1. Objectives:
Examine the different types of solutions:
unsaturated, saturated, and supersaturated.
Describe the composition of a solution using
percent by mass and percent by volume.
Identify and explain the factors that affect
solubility.
10. Soluble – water-loving solute
(can be dissolved in water)
Example:
sugar, salt, coffee, milk powder
11. Insoluble – water Fearing
solute (cannot be dissolved
in water)
Example:
Sand, paper, oil, rubber,
wood
12. Saturated
• Saturated solutions contain the
maximum amount of solute that can
dissolve
• All of the spaces between the solute
and the solvent are filled.
• Any more solute that is added will
not dissolve.
13. 13
* soda is a saturated solution of carbon
dioxide in water.
* Adding chocolate powder to milk so
that it stops dissolving forms a
saturated solution.
* salt can be added to melted butter or
oil to the point where the salt grains
stop dissolving, forming a saturated
solution
14. Unsaturated
* Are still able to dissolve more solute.
* The solution still contains empty spaces
between the solute and solvent particles.
* Both dilute and concentrated solutions
are unsaturated.
Example: Tea and the sugar solution
15. Supersaturated
* Solutions contain more solute that can
dissolve in a solution
* Solute will remain undissolved in a
supersaturated solution.
Example: Honey is technically a
supersaturated solution,
meaning it contains more sugar than
would normally dissolve at that temperature
16. Composition of solutions by specifying the
amount of solute in a given quantity of
solution.
Percent by mass, sometimes referred to
mass percent or weight percent is one way of
numerically describing the concentration of a
solution. This is expressed in %
𝒎
𝒎
. Mass
percent is usually applied to solutions
containing a solid solute dissolved in water.
18. Reiza prepared a solution by adding 2.0g of
salt (sodium chloride) in 48.0 g of water, the
total mass of the solution is 2.0g salt + 48.0
g water = 50.0 g solution. The mass percent
of the solute (salt) is then given by:
2.0 𝑔
50.0 𝑔
𝑥 100% = 4.0%
19. Percent by volume
This concentration is usually used with
solutions containing two liquids, where it is
easier to measure the volume instead of the
mass. This is expressed in %
𝑽
𝑽
.
%
𝑉
𝑉
=
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑥 100% 𝑜𝑟
20. Reiza found out that the rubbing alcohol
(ethanol) that she uses has a concentration of
70% by volume. What would be the original
volume of ethanol needed to make 100 mL of
70%
𝑉
𝑉
ethyl alcohol?
70%
𝑉
𝑉
=
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
100𝑚𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑥 100%
21. Cross multiplying %
𝑉
𝑉
with
100mL and then divide it
100, and you will get 70mL.
This means that 70mL of
pure ethanol dissolved in
100mL of solution gives a
70% by volume
concentration of alcohol.
