12
GENERAL
CHEMISTRY 2
QUARTER 1
LEARNING ACTIVITY SHEET
Republic of the Philippines
Department of Education
COPYRIGHT PAGE
Learning Activity Sheet in GENERAL
CHEMISTRY 2 (Grade 12)
Copyright © 2020
DEPARTMENT OF EDUCATION
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: ORLANDO E. MANUEL, PhD, CESO V
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CHELO C. TANGAN, PhD., CESE
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: ROGELIO H. PASINOS, PhD.
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NOTE: Practice personal hygiene protocols at all times
i
Table of Contents
Code
Page
number
STEM_GC11IMFIIIa-c-99
1 - 13
Describe and differentiate the types of
intermolecular forces
STEM_GC11IMFIIIa-c-100
14 – 29
Describe the following properties of liquids,
and explain the effect of intermolecular
forces on these properties: surface tension,
viscosity, vapor pressure, boiling point, and
molar heat of vaporization
STEM_GC11IMFIIIa-c-102
30 – 50
Explain the properties of water with its
molecular structure and intermolecular
forces
STEM_GC11IMFIIIa-c-103
51 - 67
Describe the difference in structure of
crystalline and amorphous solids
STEM_GC11IMFIIIa-c-104
68 – 91
Interpret the phase diagram of water and
carbon dioxide
STEM_GC11IMFIIIa-c-107
92 – 110
Determine and explain the heating and
cooling curve of a substance
STEM_GC11IMFIIIa-c-109
111- 119
Use different ways of expressing
concentration of solutions: percent by mass,
mole fraction, molarity, molality, percent by
volume, percent by mass, ppm
STEM_GC11PPIIId-f-111
120 – 142
Perform stoichiometric calculations for
reactions in solution
STEM_GC11PPIIId-f-112
143 – 159
Describe the effect of concentration on the
colligative properties of solutions
STEM_GC11PPIIId-f-115
160 – 170
Differentiate the colligative properties of
nonelectrolyte solutions and of electrolyte
solutions
STEM_GC11PPIIId-f-116
171 – 187
Calculate boiling point elevation and
freezing point depression from the
concentration of a solute in a solution
STEM_GC11PPIIId-f-117
189- 198
Calculate molar mass from colligative
property data
STEM_GC11PPIIId-f-118
199 – 209
Competency
Use the kinetic molecular model to explain
properties of liquids and solids
NOTE: Practice personal hygiene protocols at all times
ii
Describe laboratory procedures in
determining concentration of solutions
STEM_GC11PPIIId-f-119
210 – 224
Explain the first law of thermodynamics
STEM_GC11TCIIIg-i-124
225 – 233
Explain enthalpy of reaction
STEM_GC11TCIIIg-i-125
234 – 244
Calculate the change in enthalpy of a given
reaction using Hess Law
STEM_GC11TCIIIg-i-127
245 – 255
Describe how various factors influence the
rate of reaction
STEM_GC11CKIIIi-j-130
256 – 268
Differentiate zero, first- , and second-order
reactions
STEM_GC11CKIIIi-j-132
269 – 293
Explain reactions qualitatively in terms of
molecular collisions
STEM_GC11CKIIIi-j-136
294 – 313
Explain activation energy and how a catalyst
STEM_GC11CKIIIi-j-137
affects the reaction rate
314 – 229
Cite and differentiate the types of catalysts
230 – 249
STEM_GC11CKIIIi-j-138
NOTE: Practice personal hygiene protocols at all times
iii
GENERAL CHEMISTRY 2
Name: _
Grade Level:
Date:
Score:
__
LEARNING ACTIVITY SHEET
THE STRUCTURE OF CRYSTLLINE AND AMORPHOUS SOLIDS
Background Information for the Learners (BIL)
A solid interface is defined as a small number of atomic layers that separate
two solids in intimate contact with one another, where the properties differ significantly
from those of the bulk material it separates.
Based on their crystal structures, solids can be classified into the following categories:
1. Crystalline solids
2. Amorphous solids
However, crystalline solids can be further classified into molecular, ionic, metallic, and
covalent solids.
An
illustration
detailing
the
classification
of
solids
is
provided
below
.
Source:https://byjus.com/chemistry/classification-of-solids-based-on-crystal-structure/
5
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Crystalline are solids featuring highly ordered arrangements of their particles
(atoms, ions, and molecules) in microscopic structures.
These ordered microscopic structures make up a crystal lattice that accounts
for the structure of the solid at any given point. Examples of crystalline solids include
salt (sodium chloride), diamond, and sodium nitrate.
Ionic solids, such as sodium chloride and nickel
oxide, are composed of positive and negative ions
Figure 1. Sodium chloride is an
ionic solid.
that are held together by electrostatic attractions,
which can be quite strong Figure 1. Many ionic
crystals also have high melting points. This is due to
the very strong attractions between the ions—in ionic
compounds, the attractions between full charges are
(much) larger than those between the partial charges
in polar molecular compounds. This will be looked at
in more detail in a later discussion of lattice energies.
Although they are hard, they also tend to be brittle,
https://opentextbc.ca/chemistry/chapter/10-5-
and they shatter rather than bend. Ionic solids do not
conduct electricity; however, they do conduct when molten or dissolved because their
ions are free to move. Many simple compounds formed by the reaction of a metallic
element with a nonmetallic element are ionic.
Metallic solids such as crystals of copper,
aluminum, and iron are formed by metal atoms 2.
Figure 2. Copper is a metallic solid.
The structure of metallic crystals is often described
.
as a uniform distribution of atomic nuclei within a
“sea” of delocalized electrons. The atoms within
such a metallic solid are held together by a unique
force known as metallic bonding that gives rise to
many useful and varied bulk properties. All exhibit
high thermal and electrical conductivity, metallic
luster, and malleability. Many are very hard and quite
strong. Because of their malleability (the ability to
https://opentextbc.ca/chemistry/chapter/10-5-the-
deform under pressure or hammering), they do not
shatter and, therefore, make useful construction materials. The melting points of the
metals vary widely. Mercury is a liquid at room temperature, and the alkali metals melt
6
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below 200 °C. Several post-transition metals also have low melting points, whereas
the transition metals melt at temperatures above 1000 °C. These differences reflect
differences in strengths of metallic bonding among the metals.
Covalent
network
solids
include
crystals
of
Figure 3.
diamond, silicon, some other nonmetals, and some
covalent compounds such as silicon dioxide (sand)
and silicon carbide (carborundum, the abrasive on
sandpaper). Many minerals have networks of covalent
bonds. The atoms in these solids are held together by
a network of covalent bonds, as shown in Figure 3. To
break or to melt a covalent network solid, covalent
bonds must be broken. Because covalent bonds are
relatively strong, covalent network solids are typically
characterized by hardness, strength, and high melting
points. For example, diamond is one of the hardest
https://opentextbc.ca/chemistry/chapter/10-5-thesolid- state-of-matter
substances known and melts above 3500 °C.
A covalent crystal contains a three-dimensional network of covalent bonds, as
illustrated by the structures of diamond, silicon dioxide, silicon carbide, and graphite.
Graphite is an exceptional example, composed of planar sheets of covalent crystals
that are held together in layers by noncovalent forces. Unlike typical covalent solids,
graphite is very soft and electrically conductive.
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Molecular solids, such as ice, sucrose (table sugar),
and iodine, as shown in Figure 4, are composed of
Figure 4. Carbon dioxide.
neutral molecules. The strengths of the attractive
forces between the units present in different crystals
vary widely, as indicated by the melting points of the
crystals. Small symmetrical molecules (nonpolar
molecules), such as H2, N2, O2, and F2, have weak
attractive forces and form molecular solids with very
low melting points (below −200 °C). Substances
consisting of larger, nonpolar molecules have larger
attractive forces and melt at higher temperatures.
Molecular
solids
composed
of
molecules
with
permanent dipole moments (polar molecules) melt at
https://opentextbc.ca/chemistry/chapter/10-5-the-solid-
still higher temperatures. Examples include ice (melting point, 0 °C) and table sugar
(melting point, 185 °C).
Figure 4. Carbon dioxide (CO2) consists of small, nonpolar molecules and forms a
molecular solid with a melting point of −78 °C. Iodine (I2) consists of larger, nonpolar
molecules and forms a molecular solid that melts at 114 °C.
Type of Solid
Type of
Types of
Particles
Attractions
Properties
hard,
Examples
brittle,
conducts
electricity as a
ionic
Ions
ionic bonds
liquid but not
NaCl, Al2O3
as a solid, high
to
very
high
melting points
shiny,
atoms
metallic
of
electropositive
malleable,
metallic bonds ductile,
elements
conducts heat
Cu, Fe, Ti, Pb,
U
and electricity
8
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well,
variable
hardness and
melting
temperature
covalent
atoms
of covalent
network
electronegative
bonds
elements
very hard, not
C (diamond),
conductive,
SiO2, SiC
very
high
melting points
molecular
molecules
(or MFs
atoms)
variable
H2O, CO2, I2,
hardness,
C12H22O11
variable
brittleness, not
conductive,
low
melting
points
https://opentextbc.ca/chemistry/chapter/10-5-the-solid-state-of-matter
Amorphous are solids in which the particles are not arranged in any specific
order or the solids that lack the overall order of a crystal lattice.
The term ‘amorphous’, when broken down into its Greek roots, can be roughly
translated to “without form”. Many polymers are amorphous solids. Other examples
of such solids include glass, gels, and nanostructured materials.
An ideal crystal is defined as an atomic arrangement that has infinite
translational symmetry in all the three dimensions, whereas such a definite definition
is not possible for an ideal amorphous solid (a-solid).
Features of Crystalline and Amorphous Solids
CRYSTALLINE
NATURE
AMORPHOUS
Pseudo – Solids or super-
True Solids.
cooled liquids.
Geometry
Particles are arranged in a Particles
are
arranged
repeating pattern. They randomly. They do not
have
a
regular
and have
an
ordered
9
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ordered
arrangement
result
arrangement resulting in
irregular shapes.
ing in a definite shape.
Melting
They have a sharp melting

point.
They do not have
sharp
melting
points.
The
tends
to
solid
soften
gradually over a
temperature range.
Heat of Fusion (The
change in enthalpy when
They have definite heat of

fusion.
They do not have
definite
a substance is heated to
heat
of
fusion.
change its state from
solid to liquid.)
Isotropism
Anisotropic in nature. i.e.,

Isotropic in nature.
the magnitude of physical
i.e., the magnitude
properties
of
(such
as
the
physical
refractive index, electrical
properties
conductivity,
same along with all
conductivity
different
thermal
etc)
along
is
directions
with
is
of
the
the
solid.
different directions of the
crystal.
Cleavage
When cutting with a sharp

When cutting with a
edge, the two new halves
sharp edge, the two
will have smooth surfaces.
resulting halves will
have
irregular
surfaces.
Rigidity
They are rigid solids and
They are not rigid, so mild
applying mild forces will
effects may change the
not distort its shape.
shape.
https://byjus.com/chemistry/classification-of-solids-based-on-crystal-structure
10
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Learning Competency:
Describe the difference in structure of crystalline and amorphous solids.
(STEM_GC11IMF-IIIa-c-104)
Activity 1: CRYSTALLINE SOLID
Objective: Identify the type of crystalline solid formed by a substance.
Materials: Paper and pen
Direction: Identify the type of crystalline solid (metallic, network covalent, ionic, or
molecular) formed by each of the following substances.
_1. CaCl2
6. CH3CH2CH2CH3
_2. SiC
7. HCl
_3. N2
8. NH4NO3
_4. Fe
9. K3PO4
_5. C (graphite)
10. SiO2
Q1. Explain why ice, which is a crystalline solid, has a melting temperature of 0 °C,
whereas butter, which is an amorphous solid, softens over a range of temperatures.
_
_
_
11
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Activity 2: CONCEPTUAL PROBLEMS
Objective: Determine the difference in the structure of crystalline and amorphous
solids.
Materials: Paper and pen
Directions: Read and answer the questions briefly but substantially. Write your
answer on the space provided.
a. Why is the arrangement of the constituent atoms or molecules more important
in determining the properties of a solid than a liquid or a gas?
b. A student obtained a solid product in a laboratory synthesis. To verify the
identity of the solid, she measured its melting point and found that the material
melted over a 12°C range. After it had cooled, she measured the melting point
of the same sample again and found that this time the solid had a sharp melting
point at the temperature that is characteristic of the desired product. Why were
the two melting points different? What was responsible for the change in the
melting point?
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Activity 3: CRYSTAL SYSTEMS
Objectives:

describe the main points of difference between a crystalline solid and an
amorphous solid;

recognize and identify at least 3 of the 7 crystal systems;
Introduction/background
Traditional ceramics are clay-based. Clays have a mineral composition and
minerals have a crystalline structure. A mineral is defined as a naturally occurring
inorganic substance with a certain chemical composition and set of physical
properties. Many minerals occur in characteristic crystal shapes.
A crystalline solid is made up of an orderly repeating pattern of constituent
atoms, molecules or ions extending in all 3 spatial dimensions.
A limited number of crystal shapes have been found in nature. There are only
7 groups, or crystal systems, into which all naturally occurring crystals can be
placed. Careful observation of crystal shapes is one of the best ways to classify and
distinguish between different minerals. This activity focuses on three of these crystal
systems – cubic, triclinic and rhombohedral.
What you need

Crystal systems diagram

Copies of the student worksheet

Small dropper bottles of 1 molL-1 solutions of sodium chloride (NaCl) and copper
sulfate (CuSO4)

Clean ‘golden’ beach sand

Simple light microscope plus microscope slides

Electric hot plate

Templates to construct models of cubic, triclinic and rhombohedral crystal
systems

Paper glue
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What to do
1. Hand out copies of the crystal systems diagram and discuss with the class.
Explain that they will be investigating 3 of these crystal systems – cubic, triclinic
and rhombohedral.
2. Make sure each student has the necessary materials and equipment and a copy
of the student worksheet and templates.
Student worksheet – Studying crystal systems
Note: Please refer to the figure below for the reference of cubic, triclinic and
rhombohedral crystals.
1. Cubic crystals:

Place a drop of the sodium
chloride solution supplied in
the center of a microscope
slide.

Gently heat the slide by
placing it on a hot plate (low
setting).

When all the water has
evaporated, view the sodium
chloride crystals that remain
under the low power of a
microscope.

Note the shape of the
crystals and sketch what
you see.
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2. Triclinic crystals:

Place a drop of the copper
sulfate solution supplied in
the center of a microscope
slide.

Gently heat the slide by
placing it on a hot plate (low
setting).

When all the water has
evaporated, view the copper
sulfate crystals that remain
under the low power of a
microscope.

Note the shape of the
crystals and sketch what
you see.
3. Rhombohedral crystals:

Place a small sample of
beach sand in the center of
a microscope slide and
spread out the grains.

View under the low power of
a microscope.

Note the shape of the grains
with a clear or whitish
appearance – these are
grains of the mineral quartz.
Sketch what you see.
4. Compare the sketches you have drawn to the crystal systems diagram.
15
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5. The mineral halite, a naturally occurring form of sodium chloride, has a cubic
crystal structure. Use the cubic crystal template to construct a model of a halite
crystal. Fold all edges. Glue the tabs and stick together.
6. The feldspar minerals plagioclase and orthoclase have a triclinic crystal structure.
Copper sulfate crystallizes out of solution as triclinic crystals just like the
feldspars. Use the triclinic crystal template to construct a model of a feldspar
mineral crystal. Fold all edges. Glue the tabs and stick together.
7. Quartz minerals are commonly found in beach sand. These tiny grains have a
rhombohedral shape (cubic system stretched along a body diagonal). Use the
rhombohedral crystal template to construct a model of a quartz crystal.
16
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Crystal systems
17
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Activity 4: BUILD ME UP
Objectives:

use models to point out the angular and side length differences that
characterize the cubic, triclinic and rhombohedral crystal systems.
Materials: template of cubic, triclinic and rhombohedral crystal system
Directions: Use the given template to point out the angular and side length differences
that characterize the cubic, triclinic and rhombohedral crystal systems
Cubic crystal template
All axes are of equal length. All axes are at 90° to one another.
All axes are of variable lengths. All axes are at variable angles.
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Rhombohedral crystal template
All the axes are equal. All axes are at angles other than 90°.
*
Activity 5: CRYSTALLINE AND AMORPHOUS SOLID (Pre-lab)
Objective: Distinguish between crystalline and amorphous substances.
Materials: Worksheet
Students create
patterns using
Altair
designs.
Background: The atoms in crystalline solid matter
are arranged in regular, repeating .patterns. All other
types of solid matter are amorphous or without a
regular atomic arrangement. Metals and minerals are
crystalline. Glass is amorphous. Depending upon its
composition, the crystalline pattern of a mineral may
not be visible in a hand sample. In this case minerals
are studied using X-ray diffraction, a technique that
uses the reflection of X-rays to determine crystal
structure and composition.
Electron level picture of tin
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Procedure:
1. Observe the following diagram below illustrating crystalline versus a noncrystalline (amorphous) patterns.
Crystalline
Amorphous
2. On the worksheet, outline or fill in spaces on the Altair designs sheet to
create patterns. Your patterns are examples of order within the overall
structure of the design. This same type of organization generates crystalline
structures in minerals. The Altair designs sheet will naturally guide your
imagination through the maze of lines. Since no two students are alike, none
of you will see the same shapes, forms or patterns hidden in these designs.
You may create some very interesting artwork.
3. After finishing your patterns pair up with your seatmate and see if there are
any similar patterns. The similarities and differences means that there are
many types of minerals, and hence many different crystal patterns.
20
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Activity 6: MULTIPLE CHOICE
Directions: Read each item carefully. Write the letter that corresponds to the correct
answer on the space provided.
_1. In amorphous solid, the atoms or molecules are held together in a completely
random formation.
A. True
B. False
_2. Which of the following is true of solids?
A. Solids maintain a defined shape and size under all conditions.
B. All solids maintain a defined shape and size if conditions remain constant.
C. All solids have a lattice structure at atomic level.
D. All solids have a crystalline structure.
_3. One major difference between crystalline and amorphous solids is that
A. Crystalline solids have precise melting point.
B. Amorphous solids have a lattice structure.
C. Crystalline solids break unpredictably and can produce curved fragments.
D. Amorphous solids always behave consistently and uniformly.
_4. A friend in your chemistry class is struggling to understand why crystalline
solids are grouped into four main types: network, molecular, ionic, and metallic.
Which explanation below will best help him begin to understand why chemists
might have these groups?
A. Crystalline solids all share a lattice structure, but have different densities.
Chemists use the groups to organize the solids by density.
B. Crystalline solids all share a lattice structure and the same types of bonds,
but are composed of different elements. These elements affect the way the
solid conducts heat and electricity.
C. Crystalline solds all share a lattice structure, but behave differently under
similar conditions.
D. Crystalline solids all share a lattice structure, but the bonds that hold them
together at the atomic level differ. The elements that make up the solids also
differ. These differences affect how a solid conducts heat and electricity,
and its density.
21
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_5. Solids have many different properties. _
solids are known for their ability
to be flattened into a sheet, stretched into a wire, and to conduct energy well.
A. Molecular
B. Metallic
C. Network
D. Ionic
_6. It is possible to tell the difference between a solid with a crystalline structure
and one with an amorphous structure just by looking at it.
A. True
B. False
_7. An engineer is designing an electrical system and is looking for a material to
transmit energy. She has four solids available, each made with different
materials. To conduct energy most efficiently and effectively, she should use
material
A. Whose electrons are held with ionic bonds.
B. Whose electrons are held with covalent bonds.
C. Whose electrons are held with metallic bonds.
D. That is an electrical insulator.
_8. Which statement is true about the properties of solids?
A. Metallic solids have a high melting point.
B. Network solids are generally not soluble in water.
C. Molecular solids do not dissolve easily in water.
D. All ionic solids are similar in density.
22
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Reflection
1. I learned that
_
_
2. I enjoyed most on
_
_
_
_
.
3. I want to learn more on
_
_
_.
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References:
https://opentextbc.ca/chemistry/chapter/10-5-the-solid-state-of-matter
https://byjus.com/chemistry/classification-of-solids-based-on-crystal-structure/
https://www.sciencelearn.org.nz/resources/1784-crystal-systems
Prepared by:
DIVINA S. RIBIACO
Baua National High School
24
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GENERAL CHEMISTRY 2
Name: _
Grade Level:
Date:
Score:
__
LEARNING ACTIVITY SHEET
PHASE DIAGRAM OF WATER AND CARBON DIOXIDE
Background Information for the Learners
A typical phase diagram consists of discrete regions that represent the different
phases exhibited by a substance (Figure 12.4.1). Each region corresponds to the
range of combinations of temperature and pressure over which that phase is stable.
The combination of high pressure and low temperature (upper left of Figure 12.4.1)
corresponds to the solid phase, whereas the gas phase is favored at high temperature
and low pressure (lower right). The combination of high temperature and high pressure
(upper right) corresponds to a supercritical fluid.
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/12%3A_Intermolec
ular_Forces%3A_Liquids_And_Solids/12.4%3A_Phase_Diagrams
Figure 12.4.1: A Typical Phase Diagram for a Substance That Exhibits Three
Phases—Solid, Liquid, and Gas—and a Supercritical Region
The solid phase is favored at low temperature and high pressure; the gas phase
is favored at high temperature and low pressure.
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The lines in a phase diagram correspond to the combinations of temperature
and pressure at which two phases can coexist in equilibrium. In Figure 12.4.1, the line
that connects points A and D separates the solid and liquid phases and shows how
the melting point of a solid varies with pressure. The solid and liquid phases are in
equilibrium all along this line; crossing the line horizontally corresponds to melting or
freezing. The line that connects points A and B is the vapor pressure curve of the
liquid, which we discussed in Section 11.5. It ends at the critical point, beyond which
the substance exists as a supercritical fluid. The line that connects points A and C is
the vapor pressure curve of the solid phase. Along this line, the solid is in equilibrium
with the vapor phase through sublimation and deposition. Finally, point A, where the
solid/liquid, liquid/gas, and solid/gas lines intersect, is
the triple point;
it is
the only combination of temperature and pressure at which all three phases (solid,
liquid, and gas) are in equilibrium and can therefore exist simultaneously. Because no
more than three phases can ever coexist, a phase diagram can never have more than
three lines intersecting at a single point.
Remember that a phase diagram, such as the one in Figure 12.4.1, is for a
single pure substance in a closed system, not for a liquid in an open beaker in contact
with air at 1 atm pressure. In practice, however, the conclusions reached about the
behavior of a substance in a closed system can usually be extrapolated to an open
system without a great deal of error.
The Phase Diagram of Water
Figure 12.4.2 shows the phase diagram of water and illustrates that the triple
point of water occurs at 0.01°C and 0.00604 atm (4.59 mmHg). Far more reproducible
than the melting point of ice, which depends on the amount of dissolved air and the
atmospheric pressure, the triple point (273.16 K) is used to define the absolute (Kelvin)
temperature scale. The triple point also represents the lowest pressure at which a
liquid phase can exist in equilibrium with the solid or vapor. At pressures less than
0.00604 atm, therefore, ice does not melt to a liquid as the temperature increases; the
solid sublimes directly to water vapor. Sublimation of water at low temperature and
pressure can be used to “freeze-dry” foods and beverages. The food or beverage is
first cooled to subzero temperatures and placed in a container in which the pressure
is maintained below 0.00604 atm. Then, as the temperature is increased, the water
26
NOTE: Practice personal hygiene protocols at all times
sublimes, leaving the dehydrated food (such as that used by backpackers or
astronauts) or the powdered beverage (as with freeze-dried coffee).
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/12%3A_Intermolec
ular_Forces%3A_Liquids_And_Solids/12.4%3A_Phase_Diagrams
Figure 12.4.2: Two Versions of the Phase Diagram of Water. (a) In this graph
with linear temperature and pressure axes, the boundary between ice and liquid
water is almost vertical. (b) This graph with an expanded scale illustrates the
decrease in melting point with increasing pressure. (The letters refer to points
discussed in Example 12.4.1).
The phase diagram for water illustrated in Figure 12.4.2b shows the boundary
between ice and water on an expanded scale. The melting curve of ice slopes up and
slightly to the left rather than up and to the right as in Figure 12.4.1; that is, the melting
point of ice decreases with increasing pressure; at 100 MPa (987 atm), ice melts at
−9°C. Water behaves this way because it is one of the few known substances for
which the crystalline solid is less dense than the liquid (others include antimony and
bismuth). Increasing the pressure of ice that is in equilibrium with water at 0°C and 1
atm tends to push some of the molecules closer together, thus decreasing the volume
of the sample. The decrease in volume (and corresponding increase in density) is
smaller for a solid or a liquid than for a gas, but it is sufficient to melt some of the ice.
In Figure 12.4.2b point A is located at P = 1 atm and T = −1.0°C, within the solid
(ice) region of the phase diagram. As the pressure increases to 150 atm while the
temperature remains the same, the line from point A crosses the ice/water boundary
to point B, which lies in the liquid water region. Consequently, applying a pressure of
150 atm will melt ice at −1.0°C. We have already indicated that the pressure
dependence of the melting point of water is of vital importance. If the solid/liquid
27
NOTE: Practice personal hygiene protocols at all times
boundary in the phase diagram of water were to slant up and to the right rather than
to the left, ice would be denser than water, ice cubes would sink, water pipes would
not burst when they freeze, and antifreeze would be unnecessary in automobile
engines.
Referring to the phase diagram of water in Figure 12.4.2:
a. Predict the physical form of a sample of water at 400°C and 150 atm.
b. Describe a changes that occur as the sample in part (a) is slowly allowed
to cool to -50°C at a constant pressure of 150 atm
Given: phase diagram, temperature, and pressure
Asked for: physical form and physical changes
Strategy:

Identify the region of the phase diagram corresponding to the initial conditions
and identify the phase exist in this region.

Draw a line corresponding to the given pressure. Move along that line in the
appropriate direction (in this case cooling) and describe the phase changes.
Solution:
a. Locate the starting point on
the
phase
diagram
in
part
(a)
in
Figure 12.4.212.4.2. The initial conditions correspond to point A, which lies in
the region of the phase diagram representing water vapor. Thus water at T =
400°C and P = 150 atm is a gas.
b. Cooling the sample at constant pressure corresponds to moving left along the
horizontal line in part (a) in Figure 12.4.212.4.2. At about 340°C (point B), we
cross the vapor pressure curve, at which point water vapor will begin to
condense and the sample will consist of a mixture of vapor and liquid. When all
of the vapor has condensed, the temperature drops further, and we enter the
region corresponding to liquid water (indicated by point C). Further cooling
brings us to the melting curve, the line that separates the liquid and solid phases
at a little below 0°C (point D), at which point the sample will consist of a mixture
of liquid and solid water (ice). When all of the water has frozen, cooling the
sample to −50°C takes us along the horizontal line to point E, which lies within
the region corresponding to solid water. At P = 150 atm and T = −50°C,
therefore, the sample is solid ice.
28
NOTE: Practice personal hygiene protocols at all times
The Phase Diagram of Carbon dioxide
In contrast to the phase diagram of
water,
the
phase
diagram
of
CO2 (Figure 12.4.312.4.3) has a more typical melting curve, sloping up and to the
right. The triple point is −56.6°C and 5.11 atm, which means that liquid CO2 cannot
exist at pressures lower than 5.11 atm. At 1 atm, therefore, solid CO2 sublimes directly
to the vapor while maintaining a temperature of −78.5°C, the normal sublimation
temperature. Solid CO2 is generally known as dry ice because it is a cold solid with no
liquid phase observed when it is warmed.
Dry ice (CO2(s)CO2(s)) sublimed in air under room
temperature and pressure.
Also notice the critical point at 30.98°C and 72.79 atm. Supercritical carbon dioxide is
emerging as a natural refrigerant, making it a low carbon (and thus a more
environmentally friendly) solution for domestic heat pumps.
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/12%3A_Intermolec
ular_Forces%3A_Liquids_And_Solids/12.4%3A_Phase_Diagrams
Figure 12.4.3: The Phase Diagram of Carbon Dioxide. Note the critical point, the
triple point, and the normal sublimation temperature in this diagram.
29
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The Critical Point
As the phase diagrams above demonstrate, a combination of high pressure and
low temperature allows gases to be liquefied. As we increase the temperature of a
gas, liquefaction becomes more and more difficult because higher and higher
pressures are required to overcome the increased kinetic energy of the molecules. In
fact, for every substance, there is some temperature above which the gas can no
longer be liquefied, regardless of pressure. This temperature is the critical temperature
(Tc), the highest temperature at which a substance can exist as a liquid. Above the
critical temperature, the molecules have too much kinetic energy for the intermolecular
attractive forces to hold them together in a separate liquid phase. Instead, the
substance forms a single phase that completely occupies the volume of the container.
Substances with strong intermolecular forces tend to form a liquid phase over a very
large temperature range and therefore have high critical temperatures. Conversely,
substances with weak intermolecular interactions have relatively low critical
temperatures. Each substance also has a critical pressure (Pc), the minimum pressure
needed to liquefy it at the critical temperature. The combination of critical temperature
and critical pressure is called the critical point. The critical temperatures and pressures
of several common substances are listed in Figure 12.4.1
Figure 12.4.1: Critical Temperatures and Pressures of Some Simple Substances
Substance
Tc (°C)
Pc (atm)
NH3
132.4
113.5
CO2
31.0
73.8
CH3CH2OH (ethanol)
240.9
61.4
He
−267.96
2.27
Hg
1477
1587
CH4
−82.6
46.0
N2
−146.9
33.9
H2O
374.0
217.7
*High-boiling-point, nonvolatile liquids have high critical temperatures and vice versa.
30
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https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/12%3A_Intermolec
ular_Forces%3A_Liquids_And_Solids/12.4%3A_Phase_Diagrams
Learning Competency:
Interpret the phase diagram of water and carbon dioxide (STEM_GC11IMFIIIa-c107)
Activity 1: WATER AND CARBON DIOXIDE
Objective: determine the state of water at each given temperature and pressure.
Materials: Paper and pen
Figure A.
Figure B
courses.lumenlearning.com/wsu-sandbox2/chapter/phase-diagram-2/
A. Directions: Using the phase diagram (fig. a) for water, determine the state of H 2O
at the following temperatures and pressures. Write your answer on the space
provided.
_1. -10 °C and 50 kPa
_2. 25°C and 90 kPa
_3. 50°C and 40 kPa
_4. 80°C and 5 kPa
_5. -10°C and 0.3 kPa
31
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A. Directions: Using the phase diagram for carbon dioxide, determine the state of
CO2 at the following temperatures and pressures. Write your answer on the space
provided.
6. −30 °C and 2000 kPa
_7. −60 °C and 1000 kPa
_8. −60 °C and 100 kPa
_9. 20 °C and 1500 kPa
_10. 0 °C and 100 kPa
Activity 2: CRITICAL THINKING (H2O and CO2)
Objective: Interpret the phase diagram of water and carbon dioxide.
Materials: Paper and pen
Directions: Read and analyze the given problem, then answer the questions below.
Write your answer on the space provided.
Problem: Imagine a substance with the following points on the phase diagram: a triple
point at .5 atm and -5ºC; a normal melting point at 20ºC; normal boiling point at 150ºC;
and a critical point at 5 atm and 1000ºC. The solid liquid line is “normal” (meaning
positive sloping). For this, complete the following:
1. Describe what one would see at pressures and temperatures above 5 atm and
1000ºC.
2. Describe what will happen to the substance when it begins in a vacuum at -15 ºC
and is slowly pressurized.
3. Describe the phase changes from -80ºC to 500ºC at 2 atm.
32
NOTE: Practice personal hygiene protocols at all times
Activity 3: THE COOL CHEMISTRY OF DRY ICE
Objective: Interpret the phase diagram of water.
Materials: Paper and pen
Directions: Read and analyze the given problem, then answer the question below.
Write your answer on the space provided.
Problem: Referring to the phase diagram of water in figure 12.4.2, predict the
physical form of a sample of water at -0.0050ºC as the pressure is gradually
increased from 1.0 mmHg to 218 atm. Write your answer on the space provided.
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/12%3A_Intermolec
ular_Forces%3A_Liquids_And_Solids/12.4%3A_Phase_Diagrams
_
_
_
.
33
NOTE: Practice personal hygiene protocols at all times
Activity 4: DIHYDROGEN MONOXIDE
Objective: Interpret the phase diagram of water.
Materials: Paper and pen
A phase diagram of water is shown below
B
4
C (374 °C, 218 atm)
2
1
3
A (0.01 °C, 0.00603 atm)
Temperature
https://scilearn.sydney.edu.anu...
1.
Identify the four phases shown as 1-4 in the phase diagram.
a.
c.
b.
d.
2. What names are given to the points A and C?
a.
b.
3. The boundary line A-B is slightly tilted to the left. What are the physical and
biological significances of this?
34
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Activity 5: CARBON DIOXIDE
Objective: determine the state of water at each given temperature and pressure.
Materials: Paper and pen
Directions: Answer the following questions based on the P-T phase diagram of
carbon dioxide.
Write your answer on the space provided.
Phase diagram of carbon dioxide
https://www.toppr.com/ask/question/answer-the-following-questions-based-on-the-pt-phase-diagram/
1. At what temperature and pressure can the solid, liquid and vapor phases of
CO2 co-exits in equilibrium?
_
.
2. What is the effect of decrease of pressure on the fusion and boiling point of
CO2?
.
3. What are the critical temperature and pressure for CO2?
_.
35
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Reflection
1. I learned that
_
_
2. I enjoyed most on
_
_
_
.
3. I want to learn more on
_
_
_.
36
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References:
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Che
mistry_(Petrucci_et_al.)/12%3A_Intermolecular_Forces%3A_Liquids_And_Solids/12
.4%3A_Phase_Diagrams
https://msnucleus.org/membership/html/k-6/rc/minerals/3/rcm3_4a.html
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textb
ook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical
_Properties_of_Matter/States_of_Matter/Phase_Transitions/Phase_Diagrams
https://www.toppr.com/ask/question/answer-the-following-questions-based-on-the-ptphase-diagram/
courses.lumenlearning.com/wsu-sandbox2/chapter/phase-diagram-2/
Prepared by:
DIVINA S. RUBIACO
Baua National High School
37
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GENERAL CHEMISTRY 2
Name:
Grade Level:
Date:
Score:
LEARNING ACTIVITY SHEET
HEATING AND COOLING CURVES
Background Information for the Learners (BIL)
Heating Curve
Imagine that you have a block of ice that is at a temperature of -30°C, well
below its melting point. The ice is in a closed container. As heat is steadily added to
the ice block, the water molecules will begin to vibrate faster and faster as they absorb
kinetic energy. Eventually, when the ice has warmed to 0°C, the added energy will
start to break apart the hydrogen bonding that keeps the water molecules in place
when it is in the solid form. As the ice melts, its temperature does not rise. All of the
energy that is being put into the ice goes into the melting process and not into any
increase in temperature. During the melting process, the two states – solid
and liquid – are in equilibrium with one another. If the system was isolated at that
point and no energy was allowed to enter or leave, the ice-water mixture at 0°C would
remain. Temperature is always constant during a change of state.
Continued heating of the water after the ice has completely melted will now
increase the kinetic energy of the liquid molecules and the temperature will rise.
Assuming that the atmospheric pressure is standard, the temperature will rise steadily
until it reaches 100°C. At this point, the added energy from the heat will cause the
liquid to begin to vaporize. As with the previous state change, the temperature will
remain at 100°C while the water molecules are going from the liquid to the gas or
vapor state. Once all the liquid has completely boiled away, continued heating of the
steam (remember the container is closed) will increase its temperature above 100°C.
38
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The experiment described above can be summarized in a graph called a
heating curve:
F
D
B
E
C
A

Between A & B, the material is a solid. The heat supplied to the material is used
to increase the kinetic energy of the molecules and the temperature rises.

Between B & C, the solid is melting. Heat is still being supplied to the material
but the temperature does not change. Heat energy is not being changed into
kinetic energy. Instead, the heat is used to change the arrangement of the
molecules.

At point C, all of the material has been changed to liquid.

Between C & D, the heat supplied is again used to increase kinetic energy of
the molecules and the temperature of the liquid starts to rise.

Between C & D, the liquid is heated until it starts to boil.

Between D & E, the liquid is still being heated but the extra heat energy does
not change the temperature (kinetic energy) of the molecules. The heat energy
is used to change the arrangement of the molecules to form a gas.

At point E, all of the liquid has been changed into gas.

Between E & F, the gas is heated and the heat energy increases the kinetic
energy of molecules once more, so the temperature of the gas increases.
39
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When a system contains only one phase (solid, liquid, or gas), the temperature
will increase when it receives energy. The rate of temperature increase will be
dependent on the heat capacity of the phase in the system. When the heat capacity is
large, the temperature increases slowly, because much energy is required to increase
its temperature by one degree. Thus, the slopes of temperature increase for the solid,
liquid, and gases are different.
In the heating curve of water, the temperature is shown as heat is continually
added. Changes of state occur during plateaus because the temperature is constant.
The change of state behavior of all substances can be represented with a
heating curve of this type. The melting and boiling points of the substance can be
determined by the horizontal lines or plateaus on the curve. Other substances would
of course have melting and boiling points that are different from those of water. One
exception to this exact form for a heating would be for a substance such as carbon
dioxide which sublimes rather than melts at standard pressure. The heating curve for
carbon dioxide would have only one plateau, at the sublimation temperature of CO2.
Cooling Curves
Heating curves show how the temperature changes as a substance is heated
up. Cooling curves are the opposite. They show how the temperature changes as a
substance is cooled down. Just like heating curves, cooling curves have horizontal flat
parts where the state changes from gas to liquid, or from liquid to solid. These are
mirror images of the heating curve.
You will use lauric acid in a school lab to make your own cooling curve. Lauric
acid has a melting point of about 45°C and is easily melted in a test tube placed in a
beaker of hot water. The temperature can be followed using a thermometer or
temperature probe connected to a data logger. The liquid may be cooled by putting
the boiling tube in a beaker of cold water or just leaving it in the air.
40
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Note: The melting and freezing occur at the same temperature. During freezing,
energy is removed and during melting, energy is absorbed.
Energy Changes
Since Temperature is a measure of "Average Kinetic Energy", any change in
temperature is a change in Kinetic Energy. All of the diagonal line segments on a
heating or cooling curve show a temperature change and therefore a change in kinetic
energy.
During these regions, a single state of matter exists and the sample is either
getting hotter or cooler. During the horizontal line segments, there is no change in
temperature, so kinetic energy remains constant. However, all the energy that is
absorbed or released is related to changes in potential energy.
Remember the 3 Ps: Plateau, Phase change and Potential Energy Change.
Source:
https://www.rcboe.org/cms/lib/GA01903614/Centricity/Domain/1951/Heating%20and
%20Cooling%20Curves%20new.pdf
41
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Learning Competency
Determine
and
explain
the
heating
and
cooling
curve
of
a
substance
(STEM_GC11IMFIIIa-c-109)
Activity 1: THE COOLING CURVE OF WATER
Directions: Using the curve below describe what is happening between each of
the points:
i.
A-B
ii.
B-C
iii.
C-D
iv.
D-E
v.
E-F
42
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Activity 2: THE HEATING CURVE OF WATER
Directions: Use the cooling curve below to answer the following questions.
Photo credit: https://sites.google.com/site/heatingandcoolingcurves/_/rsrc/1299042706797/curveexplanation/hEATING%20CURVE.png
1. In which region(s) does temperature remain constant?
2. In which region(s) does temperature increase?
3. In which region(s) of the graph does a phase change occur?
4. In which region(s) of the graph would the substance only be in one phase?
5. In which region(s) of the graph would the substance be a solid only?
6. In which region(s) of the graph would the substance be a solid and a liquid?
7. In which region(s) of the graph would the substance be a liquid and a gas?
8. In which region(s) of the graph would the substance be a gas only?
9. In which region(s) of the graph does boiling take place?
10. In which region(s) of the graph does melting take place?
43
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Reflection:
1. I learned that
2. I enjoyed most on
3. I want to learn more on
44
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References:

https://www.quora.com/How-do-you-determine-the-freezing-point-of-asolution-do-you-follow-this-process-for-every-solution

Curriculum Guide and Teaching Guide. K to 12 Basic Education Curriculum
Senior High School – Science, Technology, Engineering and Mathematics
(STEM) Specialized Subject

http://teachtogether.chedk12.com/teaching_guides/view/499

courses.lumenlearning.com/cheminter/chapter/heating-and-cooling-curvesalso-called-temperature-curves/

https://www.rcboe.org/cms/lib/GA01903614/Centricity/Domain/1951/Heating%
20and%20Cooling%20Curves%20new.pdf

https://www.tes.com/teaching-resource/graphs-and-heating-cooling-curvesworksheet-6064146

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemis
try_The_Central_Science_(Brown_et_al.)
Prepared by:
JACKIE B. UBINA
Solana National High School
45
NOTE: Practice personal hygiene protocols at all times
GENERAL CHEMISTRY 2
Name:
__
Grade Level:
_
Date:_
__
Score:
_
LEARNING ACTIVITY SHEET
WAYS OF EXPRESSING CONCENTRATION OF SOLUTIONS
Background Information for the Learners (BIL)
The term solution is used in Chemistry to describe a homogeneous mixture in
which at least one substance (the solute) is dissolved in another substance (the
solvent). The solvent is the substance in greater quantity and the name of the of the
solution is taken from the name of the solute. For example, when sodium chloride is
dissolved in water, sodium chloride is the solute, and water is the solvent, and the
solution is called a sodium chloride solution.
There are different methods of expressing solution concentrations namely;
Molarity, Molality, Percent by Mass, Percent by Volume, Mole fraction and Parts Per
Million. These methods are used to express relative amounts of solute and solvent in
a solution. In other words, the concentration of a solution is the amount of solute
present in a given amount of solvent, or a given amount of solution.
Percent by Mass
The Percent by Mass (also called percent by weight or weight percent) is the ratio
of the mass of a solute to the mass of the solution, multiplied by 100 percent:
Mass of solute
Percent by Mass =
Mass of solute + Mass of solvent
X 100%
46
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Or,
Mass of solute
Percent by Mass =
X 100%
Total mass of Solution
Let us consider the examples below;
Example 1: In a solution prepared by dissolving 24g of Sodium Chloride
(NaCl) in 152g of water, what is the mass percent of Sodium
Chloride (NaCl)?
Solution: First, identify the given.
Given: Solute = 24g of NaCl
Solvent = 152g of Water
Second, identify the unknown or what is being asked in the
problem.
Unknown = Percent by Mass of NaCl.
Third, write the formula and calculate the unknown.
Mass of solute
Percent by Mass =
X 100%
Mass of solute + Mass of solvent
Percent by Mass of NaCl=
24g NaCl
X 100%
24g NaCl + 152g Water
Percent by Mass of NaCl=
24g
X 100%
= 14%
176g
47
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Example 2: A sample of 0.892 g of potassium chloride (KCl) is dissolved
in 54.6 g of water. What is the percent by mass of KCl in the
solution?
Solution: First, identify the given.
Given: Solute = 0.892g of KCl
Solvent = 54.6g of water
Second, identify the unknown or what is being asked in the
problem.
Unknown = Percent by Mass of KCl.
Third, write the formula and calculate the unknown.
Mass of solute
Percent by Mass =
Percent by Mass =Mass of solute + Mass of solvent
Percent by Mass of KCl=
X 100%
0.892g of KCl
X 100%
0.892g of Cl + 54.6g of Water
Percent by Mass of KCl=
0.892g
X 100%
= 1.61%
55.492g
Percent by Volume
Percent by Volume or Volume Percent is a common expression used for
expressing concentration. It is related to the molar concentration but the difference is
that the volume percent is expressed with a denominator of 100. It is used for reporting
concentration of liquids solutes in solution. It is also called %V/V and it is always
expressed as percentage (%) and the units of the volume should be in mL. . Percent
48
NOTE: Practice personal hygiene protocols at all times
by volume is also widely use in pharmaceutical field for expressing the concentration
of different components in solution.
Percent Volume =
Volume of solute
X 100%
Volume of solute +Volume of solvent
Or,
Percent Volume =
Volume of solute
X 100%
Total Volume of Solution
Let us consider the examples below;
Example 1: A solution of propanol (CH3CH2CH2OH) is prepared by
dissolving 67 mL in enough water to have a final volume of 250
mL. What is the volume percent of the propanol?
Solution: First, identify the given.
Given: Solute = 67mL propanol
Solution = 250mL
Second, identify the unknown or what is being asked in the
problem.
Unknown = Percent Volume of Propanol
49
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Third, write the formula and calculate the unknown.
Volume of solute
Percent Volume =
Percent Volume =
Total Volume of Solution
67 mL
X 100%
X 100% = 26.8%
250mL
Example 2: How many mL of HNO3 concentrate are needed to prepare
250 mL of solution 4%?
Solution: First, identify the given
Given: solution = 250mL
Volume = 4 percent by
Second, identify the unknown or what is being asked in the
problem.
Unknown = Volume of solute (HNO3)
Third, write the formula and calculate the unknown. But in this
case we have to derive the formula.
From the mother formula;
Volume of solute
Percent Volume =
Total Volume of Solution
X 100%
50
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To derived formula to get the volume of solute.
(Percent by Volume) (Volume of Solution)
Volume of Solute =
100%
Thus,
Volume of Solute =
(4%) (250mL)
= 10mL
100%
Parts per Million
When the amount of solute is very small, as with trace impurities in water,
concentration is often expressed in parts per million.
PPM is a term used in chemistry to denote a very, very low concentration of a
solution. One gram in 1000 ml is 1000 ppm and one thousandth of a gram (0.001g) in
1000 ml is one ppm. Parts Per Million (ppm) is a measurement of the concentration of
a solution.
Parts per Million =
Gram of Solute
X 106
Gram of Solution
Let us consider the examples below;
Example 1: What is the concentration of a solution in parts per million, if
0.02 grams of NaCl is dissolved in 1000 grams of solution?
Solution: First, identify the given.
Given: Solute = 0.02g of NaCl
Solution = 1000g
51
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Second, identify the unknown or what is being asked in the
problem.
Unknown = Concentration in parts per million
Third, write the formula and calculate the unknown
Parts per Million =
Gram of Solute
X 106
Gram of Solution
Parts per Million =
0.02 grams
X 106 = 20 ppm
1000 grams
Example 2: What is the total mass of solute in 1000g of a solution having
a concentration of 5ppm?
Solution: First, identify the given
Given: Solution = 1000 grams
Concentration in ppm = 5ppm
Second, identify the unknown or what is being asked in the
problem.
Unknown = Mass of Solute
Third, write the formula and calculate the unknown. But in this
case we have to derive the formula.
52
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From the mother formula;
Gram of Solute
Parts per Million =
X 106
Gram of Solution
To derived formula to get the unknown which is the mass of solute.
Gram of Solute =
(ppm) (gram of solution)
1000000
Thus,
Gram of Solute =
( 5 ppm) (1000 grams)
= .005g
1000000
Mole Fraction
It is a dimensionless quantity that expresses the ratio of the number of moles
of one component to the number of moles of all components present. For a mixture of
two substances, A and B, the mole fractions of each would be written as follows:
Mole fraction of component A:
mol A
XA =
mol A + mol B
mol B
Mole fraction of component B:
XB =mol A + mol B
In general, the mole fraction of component “i” in a mixture is given by;
Xi =
ni
nT
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NOTE: Practice personal hygiene protocols at all times
where ni and nT are the number of moles of component i and the total number of
moles present, respectively. The mole fraction is always smaller than 1.
*The mole fraction is unitless or dimensionless because it is a ratio of two similar
quantities*
Let us consider the examples below;
Example 1: 0.100 mole of NaCl is dissolved into 100.0 grams of pure H 2O.
What is the mole fraction of NaCl? What is the mole fraction of
H2O?
Solution: First, identify the given
Given: 0.100 mole of NaCl
100.0 grams of pure H2O
Second, identify the unknown or what is being asked in the
problem.
Unknown: mole fraction of H2O.
*Note that the component being asked in the problem is the water (H 2O) component,
but as you may notice, the unit of water as stated in the problem is in grams. Before
you can finally input the all the given in the formula you have to make sure that units
to be used are appropriate. Since we are dealing with mole fraction, we have to
convert 100 grams of H2O into moles using the molar mass of H2O (18g/mol).*
Converting 100 grams of water into moles:
(100 grams of H2O) X
(1 mol H2O)
= 5.56mol H2O
(18.0g H2O)
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Third, write the formula and calculate the unknown.
Xi =
nT
ni
5.56 mol
Xi =
5.66 mol
= 0.982
Example 2: A solution is prepared by mixing 25.0 grams of water and 25.0
grams of ethanol (C2H5OH). Determine the mole fractions of
each substance.
Solution: First, identify the given
Given: 25 grams of water
25 grams of ethanol
*As you may notice, all the given are in grams. You may think that you could solve
right away for the mole fraction since you will arrive in a unit less answer. But that is
not how it works in mole fraction because mole fraction deals with moles, and so we
need to convert this grams into moles first before we can be able to get the mole
fraction. In converting the given grams to moles, refer to the method shown in
example 1 and the molar mass of the substance can be summed up using the mass
of the atoms in that given substance.*
Thus, 25 grams of water = 1.34 mol of water
25 grams of ethanol = 0.543 mole of ethanol
Second, identify the unknown or what is being asked in
the
problem.
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Unknown: Mole fractions of each substance.
Therefore, for this
problem
we
have
to
treat
water
as
component A and ethanol as component B.
Third, write the formula and calculate the unknown.
Water as component A
XA =
Ethanol as component B
mol A
mol A + mol B
XA =
1.34 mol
mol B
XB =
mol A + mol B
= 0.71
0.543 mol
XB =
1.34 mol+ 0.543mol
= 0.29
1.34 mol + 0.543mol
Molarity
Otherwise known as “molar concentration”. It is defined as the number of moles of
solute per liter of solution. The SI unit for molarity is mol/m 3; however, you will almost
always encounter molarity with the units of mol/L. A solution of concentration 1 mol/L
is also denoted as “1 molar” (1 M). Mol/L can also be written in the following ways
(however, mol/L, or simply M, is most common)
It is important to keep in mind that molarity refers only to the amount of solute
originally dissolved in water and does not take into account any subsequent
processes, such as the dissociation of a salt or the ionization of an acid.
In equation form it is written as;
Moles of solute
M=
Liter of Solution
and can be expressed algebraically as;
M=
n
v
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Where n, denotes the number of moles of solute. And v is the volume of solution
in liters. Note that the volume in the definition of Molarity refers to the volume of
solution and not the volume of the solvent. The reason for this is because one liter of
solution usually contains either slightly more or slightly less than 1 liter of solvent, due
to the presence of the solute.
Let us consider the examples below;
Example 1: How many grams of potassium dichromate (K2Cr2O7) are required to
prepare a 250-mL solution whose concentration is 2.16 M ?
Solution: First, identify the given
Given: Molarity = 2.16M
Solution = 250mL
*Note that the solution must be converted into Liters and so
250 mL is equal to 0.250L.*
Second, identify the unknown or what is being asked in
the
problem.
Unknown = Potassium dichromate in grams
Third, write the formula and calculate the unknown. But in this
case we have to derive the formula.
From the mother formula;
n
M= v
M=
To derived formula to get the moles of potassium dichromate. (which later on be
converted into grams since the problem asks for the quantity of potassium dichromate
in grams).
N = (M) (v)
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Thus,
N = (2.16M) (0.250L) = 0.54 mol of potassium dichromate.
The 0.54 mol potassium dichromate is not yet the final answer because we still must
convert it into grams. Using the molar mass of K2Cr2O7 which is 294.2 g.
Converting 0.540mol of K2Cr2O7 to grams we have;
294.2 g of K2Cr2O7
(0.540mol of K2Cr2O7) X
= 159g of K2Cr2O
1mol of K2Cr 2O
Example 2: In a biochemical assay, a chemist needs to add 3.81 g of
glucose to a reaction mixture. Calculate the volume in milliliters
of a 2.53 M glucose solution she should use for the
addition.
Solution: First, identify the given
Given: 3.81 grams of glucose
2.53 M of glucose
*Glucose is given in grams and we must first convert it into moles using its molar mass
which is equal to 180.2 grams. Thus, 3.81 grams of glucose is equal to 2.114x10-2 mol
of glucose.*
Second, identify the unknown or what is being asked in
the
problem.
Unknown: Volume in mL of a 2.53 M glucose solution
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Third, write the formula and calculate the unknown. But in this
case we have to derive the formula.
From the mother formula;
n
M=
v
To derived formula;
n
V=
M
Thus,
2.114x10-2mol of glucose
V=
= 8.36x10-3 L
2.53M
*Notice that our units is in Liter and the problem is asking for the uni to be in mL, that
is why we must convert 8.36x10-3 L into mL which is equivalent to 8.36 mL solution.
And 8.36 mL is the final answer for this problem.*
Molality
It is an intensive property of solutions, and it is calculated as the moles of a solute
divided by the kilograms of the solvent. Unlike molarity, which depends on the volume
of the solution, molality depends only on the mass of the solvent. Molality is the
number of moles of solute dissolved in 1 kg (1000 g) of solvent—that is,
Moles of solute
m=
Mass of solvent
The SI unit for molality is mol/kg. A solution with a molality of 3 mol/kg is often
described as “3 molal” or “3 m.” However, following the SI system of units, mol/kg or
a related SI unit is now preferred.
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Let us consider the examples below;
Example 1: Calculate the molality of a sulfuric acid solution containing
24.4 g of sulfuric acid in 198 g of water. The molar mass of
sulfuric acid is 98.09 g.
Solution: First,identify the given
Given: Solute = 24.4 grams of sulfuric acid
Solvent = 198 grams of water
*Always be mindful with the units, the solute given is in grams and it should be first
converted into moles (that is 0.249 molH2SO4) and also the solvent is expressed in
grams that should be in kilograms so, solvent must be 0.198 kg).*
Second, identify the unknown or what is being asked in the problem.
Unknown: Molality of sulfuric acid
Third, write the formula and calculate the unknown
mm== Moles of solute
m = Mass of solvent
Thus,
0.249 molH2SO4
m=
0.198 kg
= 1.26m
Example 2: 80.0 grams of glucose (C6H12O6 ) is dissolved in 1.00 kg of
solvent. What is its molality? Molar mass of (C6H12O6 ) is
equal to 180g/mol.
Solution: First, identify the given;
Given: Solute = 80.0 grams of glucose
Solvent = 1.00kg
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*Remember that, molality should be in the units of moles and kilogram. So, 80.0 grams
of glucose should be converted first into moles. Thus, glucose is equal to 0.444 mol.*
Second, identify the unknown or what is being asked in
the
problem.
Unknown: Molality of Glucose solution
Third, write the formula and calculate the unknown
m=
Moles of solute
Mass of solvent
Thus,
m=
0.444 mol glucose
= 0.444m
1.00kg
Learning Competency:
Use different ways of expressing concentration of solutions: Molarity, Molality, Percent
by mass, Percent by volume, mole fraction and ppm. (STEM_GC11PP-IIId-f-111)
Activity 1: Choose The “RIGHT” One
Directions: Read and analyze the following questions and choose from the given
options the best correct answer.
1. Which of the following is not the unit of concentration?
a. Mole/m3
b. Molar
c. N/m3
d. ppm
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2. Which of the following material present in a solution is largest in amount?
a. Salt
b. Solute
c. Solvent
d. Molecules
3. Which of the following is defined as the relative amount of solute and solvent in a
solution?
a. Polarity
b. Solubility
c. Miscibility
d. Concentration
4. Which of the following describes a solvent in a solution?
a. Always a water
b. Always a liquid
c. The substance being dissolved
d. The substance present in the greatest amount
5. Which of the following is defined as the quantity of solute per unit volume?
a. Density
b. Concentration
c. Mole
d. None of the above mentioned
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Activity 2: CONCEPTUAL ANALYSIS
Directions: Base on what you have learned from this lesson and from other previous
lessons. Analyze the given statement and scientifically discuss your claim.
A solution is prepared at 20oC and its concentration is expressed in two different
units; Molarity and Molality. The solution is then heated to 88 oC. Which of the
concentration units will change?
Activity 3: MATCH ME!
Directions: Read and analyze the following questions and compute for what is
unknown in the given problem. Choose the correct numerical value from the response
list on the right. Responses on the right may be used more than once or need not be
used at all.
B. 58.44g
1. What is the percent by volume concentration of a
solution in which 75.0mL of ethanol is diluted to a volume
of 250mL?
A. 1.43mL
D. 12.39%
2. What volume of acetic acid is present in a bottle
containing 350.0mL of a solution which measures 5.00%
concentration.
C. 1gram
E.17.5mL
3. Find the percent by mass in which 41.0g of NaCl is
dissolved in 331g of water.
4. How many grams of NaCl would you need to prepare
F.2.0x1010ppm
G. 30%
200.0mL of a 5M solution.
5. What is the ppm concentration of 6.00 mL sample of
solution that has 3.6 x 10-4 g of sodium ions?
J. 60ppm
I. 8.07%
H. 33.3%
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Activity 4: Calculate The Unknown
Directions: Read and analyze the following questions and compute for what is
unknown in the given problem. Show complete solution by stating the given and
unknown, and show the process from writing the formula to unit conversion (if
applicable).
1. Suppose you added 4.0 moles of sugar to 10.0 L of solution. Calculate the molar
concentration of the solution.
2. A sample of water taken from a nearby lake is found to have 0.0035 mol of salt in a
100mL solution. Determine the molar concentration of the solution in the lake.
3. You dissolve 30.0g of sodium sulfate (Na2SO4(s)) into 300mL of water. Calculate the
molar concentration of the solution.
4. What is the Molality of a solution containing 7.78g of Urea [(NH 2)2CO2] in 203g of
water?
5. Lead is a poisonous metal that especially affects children because they retain a
larger fraction of lead than adults do. Lead levels of 0.250ppm in a child cause delayed
cognitive development. How many moles of lead present in 1.00g of child’s blood
would 0.250ppm represent.
6. Acetone, C3H6, is the main ingredient of nail polish remover. A solution is made up
by adding 35.0mL of acetone (d=0.790g/mL) to 50.0mL of ethyl alcohol, C2H6O
(d=0.789g/mL). Assuming volumes are additive, calculate (a) the mass percent of
acetone in the solution. (b) the volume percent of ethyl alcohol in the solution. (c) the
mole fraction of acetone in the solution.
Activity 5: Expressing Concentration in Different Units
Directions: Read and internalize the short story below and answer the questions that
follow. Complete solution is required.
“The coronavirus disease (COVID-19) is an infectious disease caused by a new strain
of coronavirus. This new virus and disease were unknown before the outbreak began
in Wuhan, China, in December 2019.
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On 30 January 2020, the Philippine Department of Health reported the first case of
COVID-19 in the country with a 38-year-old female Chinese national. On 7 March, the
first local transmission of COVID-19 was confirmed. WHO is working closely with the
Department of Health in responding to the COVID-19 outbreak.” Aki and her family
were alarmed with the news they watched and so first thing in the morning they rush
to the nearest convenient store to secure disinfectant and sanitizers but unfortunately
the store already had empty shelves of the essentials they needed. They went to other
stores searching and to their dismay they acquired nothing. They were on their way
home when she suddenly remembered her past lesson on “solutions”, and so she
immediately ran back to the store and purchase the things she needed for her simple
experiment. She bought a bleach (Zonrox), gloves and measuring spoon and cups.
Arriving at home she then put her gloves on and prepared the things she needed such
as; 5tbsp bleach (0.0739L), 1 gallon of water (3.8L), pail and stirring rods. Using the
pail with 3.8L of water, she carefully poured the 5tbsp bleach solution and then mixed
it with the stirring rod. And they now have a disinfectant.
In connection to her home made disinfectant and with our lesson, we will express her
solutions’ concentration into different units; Molarity, Molality, Percent by mass,
Percent by volume, mole fraction and ppm.
Questions:
1. What is the molar concentration of Aki’s disinfectant if she dissolved 5Tbsp. of
NaClO (sodium hypochlorite) in 3.8 liters of H2O (water)?
2. Compute for the molality of her disinfectant if she dissolved 5Tbsp. of NaClO
(sodium hypochlorite) in 3.8 liters of H2O (water).
3. Calculate the percent by mass of sodium hypochorite in her disinfectant solution. (
5Tbsp. of NaClO (sodium hypochlorite) and 3.8 liters of H2O (water). In units of grams
for both of the solute and solvent.
4. Calculate the mole fraction of sodium hypochorite and water in Aki’s solution. (
5Tbsp. of NaClO (sodium hypochlorite) and 3.8 liters of H2O (water).
5. Calculate the percent by volume of the disinfectant Aki made. Units in mL.
6. What is the concentration of her solution in parts per million? ( 5Tbsp. of NaClO
(sodium hypochlorite) and 3.8 liters of H2O (water).
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Reflection
1. I learned that
_
_
2. I enjoyed most on
_
_
_
_.
3. I want to learn more on
_
_
_.
References:
Masterton, William and Cecille Hurley. Chemistry Principles and Reactions.
Solutions.Fifth Edition, Thomson Books/Cole,2004.
Hein,Morris et.al. Foundations of Chemistry in the Laboratory for Sciences. Properties
of Solutions. Twelfth Edition, John Wiley and Sons Inc.,2007.
Whitten, Kenneth et.al. General Chemistry. Solutions. Seventh Edition, Thomson
Books/Cole,2004.
Chang, Raymond. Chemistry. Concentration Units. Tenth Edition, McGraw-Hill, 2010.
Prepared by:
ANGELIKA TORRES
Sta Ana Fishery National High School
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GENERAL CHEMISTRY 2
Name: _
Grade Level:
Date:
Score:
__
LEARNING ACTIVITY SHEET
STOICHIOMETRIC CALCULATIONS FOR REACTIONS IN SOLUTION
Background Information for the Learners (BIL)
Stoichiometry is the calculation of reactants and products in a certain chemical
reactions. It applies the law of conservation of mass wherein the total mass of the
reactants is always equal the total mass of the products, leading to the insight that the
relations among the value or amount of reactants and products typically produce a
ratio of positive numbers. This implies that if the amounts of the separate reactants
are known, then the amount of the product can be calculated and vice versa.
This image here, shows that the chemical reaction is balanced.
Source: https://en.wikipedia.org/wiki/Stoichiometry#/media/File:Combustion_reaction_of_methane.jpg
It shows that one molecule of methane, CH4 reacts with two molecules
of oxygen gas, O2 to produce one molecule of carbon dioxide, CO2 and two molecules
of water, H2O. This chemical reaction is an example of complete combustion.
Stoichiometry measures these numerical relationships and is used to calculate the
amount of products and reactants that are produced or needed in a given reaction.
Describing the mathematical relationships of the substances that contributed in
chemical reactions is what we call reaction stoichiometry. It measures the
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relationship between the amount of methane and oxygen that react to form carbon
dioxide and water.
Elements in the periodic table have a different atomic mass, and as collections
of single atoms or molecules have a fixed molar mass, measured with the unit mole
(6.02 × 1023 individual molecules, Avogadro's constant). Carbon-12 has a molar mass
of 12 g/mol. Thus, to compute the stoichiometry by mass, the number of molecules
needed for each reactant is expressed in moles multiplied by the molar mass of each
to give the mass of each reactant per mole of reaction. The mass ratios can be
computed by dividing each by the total number in the whole reaction.
Stoichiometry is often used to balance chemical equations. For example, the
two diatomic gases, hydrogen and oxygen, when it combine H2 and O2, it produce a
liquid, water, in an exothermic reaction, as described by the following equation:
2 H2 + O2 →2 H2O
It shows the 2:1:2 ratio of hydrogen, oxygen, and water molecules in the above
equation.
The molar ratio permits for conversion between moles of one substance and
moles of another. For example,
2 CH3OH +3 O2 →2 CO2 +4 H2O
the amount of water that formed by the combustion of 0.27 moles of CH 3OH is
obtained using the molar ratio between CH3OH and H2O of 2 to 4.
Stoichiometry is also used for determining the molar proportions of elements in
stoichiometric compounds. For example, the stoichiometry of hydrogen, H 2 and
oxygen, O2 in H2O is 2:1. In stoichiometric compounds, the molar proportions should
be whole numbers.
Determining the Amount of Product
The term stoichiometry can be used to find the quantity of a product produced
by a reaction. If a piece of solid copper (Cu) were added to an aqueous solution
of silver nitrate (AgNO3), the silver (Ag) would be substituted in a single displacement
reaction forming aqueous copper(II) nitrate (Cu(NO3)2) and solid silver. How many
silver, Ag is formed if 16.00 grams Cu is added to the solution of excess silver nitrate,
AgNO3?
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The following steps would be used:
1. Write and balance the chemical equation
2. Mass to moles conversion: Convert grams of Cu to moles of Cu
3. Mole ratio determination: Convert moles of Cu to moles of Ag produced
4. Mole to mass conversion: Convert moles of Ag to grams of Ag produced
The complete balanced equation would be:
Cu +2 AgNO3 → Cu(NO3)2 + 2 Ag
For the mass to mole conversion, the mass of copper (16.00 g) would be
converted to moles of copper by dividing the mass of copper to its molecular mass:
63.55 g/mol.
Now that the amount of Cu in moles (0.2518) is form, we can set up the mole
ratio. This is done by looking at the coefficients in the balanced equation: Cu and Ag
are in a 1:2 ratio.
Now that the moles of Ag produced is known to be 0.5036 mol, this amount can
be converted into grams of Ag produced to determine the final answer:
This set of calculations can be further shortened into a single step:
Stoichiometric Calculations
The coefficients in the balanced equation give the ratio of moles of reactants
and products.
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Source: https://www2.chemistry.msu.edu/courses/cem151/chap3lect_2009.pdf
From the mass of Substance A, you can use the ratio of the coefficients of A
and B to determine or calculate the mass of Substance B formed (if it’s a product) or
used (if it’s a reactant).
Example: 10 grams of glucose (C6H12O6) react in a combustion reaction. How many
grams of each product are produced?
C6H12O6(s) + 6 O2(g)
6 CO2(g) + 6 H2O(l)
10.g
?
+
?
Starting with 10. g of C6H12O6, we calculate the moles of C6H12O6. Use the coefficients
to find the moles of H2O & CO2 and then turn the moles to grams
C6H12O6(s) + 6 O2(g)
10.g
MW:
180g/mol
#mol:
10.g(1mol/180g)
0.055 mol
6 CO2(g) + 6 H2O(l)
?
+
44 g/mol
6(.055)
6(.055mol)44g/mol
#grams:
15g
?
18g/mol
6(.055mol)
6(.055mol)18g/mol
5.9 g
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Reaction Stoichiometry in Solutions
We can perform stoichiometric calculations for aqueous phase reactions just as
we can for reactions in solid, liquid, or gas phases. Much of chemistry takes place in
solution. Stoichiometry allows us to work in solution by giving us the concept of solution
concentration, or molarity. Molarity is a unit that is often abbreviated as capital M. It is
defined as the moles of a substance contained in one liter of solution. Almost always,
we will use the concentrations of the solutions as conversion factors in our
calculations. For instance, if a solution has a concentration of 1.20 M NaCl, this means
that there are 1.20 moles of NaCl per liter of solution.
Example 1: What mass of Aluminum (Al) is needed to react completely with 35.0 mL
of 2.0 M Hydrochloric acid?
Solution:
6 HCl + 2 Al
2 AlCl3 + 3 H2
35.0 mL HCl x 1L HCl x 2 mol HCl x 2 mol Al x 26.98 g Al
1000 mL
1L HCl
6 mol HCl
1 mol Al
Al = 0.63 g
Example 2. What volume (mL) of 0.75 M calcium nitrate would react completely with
148g of carbonate?
Solution:
Ca(NO3)2 + Na2CO3
CaCO3 + 2 NaNO3
148 g Na2CO3 x 1mol Na2CO3 x 1 mol Ca(NO3)2 x 1L Ca(NO3)2 x 1000 mL
105.99 g Na2CO3
1mol Na2CO3
0.75mol Ca(NO3)2
1L
CaCO3 = 1.900 mL
Learning Competency:
Perform stoichiometric calculations for reactions in solution (STEM_GC11PP-IIId-f112)
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Activity 1: FILL THE EMPTY LINE
Directions: Read the following statement below and solve the problem. In the
equation that follows each problem, write on the space provided for the mole ratio that
can be used to solve the problem. Write the correct answer on the space provided for.
The reaction of sodium peroxide and water produces sodium hydroxide
and oxygen gas. The following balanced chemical equation represents the
reaction.
2 Na2O2(s) + 2 H2O(l) → 4NaOH(s) + O2(g)
1. How many moles of NaOH are produced when 1.00 mol sodium peroxide reacts
with water?
1mol Na2O2 x
_
=
mol NaOH
2. How many moles of oxygen gas are produced when 0.500 mol sodium peroxide
reacts with water?
0.5 mol Na2O2 x
_
=
_ mol O2
3. How many moles of sodium peroxide are needed to produce 1.00 mol NaOH?
1 mol NaOH x
=
_ mol NaOH
4. How many moles of water are required to produce 2.15 mol oxygen gas?
2.15 mol O2 x
=
_ mol H2O
5. How many moles of water are needed for 0.100 mol of sodium peroxide to react
completely?
0.100mol Na2O2 x
_ =
_ mol H2O
Activity 2: SIMPLE STOICHIOMETRY
Directions: Solve the following stoichiometry grams – grams problems.
The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.64
grams of water.
2 C4H10 + 13O2
8CO2 + 10H2O
a. How many moles of water formed?
___
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b.How many moles of butane burned?
_
c. How many grams of butane burned?
_
d. How much oxygen was used up in moles?
_
e. How much oxygen was used up in grams?
_
_
Activity 3: THINK ABOUT IT!
Directions: Solve the following simple stoichiometry problems.
1. 123 mL of a 1.00 M solution of NaCl is mixed with 72.5 mL of a 2.71 M solution of
AgNO3. What is the mass of AgCl(s) formed in the precipitation reaction?
AgNO3(aq) + NaCl(aq)
AgCl(s) + NaNO3(aq)
_
2. What volume (mL) of 0.70 M Sodium hydroxide (NaOH) is needed to
neutralize 270 mL of 0.40 M Sulfuric acid (H2SO4)?
2 NaOH + H2SO4
Na2SO4 + 2 H2O
_
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3. Hydrogen gas can be produced through the following reaction.
Mg(s) + 2HCl(aq) 
MgCl2(aq) +
H2(g)
a. How many grams of HCl are consumed by the reaction of 2.50 moles of
magnesium?
b. What is the mass in grams of H2 gas when 4.0 moles of HCl is added to the
reaction?
4. Acetylene gas (C2H2) is produced as a result of the following reaction.
CaC2(s) +
2H2O(l) 
C2H2(g)
+ Ca(OH) 2(aq)
a. If 3.20 moles of CaC2 are consumed in this reaction, how many grams of H2O
are needed?
b. How many grams of Ca(OH)2 would be formed with 3.20 moles of CaC2?
5. Laughing gas (nitrous oxide, N2O) is sometimes used as an anesthetic in dentistry.
It is produced when ammonium nitrate is decomposed according to the following
reaction.
NH4NO3(s) 
N2O(g) + 2H2O(l)
a. How many moles of NH4NO3 are required to produce 33.0g of N2O?
b. How many moles of water are produced with 45.0g of N2O?
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Activity 4: GIVE ME MY VALUE
Directions: Complete the equation by writing the correct value on the space
provided for.
For questions 1 – 3, refer to the equation below
4 Fe + 3 O2  2 Fe2O3
1. How many moles of Fe2O3 are produced when 0.275 moles of Fe is reacted?
0.275 mol Fe
mol Fe2O3
2. How many moles of Fe2O3 are produced when 31.0 moles of O2 is reacted?
3. How many moles of O2 are needed to react with 8.9 moles of Fe?
For questions 4 – 6, refer to the equation below
2 KClO3  2 KCl + 3 O2
4. How many moles of O2 will be formed from 1.65 moles of KClO3?
1.65 mol KClO3
mol O2
mol O2
mol KClO3
5. How many moles of KClO3 are needed to make 3.50 moles of KCl?
3.50 mol KCl
_ mol KClO3
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Activity 5: GIVE ME THE SOLUTION
Directions: Solve the following problems based on the chemical reaction below
4 Fe + 3 O2  2 Fe2O3
1. How many grams of Fe2O3 are produced when 42.7 grams of Fe is reacted?
2. How many grams of Fe2O3 are produced when 17.0 grams of O2 is reacted?
3. How many grams of O2 are needed to react with 125 grams of Fe?
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Reflection:
1. I learned that
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2. I enjoyed most on
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3. I want to learn more on
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References:
Hill, Petrucci. General Chemistry: An integrated approach, second edition. New
Jersey: Prentice Hall, 1999.
https://courses.lumenlearning.com/introchem/chapter/solution-stoichiometry/
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Modules_and_Website
s_(Inorganic_Chemistry)/Chemical_Reactions/Reactions_in_Solution
https://www2.chemistry.msu.edu/courses/cem151/chap3lect_2009.pdf
http://www.calhoun.k12.al.us/teacherpages/teacherfiles/Stoichiometric
https://iasmisserica.weebly.com/uploads/4/2/6/4/42642303/escanear0094.pdf
http://www2.ucdsb.on.ca/tiss/stretton/CHEM1/stoicwk2.html
https://www.murrieta.k12.ca.us/cms/lib5/CA01000508/Centricity/ModuleInstance/868
1/Stoichiometry_-_mole_to_mass.doc
Prepared by:
SHAROLYN T. GALURA
Licerio Antiporda Sr National High School- Dalaya Annex
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GENERAL CHEMISTRY 2
Name: _
Grade Level:
Date:
Score:
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LEARNING ACTIVITY SHEET
Effects of Concentration on the Colligative Properties of Solutions
Background Information for the Learners (BIL)
Colligative properties of solutions are properties that depend upon the
concentration of solute molecules or ions, but not upon the identity of the solute.
Colligative properties include vapor pressure lowering, boiling point elevation, freezing
point depression, and osmotic pressure.
Lowering the Vapor Pressure
Vapor pressure is the pressure of a vapor in thermodynamic equilibrium with ts
condensed phase in a closed container. When non-volatile solute is dissolved in
solvent, the vapor pressure of solvent is lowered. The presence of solute decreases
the rate of escape of solvent molecules resulting to lower vapor pressure.
Boiling Point Elevation
The boiling point of a liquid is defined as the temperature at which the vapor
pressure of that liquid equals the atmospheric pressure (760mm Hg). The addition of
the solute increases the boiling point of the solution. The atmospheric pressure
remains the same while the vapor pressure of the solution is lowered resulting in the
increase of the difference in atmospheric pressure and vapor pressure of the solution.
Therefore, a higher temperature is required to boil the solution.
Freezing Point Depression
Normal freezing or melting point is the temperature at which solid and liquid
are in equilibrium under 1 atm. Addition of solute will decrease the vapor pressure
and so will decrease the freezing point. In order for a liquid to freeze it must achieve
a very ordered state that results in the formation of a crystal.
If there are impurities in the liquid, i.e. solutes, the liquid is inherently less
ordered. The presence of impurities in a liquid or in a substance makes variation in the
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freezing point by making them low or high. Therefore, a solution is more difficult to
freeze than the pure solvent so a lower temperature is required to freeze the liquid.
Osmotic Pressure
This is the external pressure that must be applied to the solution in order to
prevent it being diluted by the entry of solvent via osmosis. Diffusion in liquids,
substance tend to move or diffuse from regions of higher concentration to region of
lower concentration. The overall effect is to equalize concentration throughout the
medium.
Osmosis, on the other hand is the movement of solvent particles across a
semipermeable membrane from a dilute solution into concentrated solution. The
solvent moves to dilute the concentrated solution and equalize the concentration on
both sides of the membrane.
Osmotic pressure is directly proportional to the concentration of the solution.
Therefore, doubling the concentration will also double the osmotic pressure. The
osmotic pressure of two solutions having the same molal concentration are identical.
Learning Competency
Describe the effect of concentration on the colligative properties of solutions (STEMGC11-PPIIId-f-115)
Activity 1: FACT OR BLUFF
Directions: Read each statement carefully. Identify whether the statement is a FACT
or BLUFF. Draw  on the space if it is a FACT and  if it is a BLUFF.
1. Colligative properties arise from the fact that solute affects the concentration of
solvent.
_
2. Vapor pressure is a colligative property.
3. Lowering of vapor pressure is not dependent on the number of species present
in the solution.
4. Colligative properties of solution depend on the nature of the solute and the
solvent.
5. Colligative molality is the molality times the number of solute particles per
formula unit.
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6. Osmotic pressure is directly proportional to the concentration of the solution.
_
7. Relative lowering of vapor pressure is a colligative property.
_
8. The boiling point of a solution decreases in direct proportion to the molality of
the solute._
9. When non-volatile solute is dissolved in solvent, the vapor pressure of solvent
is lowered.
10. The depression of the freezing point is directly proportional to the molality of the
solvent.
_
Activity 2: 1 PIC, 4 SENTENCES
Directions: The pictures below illustrate the different effects of colligative properties
to solutions. Using four (4) sentences, answer the question that each picture depicts.
Write your answer on the spaces provided.
1. Why adding salt to water increases the boiling point?
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Source:
https://www.thoughtco.com/thmb/Wlx0HpISUQfVc401y0XNpBdcms
=/768x0/filters:no_upscale():max_bytes(150000):strip_icc():for
mat( webp)/GettyImages-1166175911-
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2. Why does seawater have lower freezing point than pure water?
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https://www.google.com.ph/url?sa=i&url=https%3A%2F%2Ftwitter.co
m%2
Fmpi_meteo%2Fstatus%2F1074242734129434624&psig=AOvVaw0l
K2WkzZ
5BHXZeNI34w3fI&ust=1596870848098000&source=images&cd=vfe
&ved=0 CAIQjRxqFwoTCIDOpu7EiOsCFQAAAAAdAAAAABAD
3.
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https://www.google.com/search?sxsrf=ALeKk03ZiIKudLOeyQ1ocit_OtM
P6JIKQw:1596790694111Eq
4. Which sample has the lowest vapor pressure? Why?
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5. How can you regain the crispyness of a carrot and celery that have become limp?
What colligative property is involved in the process?
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https://opentextbc.ca/chemistry/chapter/11-4colligative-properties
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Activity 3: WHO WANTS TO BE A CHEMIST?
Directions: To become a chemist and receive a score, you need to answer the
question in each level. Your score increases as you go to a higher level.
Compare the properties of 1.0 M aqueous
sugar solution to a 0.5 M aqueous solution
of NaCl.
LEVEL 5
Why is vapor pressure lowering a
colligative property?
LEVEL 4
What is needed for a liquid to freeze?
LEVEL 3
LEVEL 2
What do colligative properties of solutions
depend on?
What are the different colligative
properties of solution?
LEVEL 1
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Reflection:
1. I learned that
_
_
2. I enjoyed most on
_
_
_.
3. I want to learn more on
_
_
_.
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References
Books
Santos, Gil Nonato S., Danac, Alfonso C., O-Chemistry III, 2009
Mortimer Charles E., Chemistry 6th Edition
Hagad, Hilda R., Phoenix Next Century Chemistry, 2003
Websites
https://chem.libretexts.org
https://opentextbc.ca
https://www.sparknotes.com
https://www.bhsu.edu
Prepared by:
LOVEJOICE L. AMBABAG
Tuao Vocational and Technical School
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