic compounds dissolve readily in water while some others
appear to not be as readily soluble?
The reasons for this are due to the relative shapes and structures of both the solute and
solvent molecules. Sometimes the molecules have a slight charge on one side of the
molecule compared to the other. This type of molecule is called POLAR. Water is a good
example of this. A NON-POLAR molecule, such as carbon tetrachloride, can also be used as
a solvent. The polarity of the molecules is the basis of dissolution, but we will not get into that
aspect of chemistry in this course.
When an ionic compound dissolves in water, it breaks into its ions.
Example :
NaCl (s) ----> Na+(aq) + Cl-(aq)
CaF2 (s) ----> Ca2+(aq) + 2 F-(aq)
We are going to assume that all of the compound breaks into its ion state. This is not
always true, but we will leave that for Chem 12.
If all of the compound ionizes, then we actually have none of the solute molecules left in
the solution. By looking at the ratios between the ions and compound (from the dissociation
equation) we can determine the concentrations of each individual ion in the solution.
(The abbreviation for concentration of an ion is to place the symbol for the ion inside square
brackets. For instance, shorthand for 'the concentration of the hydrogen ion' is [H+].)
Examples :
Find the concentration of each ion in the following solutions :
a) 12.0 M HCL b) 18.0 M H2SO4
c) 1.15 M Cr2(SO4)3
Solutions :
a) HCl ---> H+ + Cl[H+] = 12.0 M
[Cl-] = 12.0 M
b) H2SO4 ---> 2H+ + SO42-
[H+] = 36.0 M
c) Cr2(SO4)3 ---> 2 Cr3+ + 3 SO42-
[Cr3+] = 2.30 M
[SO42-] = 18.0 M
[SO42-] = 4.45 M
Dilutions
When a solution of known concentration has its volume changed, it stands to reason that
its concentration changes also. If the liquid we are adding does not contain any of the original
solute OR if a reaction does NOT take place, the amount of dissolved solute does not change only the volume changes. In fact the concentration will change by a factor equal to the change in
volume. In short, the new concentration can be found by using the following formula :
C1V1=C2V2
old volume
-------------------new volume
new concentration = old concentration X
Examples :
1) 125 mL of 3.00 M HCl has 35.8 mL of water added to it. What is the new concentration ?
Solution :
old volume
new M = old M X
-------------------new volume
= 3.00 M X
25 mL
-------------160.8 mL
= 2.33 M
2) 15.0 mL of a solution of 1.50 M HCl has some water added to it changing the concentration of
the acid to 1.15 M. What volume of water was added ?
Solution :
old volume
new M = old M X
---------------------therefore rearranging
new volume
we get the following
new volume = old volume X
= 15.0 mL X
old M
-----new M
1.50 M
------1.15 M
= 19.6 mL
If the new volume was 19.6 mL and the old volume was 15.0 mL, then I guess we added 4.6 mL
of water to the acid.
3) How much 6.00 M NaOH would we have to use to produce 275 mL of 0.500 M NaOH ?
Solution :
new M = old M X
old volume
----------------new volume
therefore rearranging
we get the following
old volume = new volume X
= 275 mL X
new M
------------old M
0.500 M
--------------6.00 M
= 22.9 mL of 6.00 M NaOH needed. (I
guess we would also need about
252.1 mL of water but that was not
asked.)
- Assignment : Begin Solution worksheet #2 (to be completed next class)
- Quiz on Molarity worksheet #1
Mixing together solutions of different concentrations
As we said last class, if we mix together two solutions of different concentrations
together, the formula we used can NOT be used. In order to find the new concentration, we must
find the amount of solute in each solution.
Example : Determine the new concentration of HCl if
225 mL of 1.50 M HCl is mixed with 115 mL of
2.50 M HCl.
Solution:
moles HCl from first solution
mol = c X V
= (1.50 M)(0.225 L)
= 0.3375 mol HCl
moles HCl from second solution
= (2.50 M)(0.115 L)
= 0.2875 mol HCl
total moles HCl = 0.3375 mol + 0.2875 mol
= 0.625 mol HCl
total volume = 225 mL + 115 mL
= 340 mL
total moles
final concentration =
----------------total volume
= 1.84 M
- complete worksheet #3
=-
0.625 mol
----------------0.340 L
Reactions in Solutions
Net Ionic Equations
When two solutions are mixed together, a reaction might occur. (Demo : 0.1 M Pb(NO 3)2
+ 0.1 M KI). The reactions that we are going to talk about here are all double replacement
reactions so if a reaction does take place, we can determine the products.
Pb(NO3)2 + 2 KI ---> PbI2 + 2 KNO3
The above equation is called a chemical equation. We have done these in the past.
However, we have recently seen that we don't actually have any Pb(NO3)2 or KI in the solution,
just their ions. Also, on the product side of the equation we have the yellow solid being PbI2 (we'll
figure out how to determine which is the precipitate later) and a solution of KNO3. Rewriting the
equation to show this we get
Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) ---> PbI2 (s) + 2K+(aq) + 2NO3-(aq)
This is called an overall ionic equation and is not used that often because of a number of reasons
- not the least of which is that it takes up too much space !!
If we look carefully at the overall ionic equation, we see that there are 2NO3-(aq) ions
and 2K+(aq) ions on either side of the equation. If we cancel them out, we get
Pb2+(aq) + 2I-(aq) ---> PbI2 (s)
This is called a NET IONIC EQUATION for the reaction shown above. It shows exactly
what is happening in the solution. The NO3-(aq) ions and the K+(aq) ions are actually doing
nothing in the reaction. They can be thought of as just watching the lead ions and the iodide ions
get together. For this reason, they are referred to as SPECTATOR IONS. Spectator ions are
never included in net ionic equations.
Another example :
CuSO4 + 2 NaOH ---> Cu(OH)2 + Na2SO4
assume that the
precipitate is Cu(OH)2
overall ionic equation
Cu2+(aq) + SO42-(aq) + 2Na+(aq) + 2OH-(aq) --->Cu(OH)2 (s) + 2Na+(aq) +
SO42-(aq)
net ionic equation
Cu2+(aq) + 2OH-(aq) ---> Cu(OH)2 (s)
If you look carefully at the net ionic equation of any precipitation reaction, you will find that
the equation is simply how we would go about making the solid. We do not always have to go
about all three equations in order to obtain the net ionic equation. All we have to do is to identify
the precipitate (don't worry, we'll get to it) and simply write how we would produce it using ions.
Example : if we knew we got a precipitate of Ca3(PO4)2 in a
reaction, the net ionic equation would be
3 Ca2+(aq) + 2 PO43-(aq) ---> Ca3(PO4)2 (s)
This would be the net ionic equation for the formation of the precipitate
Ca3(PO4)2 no matter what solutions were used in the first place. The other ions would just be
spectator ions.
There is one type of net ionic equation that occurs when an acid and a base are mixed
together. A precipitate is NOT formed but a reaction does take place.
Acids we know - HCl, H2SO4 , HNO3 etc. If we look at them carefully, all acids contain an
H atom. More specifically, when it ionizes, an H+ ion is released.
Bases can be identified by the presence of an OH- ion.
When an acid is added to a base, water and a salt are formed (review from types of
reactions). Let's see what happens in the reaction between NaOH and HCl.
NaOH + HCl ---> H2O + NaCl
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) ---> H2O (l) + Na+(aq) + Cl-(aq)
H+(aq) + OH-(aq) ---> H2O (l)
This is the net ionic equation for the reaction between any acid and any base.
- Assignment :
Prepare 2 data tables for 16-D
The tables should be 7x7 matrices - do NOT have the students write in the chemicals
before they come into the lab. They might find the chemicals are different. The data tables are
their 'ticket' into the class - no
data tables - no entry !!!!
- Lab 16-D
Formation of precipitates
We saw in the lab yesterday that some compounds are soluble in water and some are
not. We will use the definition here that soluble means that we can make at least a 0.1 M solution
of the compound at 25oC. How can we predict if a precipitate will form ? The answer is not easy.
In fact we don't have too many rules that always apply. Chemists have done thousands of
experiments similar to the one you did yesterday and have collated the results into pages and
pages of data tables. We will use a portion of those tables and rules - the data table on page 459
of the text book. (The students will not be required to memorize this table - they will be provided
with one for the exams.)
To use this table, we determine what the product will be in a reaction (remember, the
reactions here are double replacement reactions.) From here we must check out both products to
see if there is a precipitate formed.
Examples : 1) Identify the precipitate (if any) when the following solutions are mixed:
a) lead (II) nitrate and sodium chloride
b) ammonium hydroxide and copper (I) sulfate
c) potassium phosphate and cesium sulfide
Answers : the following are the products formed. The one that is underlined is the precipitate.
a) lead (II) chloride and sodium nitrate
b) ammonium sulfate and copper (I) hydroxide
c) potassium sulfide and cesium phosphate
2) Write the net ionic equation for the formation of any precipitate when solutions of the following
are mixed. If no precipitate is formed, write NO REACTION.
a) barium sulfide and ammonium carbonate
b) silver nitrate and sodium acetate
c) iron (III) chloride and copper (II) chloride
d) lead (II) acetate and lithium phosphate
Answers :
a)
b)
c)
d)
Ba2+(aq) + CO32-(aq) ---> BaCO3 (s)
Ag+(aq) + CH3COO-(aq) ---> AgCH3COO (s)
NO REACTION
3 Pb2+(aq) + 2 PO43-(aq) ---> Pb3(PO4)2 (s)
Precipitation Formation and Ionic Concentrations
As we can see from the above and from the lab, sometimes we get a precipitate formed
in a reaction. What does this do to the concentrations of the ions in solution ? It lowers those
involved. We will, for the sake of argument here, assume that one of the ions is completely used
up in the reaction while some of the other one may still remain. If this sounds familiar, it should.
This is really a stoichiometric INXS calculation. It is best set up in the following example :
Example : 25.0 mL of 1.00 M NaCl is mixed with 35.0 mL of2.20 M AgNO3. A precipitate is
formed and falls out of solution. Calculate the concentrations of all ions remaining in solution after
the precipitation stops.
Answer : It is best to first write the net ionic equation for the reaction that occurs.
Ag+(aq) + Cl-(aq) ---> AgCl (s)
Set up a table similar to the one below :
Ion
Moles Before Moles After
Volume
Concentration
Na+
0.0250 mol
0.0250 mol
0.0600 L
0.417 M
Cl-
0.0250 mol
0 mol
0.0600 L
0.000 M
Ag+
0.0770 mol
0.0520 mol
0.0600 L
0.867 M
NO3-
0.0770 mol
0.0770 mol
0.0600 L
1.28 M
Example : 20.0 mL of 0.500 M AgNO3 is mixed with 15.2 mL of0.750 M Na2S. Calculate the
concentrations of allions remaining in solution.
Answer :
Ion
2 Ag+(aq) + S2-(aq) ---> Ag2S (s)
Moles Before
Moles After
Volume
Concentration
Ag+
0.0100 mol
0 mol
0.0352 L
0.000 M
NO3-
0.0100 mol
0.0100 mol
0.0352 L
0.284 M
Na+
0.0228 mol
0.0228 mol
0.0352 L
0.648 M
S2-
0.0114 mol
0.0064 mol
0.0352 L
0.182 M
- Assignment :
Read p. 157, exercise p. 182 # 29-32, 35,37, 39, 43
Solution worksheet #4
Lab: Introduction of Acids and Bases, (lab 20-A)
read p. 158-164
Acids and Bases
From the lab, we should be able to identify an acid and a base from its properties and its
formula. In short, the chemical properties of solutions of acids and bases are as follows :
Acids :
1) taste sour
2) conduct an electrical current
3) cause certain dyes to change colour (e.g. litmus
goes red)
4) liberates hydrogen when it reacts with certain
metals
5) loses the above properties when mixed with a base
though the resulting solution conducts
electricity.
Bases :
1) taste bitter
2) conduct an electrical current
3) cause certain dyes to change colour (e.g. litmus
goes blue, phenolphthalein goes pink)
4) feels slippery
5) loses the above properties when mixed with a base
though the resulting solution conducts
electricity.
In their respective chemical formula, and more importantly when they dissolve in water,
acids have an H+ ion while bases have an OH- ion. (This is an incomplete definition of an acid
and a base but will suffice for chemistry 11.)
When an acid and a base react, the result is a salt and water. The type of reaction is
called a NEUTRALIZATION reaction because the properties of the acid and the base are
neutralized by each other. The net ionic reaction for this reaction is as follows :
H+(aq) + OH-(aq) ---> H2O (l)
We must remember that in any neutralization reaction, the moles of H+ MUST equal the moles of
OH-.
Example : What volume of 6.00 M HCl must be added to 125 mL of 1.59 M NaOH in order to
neutralize it ?
Answer :
NaOH + HCl ---> NaCl + H2O
moles OH- = (1.59 M)(0.125 L) = 0.199 mol OHmoles H+ = moles OH- = 0.199 mol H+
mol HCl = mol H+ = 0.199 mol HCl
vol HCl =
mol HCl
------- =
M HCl
0.199 mol
--------- = 0.0331 L (33.1 mL)
6.00 M
Example : What volume of 0.10 M Ca(OH)2 is needed to neutralize 115.2 mL of 0.55 M H3PO4?
Answer :
mol H3PO4 = (0.55 M)(0.1152 L) = 0.063 mol H3PO4
mol H+ = 3 x mol H3PO4 = (3)(0.063 mol)= 0.19 mol H+
mol OH- = mol H+ = 0.19 mol OHmol Ca(OH)2 = (mol OH-)/2 = (0.19 mol)/2 =
0.095 mol Ca(OH)2
mol Ca(OH)2 0.095 mol
vol Ca(OH)2 = ----------- =
--------------=
0.95 L
M Ca(OH)2
0.10 M
(950 mL)
Titrations
A titration reaction is simply a neutralization reaction using special techniques. A given
volume of an acid is placed in a flask (as well as some indicator - phenolphthalein is commonly
used). The base is placed in a burette and added drop wise until the colour of the solution in the
flask just changes. From here we can get the volumes needed for neutralization calculations.
- Assignment :
exercise p. 183 # 45, 46, 47, 49, 51 Zumdahl!
Solution worksheet #3
- net ionic equations,
- neutralization calculations
- titration calculations
- Set up lab 20-C
Day 10:
- Lab 20-C - day 1
- Lab 20-C day 2
- collect lab 20-C
- go over worksheet #3
Demonstration titration lab
Effectiveness of an Antacid
1.
Obtain several commercial antacids. (Tums, Rolaids, Maalox, Eno etc.)
2.
Place one 'dose' of an antacid in an Erlenmeyer flask.If the antacid is a
Tums), grind it up with a mortar and pestle first.
3.
Place 25 mL of distilled water in the flask with the antacid.
4.
Place 10.0 mL of 1.0 M HCl (stomach acid) in the flask.
5.
Add a few drops of phenolphthalein in the flask.
6.
Let the antacid react in the 'stomach' (flask) for a
minute or so.
7.
Titrate using 1.0 M NaOH. Record the volume of NaOH
needed to neutralize the excess stomach acid.
8.
Compare the volume needed for each antacid. The smaller
the volume of NaOH needed, the more effective the
antacid has been in neutralizing stomach acid. Note :
if the volume of NaOH needed exceeds the 10 mL of HCl
added, then the antacid actually adds acid to the
stomach. (Read package of Eno carefully - it doesn't
actually say antacid - now read contents - over 55% of
the contents are acid !!! Why ? plop plop fizz fizz.
9. A discussion might also ensue on the 'fillers' of the
tablets such as Tums. Weigh two tablets. Compare the
measured mass with the mass of the 'active
ingredients'. The discrepancy in mass is due to a
number of fillers.
- Assignment : Solution Worksheet #5 - Review
solid (like
- go over the homework - solution review
- exam next day on solution chemistry
- Begin year end review
NOTE : the cumulative exam coming up does not include all of the material covered in
the solution unit. For that reason, the cumulative exam can be given either at the end of the term
or about half way through the last unit.
- solution exam