Detailed Lesson Plan in Chemistry

1. ORPILLA, LESTER E.
Detailed Lesson Plan in Chemistry
I. Objectives: At the end of the lesson, the students will be able to:
1. define the meaning of concentration of solution in their own understanding
correctly;
2. identify the importance of concentration of solution in our daily life and;
3. solve percentage by mass and volume related problems.
II. Subject Matter:
A. Topic: Concentration of Solutions: Percentage by Mass and Volume
B. Reference: Principles of General Chemistry by Patricia Amateis p. 214-130
C. Materials: alcohol bottle, GSM bottle, iodine bottle, cartolina
D. Skills: Problem solving
E. Values: Patience, Honesty
F. Concept: The concentration of solutions may be expressed in terms of
percentage or parts solute in 100 parts solution.
III.Procedure:
Teacher’s Activity Learner’s Activity
A. Preliminary Activities
Greetings
Prayers
Checking of Attendance
Classroom Management
B. Review
Before we proceed to our lesson, let’s
have a recap from what we have
discussed last meeting.
Very good, what else?
Okay, So what is solubility?
Very well said. And what are the
factors that affect solubility?
Sir, last meeting we have discussed about
solubility.
Sir, we have also discussed about factors
that affect solubility.
Sir, solubility refers to the extent to which a
solute dissolves in a given solvent.
Sir, the factors that affect solubility are: the
nature of solute and solvent, temperature
and pressure.

2. You’re right! How does the nature of
solute and solvent affects solubility?
Very good. How about temperature?
Precisely! So, how about pressure?
Very good!
C. Motivation
Class, I have here an ethyl alcohol, a
betadine and a ginebra san miguel
containers. What can you observed
about their labels?
Yes. Aside from that, what else?
Good, but I am asking for what have
you observed about their labels. Okay,
what else?
Definitely! What do these percentages
implies?
You’ve got it!
D. Developmental Activities
So, what is concentration of solution?
What do you think is the importance of
concentration of solutions in our daily
life?
Sir, solubility of a solute in a solvent purely
depends on the nature of both solute and
solvent. A polar solute dissolved in polar
solvent. Solubility of a non-polar solute in a
solvent is large. A polar solute has low
solubility or insoluble in a non-polar solvent.
Sir, for an endothermic solution, increasing
the temperature increases the solubility of
solid and liquid solvent. However for gases,
an increase in temperature decreases
solubility.
Sir, increasing the pressure increases the
solubility of gases in liquid as it allows more
gas molecules to be intact with the solvent.
Sir, they have volumetric measurement.
Sir, they are examples of solution.
Sir, there are percentage of solutions and
proofs indicated.
Sir, it means that the solution is
concentrated.
Sir, concentration of solution refers to the
amount of solute in a given amount of
solvent or solute.

3. Very good! What else?
The amount of concentration of solute
in a given amount of solvent can be
expressed qualitatively. And what are
they?
Exactly! What is dilute?
Very good! And what is a concentrated
solution ?
You’re right!
A more accurate way of expressing
the amount of solute in a given
amount of solvent is in terms of
percentage, morality, molality,
normality, parts per million, and parts
per billion. But for today will be dealing
with percentage by mass and volume.
What is the formula in solving for the
percentage by mass/ weight?
That’s it! Where weight of solution is
equal to weight of solution plus weight
of solvent.
For example, a solution that is labeled
5% NaCl means that there are 5g of
NaCl for every 95g of solvent or 100g
of solution.
Take note that 1g is equivalent or
equal to 1 mL.
Okay, let’s have a problem solving.
Sir, it is important in the field of medicine
because it indicate the solute needed for
making medicines in the appropriate
amount needed by body of a patient
Sir, it is important because it indicates the
intensity of the solution needed. For
example in muriatic acid, the concentration
there is indicated to know where to use it.
Because the higher concentration,the very
corrosive it is?
Sir, dilute and concentrated.
Sir, dilute is a solution that contains a small
proportion of solute relative to solvent.
Sir, concentrated is a solution that contains
a large proportion of solute relative to
solvent.
wt/wt % = [(mass of solute)/(mass of
solution)] x 100%

4. What is the percentage by mass of a
salt solution prepared by dissolving
11g of salt in 80g of solution?
Check. Very good!
Let’s have another example.
Express the concentration of a solution
prepared by dissolving 40g of salt in
160g of water.
Kindly explain your work.
Check. Very good!
Can you follow class?
Given: wt. of solute = 11g
wt. of solution= 80g
Formula:
wt/wt % = [(mass of solute)/(mass of
solution)] x 100%
Solution: % by wt = 11g x 100
80g
= 0.1375 x 100
= 13.75%
Substitute the given values in the
formula, 11g over 80g times 100,
cancel the units of measure, 11
divided by 80 is 0.1375 multiplied to
100 is 13.75%.
Given: wt. of solute= 40g
wt. of solution= 40g + 160g= 200g
Formula:
wt/wt % = [(mass of solute)/(mass of
solution)] x 100%
Solution: % by wt. = 40g x 100
200g
= 0.20 x 100
= 20%
Notice that there was no weight of
solution given. So, what we are going to do
is to add the weight of the solute and the
solvent, 40g + 160g is 200g. Substitute the
given values in the formula, 40g over 160g
times 100, cancel the units of measure, 40
divided by 200 is 0.2 multiplied to 100 is
20%.
Yes sir.

5. Let’s move on to percent by volume.
This is generally used when the solute
is a liquid or gas.
What is the formula in solving the
percentage by volume?
Very good! This is widely used in
determining the alcohol content of
alcoholic drinks.
The amount of alcohol in alcoholic
beverages is expressed as…..what?
That’s right! And what is the formula
used for solving the proof number?
Let’s solve.
A solution is prepared by mixing 50 ml
of C2H5OH in 300 mL of distilled
water. What is the percent by volume
concentration and the proof number of
the solution?
Kindly explain your work.
v/v % = [(volume of solute)/(volume of
solution)] x 100%
Sir, proof number.
Sir, 2(%by volume)
Given: vol. of solute= 50 mL
vol. of solvent= 300mL
Formula:
v/v % = [(volume of solute)/(volume of
solution)] x 100%
Solution: %by vol.= 50 mL x 100
50 mL + 300mL
= 50 mL x 100
350mL
= 0.1428 x 100
= 14.28%
Proof No.= 2(14.28)
= 28.56
The volume of the solute is 50 mL
and notice that only the volume of the
solvent is given not the volume of the
solution. So, what we are going to do is to
add the volume of the solute and the
volume of the solvent to come up with the
volume of the solution, 50 mL plus 300 mL
is equal to 350 mL. Substitute the given
and computed values in the formula, 50 mL
over 350 mL times 100, cancel the units of

6. That’s Correct!
Okay class, can you follow?
Let’s have another example.
What volume of alcohol is required to
prepare 250 mL of a 10% by volume
solution?
Kindly explain how you get your
answer.
Very good. That’s the right thing to
solve when one of the values is
unknown.
E. Application
Direction: Solve the following problem
in your notebook.
1. What is the percentage by mass of
measure, 50 divided by 350 is 0.1429
multiplied to 100 is 14.28 %. To get the
proof number of the solution, we have to
multiply it by 2, so the answer is 28.58
proofs.
Yes sir.
Given: vol. of solution= 250 mL
vol. of solute= 10
% by volume= (unknown)
Formula:
v/v % = vol of solute x 100%
vol of solution
vol. of solute = % by vol. x vol. of solution
100
= 10 x 250 mL
100
= 2500 mL
100
= 25 mL
The volume of the solution is 250
mL, the volume of the solute is unknown,
and the percent by volume is 10. Since the
volume of the solute is unknown, what we
are going to do is to derive the formula we
come up with the formula; volume of solute
is equal to percent by volume multiplied to
the volume of solution divided by 100.
Substitute the given values, 10 multiplied to
250 is 2500 divided by 100 is 25 mL.

7. a sugar solution prepared by
dissolving 40g of sugar in 120g of
water?
2. A man drinks 500 mL of an 86
proof wine per week. What is his
total alcohol intake per week?
3. What weight of solute will be
required to produce 400g of a 10%
solution? How many mL be neede?
Solve on the board.
1. Given: wt. solute= 40g
wt. of solvent= 120g
Solution:
%by wt= wt. of solute . x 100
wt. of solute + wt. of solvent
= 40g x 100
40g + 120g
= 40g x 100
160g
= 0.25 x 100
= 25%
2. Given: proof= 86
wt. of volume= 500mL
vol. of solute= (unknown)
% by volume= (unknown)
Solution:
Proof= 2 (%by volume)
86= 2 (%by volume)
2 2
%by volume= 86
2
= 3
v/v % = vol. of solute x 100%
vol. of solution

8. Very good! All your answers are
correct.
F. Generalization
Okay class, who can summarize the
gist of our discussion this day?
How to solve percentage by mass and
volume in a given solution?
43 = vol. of solute x 100%
500
vol. of solute = 43 (500 mL)
100
= 21500 mL
100
= 215 mL
3. Given: wt. of solution= 400g
% by wt.= 10
wt. of solute= (unknown)
Solution:
%by wt= wt. of solute . x 100
wt. of solute + wt. of solvent
10= wt of solute x 100
400g
wt. of solute= 10 (400g)
100
= 4000g
100
= 40g
Sir, we discussed about concentration of
solutions and computation of percentage by
mass and volume.
Sir, in solving for the percentage by mass,
we have to use the formula
%by wt= wt. of solute . x 100
wt. of solute + wt. of solvent.
And to solve for the percentage by volume,
we have to use the formula

9. Very good! How about for the proof
number for alcoholic beverages?
Precisely!
v/v % = [(volume of solute)/(volume of
solution)] x 100%.
Sir, in solving the proof number of alcoholic
beverages, we have to multiply the
percentage by volume by 2.
IV. Evaluation
Direction: In a 1 whole sheet of paper, answer the following.
Teacher’s Activity Learner’s Activity
1. How many grams of NaCl are present in a
250g salt solution containing 9% NaCl?
2. How many mL of pure alcohol can be
obtained from 300mL of an 80 proof solution?
1.Given: wt. of solution= 250g
% solute in solution= 9
wt. of solute=(unknown)
Solution:
%by weight= wt. of solute x 100
wt. of solution
9= wt. of solute x 100
250g
wt. of solute= 9 x 250g
100
= 2250g
100
= 22.5g
2. Given: proof= 80
vol. of solution= 300 mL
vol. of solute= (unknown)
P= 2(% by volume)
80 = 2 (% by volume)
2 2

10. % by volume= 40
% by volume = vol. of solute x 100
vol. of solution
40= vol. of solute x 100
300mL
vol. of solute = 40 (300mL)
100
= 12000mL
100
= 120 mL
V. Assignment
Search on the definition of molarity, molality, and normality. Write it in 1 whole sheet of
paper.
Prepared by:
LESTER E. ORPILLA
BSE III- BIOSCI
Presented to:
DR. VIRGINIA O. RUDIO
Course Professor