1. SOLUBILITY AND
CONCENTRATION OF MIXTURES

2. What is the
maximum amount
of salt that will
completely dissolve
in a cup of water?

3. Procedure:
1. Put 20 mL of water in a small
clear transparent bottle. Add ½
teaspoon of salt and stir until all
the salt dissolves.
2. To the salt solution in step #1,
add ½ teaspoon salt, a small
portion at a time and stir the
solution to dissolve the salt. At this

4. 3. Add ½ teaspoon of salt to
the salt solution in step #2
and stir the solution. At this
point, you have added one
and ½ teaspoons of salt.
4. Continue adding ½
teaspoon salt to the same cup
until the added salt no longer

5. Q1. How many teaspoons of
sugar did you add to 1 cup of
water until the sugar no longer
dissolves? __________
NOTE: In this step, you will
observe that there is already
excess sugar which did not
dissolve.
Q2. What is the maximum

6.  In this activity, you have observed
that there is a maximum amount of
solute that can dissolve in a given
amount of solvent at a certain
temperature.
 What do you think is it called?
-SOLUBILITY
(of the solute)

7.  The solution that contains the
maximum amount of solute dissolved
by a given amount of solvent is called
a
-saturated solution.
 If you add more solute to the
solvent, it will no longer dissolve. The
solution has reached its saturation
point.
 The presence of an excess solid
which can no longer dissolve is an

8.  Is there any container
where all solids
dissolved? Which
container is this?

9. A solution is unsaturated
when it contains less
solute than the maximum
amount it can dissolve at a
given temperature.

10.  A more measurable way to find out
the solubility of a solute is to
determine the maximum amount
that can be dissolved in 100 g of
solvent at a specific temperature.
There are available data from
chemistry books that give the
solubility of common solutes at
particular temperatures. Figure 2
shows the solubility of table salt at

12. Concentration of
Solutions
The concentration
describes the relative
amounts of solute
and solvent in a
given volume of

13. Observe the
demonstration
.

14. Describe the concentrations
of solutions qualitatively (by
simply observing their
appearance) and
quantitatively (by comparing
the number of drops per
volume of water).

15. From Part 1 of the
demonstration, you were able to
describe the solutions as having
quantitative concentrations of
1 drop/50 mL and 10 drops/50
mL. Qualitatively, you were
able to distinguish the bottle
with 10 drops/50 mL more
concentrated (darker) than the
bottle with 1 drop/50 mL.

16. When there is a large amount
of dissolved solute for a
certain volume of solvent, the
solution is
-concentrated.
A solution that has a small
amount of dissolved solute in
comparison to the amount of

17.  Now that you have distinguished dilute from
concentrated solutions qualitatively and
quantitatively from your teacher’s
demonstration, you can express
concentration in other ways such as:
 (1) percent by volume, which is the amount
of solute in a given volume of solution
expressed as grams solute per 100 millliter of
solution (g/100 mL), and
 (2) percent by mass, which is the amount of
solute in a given mass of solvent expressed

18.  Labels of products sold often show the
concentrations of solutes expressed as
percent (%) by volume or mass. The
alcohol used as a disinfectant is a
solution of 70% ethyl or isopropyl
alcohol, meaning 70 mL alcohol. There
are also solutions sold as 40% ethyl or
isopropyl alcohol.
 Vinegar is often labeled as “5% acidity,”
which means that it contains 5 grams of
acetic acid in 100 g of vinegar. The
common antiseptic, agua oxinada is a

19.  The concentration of solid solutions, like
gold jewelry, is expressed as karat. Pure
gold is referred to as 24 karats. Jewelry
that is said to be 18 karats contains 18
grams of gold for every 24 grams of the
material, 6 grams consist of the other
metal like copper or silver. This material
has a concentration of 75% gold, that
is, [18/24(100)]. A 14 karat (14K) gold
contains 14 grams gold and 10 grams of
another metal, making it 58.3% gold.

20. The following sample problems
show you that there is a way to
know the exact ratio of solute to
solvent, which specifies the
concentration of a solution.
Sample problem 1
How many mL of ethyl alcohol are
present in a 50 mL bottle of
rubbing alcohol?

21. Calculation for sample problem
1
 Since rubbing alcohol contains 70%
ethyl alcohol, it means that 100 mL
of rubbing alcohol contains 70 mL
ethyl alcohol. So, the following
calculations show that in 50 mL of
rubbing alcohol, there is 35 mL ethyl
alcohol. The water content is most
likely more. There is no easy way to
determine but it would be incorrect

22. Sample problem 2
A one peso coin has a
mass of 5.5 grams.
How many grams of
copper are in a one
peso coin containing
75% copper by mass?

23.  Calculation for sample problem 2
 75% by mass means 75 grams of copper in 100
grams of one peso coin.
 So, a 5.5 grams coin contains,
 75 g copper x 5.5 g coin = 4.1 g copper
100 g coin