Lesson 1.

Kinetic Molecular
Theory of Matter

General Chemistry 2
Science, Technology, Engineering, and Mathematics
What does the kinetic
molecular theory state?

2
 Kinetic Molecular Theory of Matter

● The kinetic molecular theory of matter provides an

overview of the microscopic properties of molecules or
atoms and their interactions.

● It describes the microscopic properties of matter

and how they translate to the state and other
properties of matter.

3
 Kinetic Molecular Theory of Matter

It states that:
1. Matter is composed of small particles.
2. The molecules interact with one another through
attractive forces. The strength of these forces is
related to the distance between the particles.
3. These molecules are always in constant random
motion.
4. The temperature of a substance is a measure of
the average kinetic energy of the molecules.
4
 Kinetic Molecular Theory of Matter

Matter Is Composed of Small Particles

Earlier atomic models 5

 Kinetic Molecular Theory of Matter

Matter Is Composed of Small Particles

● Atoms are the building blocks of matter.

gold bar gold atoms

6
 Kinetic Molecular Theory of Matter

Matter Is Composed of Small Particles

● Matter can also exist as molecules.

drop of water water molecules

7
 Kinetic Molecular Theory of Matter

Matter Is Composed of Small Particles

● Matter can also exist as molecules.

table salt sodium and chloride ions

8
 Kinetic Molecular Theory of Matter

Molecules Interact through Attractive Forces

● The attractive forces between molecules are known as

the intermolecular forces.

● The stronger the interaction between two molecules,

the smaller their distance will be.

9
 Kinetic Molecular Theory of Matter

Molecules Interact through Attractive Forces

solid liquid gas

10
Solids

● Solids have particles with

strong intermolecular
forces such that their
particles are very close to
one another.

particles of solid
11
 Liquids

● Liquids have intermediate

intermolecular forces.
This makes liquid particles
farther from one another
compared to those in
solids.

particles of liquid
12
Gases

● Gases have particles that are

very far apart from one
another due to weak
intermolecular forces.

particles of gas
13
 Kinetic Molecular Theory of Matter

Molecules Are in Constant Random Motion

● All molecules are in constant random motion.

● The extent of their motion varies depending on the

temperature and strength of the interaction
between the particles.

14
 Kinetic Molecular Theory of Matter

Nanoscale Representation of Matter

molecules in ice molecules in water molecules in steam

15
Solids

● Solids, due to their strong

intermolecular forces, have
restricted motion.
● The particles of solids are
only able to vibrate back
and forth around a specific
point or location.

particles of solid
16
 Liquids

● Liquids, having
intermediate
intermolecular forces, are
able to move past each
other.
● Since their particles are still
close to one another, the
motion is restricted to
small distances as they
particles of liquid will collide with another
molecule.
17
Gases

● Gases, having weak

intermolecular forces, are
able to move in relatively
long distances before
colliding with another
molecule.
● This happens because the
particles are very far apart
from one another. particles of gas
18
 Kinetic Molecular Theory of Matter

Temperature Is a Measure of the Average KE

heat increases
KE increases

19
 States of Matter

Matter can be described in terms of its physical state.

It can either be solid, liquid, or gas.

solid liquid gas

(ice) (water) (steam)

20
 States of Matter

State of
Shape Volume Compressibility
Matter
solid fixed shape fixed volume virtually
incompressible
liquid depends on the fixed volume only slightly
compressible
container
gas depends on the assumes the volume very
compressible
container of the container

21
 States of Matter

Particulate Drawings of Solid, Liquid, and Gas

22
How can the kinetic
molecular theory explain
the properties of each state
of matter?

23
Based on the kinetic molecular theory of
matter, the state of a matter is determined
by two factors—temperature and
strength of intermolecular forces.
 Properties of the States of Matter

As explained by KMT:
● At lower temperatures, intermolecular forces
determine the state of a substance.

● Substances with intermediate to strong

intermolecular forces will form a condensed phase,
either solid or liquid.

● Those with weak intermolecular forces will be in the

gaseous state.
25
How does the kinetic
molecular theory explain
phase changes?

26
 Phase Changes

● Recall that a higher temperature means a higher

average kinetic energy.

● This means that the particles will have enough energy

to move around faster.

● Increasing the temperature also weakens the

intermolecular forces of attraction.

27
Melting
● The molecules in ice vibrates
back and forth to a specific
location since it is in the
solid phase.
● Increasing the temperature
will provide enough kinetic
energy to overcome their
strong intermolecular forces.
● The solid becomes a liquid in
a process called melting.
28
Vaporization

● When liquid water is

heated, its particles are able
to move past one another
in relatively longer
distances.
● The process in which liquid
is converted to a gas is
called vaporization.

29
How does the kinetic molecular theory
describe phases and phase changes?

30
 Check Your Understanding

Identify the state of matter being described below.

1. The particles vibrate at a specific location only.

2. The particles move about in relatively short distances.
3. The particles move in longer distances.

31
Lesson 1.2

Intermolecular Forces
of Attraction

General Chemistry 2
Science, Technology, Engineering, and Mathematics
Have you ever wondered why some substances boil
easier than others?
2
For example, liquid
nitrogen when
exposed to room
temperature
immediately turns
into vapor.

3
On the other hand, water needs to be heated first to be
converted to steam.
4
The kinetic molecular
theory states that
matter is composed
of tiny particles that
carry energy, interact
with one another and
are in constant
random motion.
particles in a gas
5
The interaction between particles and their strength
determines certain properties for that matter.
6
What are the different types
of intermolecular forces of
attraction?

7
 Intermolecular Forces of Attraction

● Intermolecular forces are attractive forces present in

between molecules.

● The four main types of intermolecular forces are:

○ London dispersion forces,
○ dipole-dipole forces,
○ ion-dipole forces, and
○ hydrogen bonding forces
8
 Intermolecular Forces of Attraction

London dispersion forces and dipole-dipole forces are

collectively known as van der Waals forces of attraction.

Johannes van der Waals 9

 Ion-Ion interactions

Ion-ion interaction is the interaction between two

oppositely charged particles.

anion
cation negatively
positively charged
charged sodium chloride ion (Cl–)
ion (Na+) 10
 Ion-Ion interactions

Ion-ion interaction is also known as ionic bonds.

table salt sodium and chloride ions

11
Remember

Ion-ion interactions are between

electrically charged particles.

12
 Ion-Dipole interactions

ion-dipole interactions strengths of ion-dipole interactions

13
 Ion-Dipole interactions

This type of interaction is responsible for the dissolution of most ionic

solids in polar solvents. 14
Remember

The partially positive end of the polar

molecule interacts with the anion,
whereas the partially negative end of
the polar molecule interacts with the
cation.

15
 Dipole-Dipole interactions

Present in polar molecules which are described as dipoles

Hydrochloric acid is a dipole.

16
The dipole-dipole force
exists between the
partially positive end of
one HCl molecule and
the partially negative
end of another HCl
molecule.

17
 Hydrogen Bonding

An attractive force that exists when hydrogen is bonded to

the most electronegative atoms, namely F, O, or N

Hydrogen bonding between formaldehyde and water

18
Many unusual
properties of water
are attributed to
hydrogen bonding.
In water, the
hydrogen of one
molecule is attracted
to the oxygen atom
of another molecule.
19
Remember

Hydrogen bonding can only be

exhibited when one molecule has a
hydrogen atom is directly bonded to
fluorine, oxygen or nitrogen atom.

20
 London Dispersion Forces

● The weakest type of IMFA and are present in between

all electrically neutral molecules
● Named after the German-American physicist Fritz
London

Temporary dipoles between nonpolar molecules

21
 London Dispersion Forces

A nonpolar molecule has an equal distribution of charges.

(a) (b) (c)

22
London Dispersion Forces

At any instant, an instantaneous dipole may form.

(a) (b) (c)

23
 London Dispersion Forces

The instantaneous dipole may induce the formation of

another dipole (induced dipole).

(a) (b) (c)

24
The formation of instantaneous dipole can be observed in
nonpolar molecules such as O2.
25
London Dispersion Forces

26
London Dispersion Forces

27
Remember

All electrically neutral molecules exhibit

London dispersion forces (LDF).

28
Induced Dipoles

29
How are induced dipoles
created?

30
 Predicting Intermolecular Forces of Attraction

Recall that compounds can be classified as ionic or

covalent based on the types of bonds present.

● Ionic compounds ⟶ ion-ion interactions

● Covalent compounds ⟶ depend on polarity

31
 Predicting Intermolecular Forces of Attraction

● The strength of the ion-ion interaction is governed by

Coulomb's law

where F is coulombic force, q1 and q2 are the charges of

the particles, and r is the distance between the
particles.
32
 Predicting Intermolecular Forces of Attraction

Compound Melting Compound Melting Compound Melting

Point Point Point
(OC) (OC) (OC)

NaF 993 CaF2 1423 MgO 2800

NaCl 801 Na2S 1180 CaO 2580

NaBr 747 K2S 840 BaO 1923

33
 Predicting Intermolecular Forces of Attraction

● Covalent bonds, on the other hand, involve the

sharing of electrons between two nonmetal atoms.

● Recall that polarity of the molecule can be determined

by identifying the polarity of the bonds and the
molecular geometry for the compound.

34
 Predicting Intermolecular Forces of Attraction

● Polar covalent compounds are molecules with a net

dipole moment due to unequal sharing of electrons
between the atoms.

● This causes the molecule to have a partial positive (δ+)

and a partial negative (δ-) charges, which are also
known as a dipole.

35
 Predicting Intermolecular Forces of Attraction

● Polar covalent compounds can either have dipole-

dipole interactions or hydrogen bonding, depending
on the presence of H and its connectivity to other
atoms in the compound, and London dispersion
forces.

36
Notice that H2O,
HF, and NH3
have higher
boiling points
than the rest of
their groups
because they
can form
hydrogen
bonding.
37
Remember

Some molecules have polar bonds but

are nonpolar as a whole. This is due to
the cancellation of the dipole moment
due to the molecular geometry.

38
 Predicting Intermolecular Forces of Attraction

● Nonpolar covalent compounds are molecules with

zero dipole moment due to equally shared electrons
between the atoms.

● The only intermolecular force present in these

compounds is the London dispersion forces.

39
 Predicting Intermolecular Forces of Attraction

Polarizability is the measure of how easy it is to distort

the electron distribution of a molecule.

Polarizability, Molar Mass, Boiling Point,

Compound
10–25 cm3 amu K
H2 7.9 2.02 20.35
O2 16.0 32.00 90.19
N2 17.6 28.01 77.35
40
Predicting Intermolecular Forces of Attraction

41
Predicting Intermolecular Forces of Attraction

42
How can one determine the
intermolecular force present for
a molecule?

43
Tips

In order to predict the intermolecular

forces between two molecules, you
must first determine the type of
compound present.

44
45
Let’s Sum It Up!

46
Challenge Yourself

Arrange the following in increasing

IMFA strength: ethanol, ethylene
glycol, ethane.

47
Challenge Yourself

48
GENERAL CHEMISTRY 2
Let’s Sum It Up!

2
Identify What type of IMFA will be the most prevalent in
each of the following substances.

1. C2H6
2. CO2
3. NH3
4. Na2O
5. N2O5
6. H 2O
 POWER COMPETENCY
Apply the properties of liquids and
solids to the nature of forces in
designing a simple investigation to
determine the effect on boiling point
or freezing point when a solid is
dissolved in water.
 LEARNING TARGETS

➢ Describing 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.
Properties of Liquids

General Chemistry 2
Science, Technology, Engineering, and Mathematics
In nature, there are
attractive and
repulsive forces.
Some of these
attractive forces also
exist among very
small particles of
matter.

7
 States of Matter

● Matter can be described in terms of its physical state.

● The three basic states of matter are solid, liquid and

gas.

8
 Properties of Liquids

● The properties of liquids can be attributed to the

intermolecular forces of attraction that hold the
molecules together in order to occupy a volume of a
certain space of a container.

● These intermolecular forces of attraction are directly

related to the surface tension, viscosity and vapor
pressure of the liquid.

9
 What is surface tension, and
how is it related to
intermolecular forces in liquids?

10
 Surface Tension

● Surface tension is the tendency of a fluid to acquire

the least possible surface area.

● A drop of liquid is assumed to be spherical in shape

due to the intermolecular forces of attraction present
in the molecules of the liquids.

11
Surface Tension

Water striders can walk on water because of surface tension. 12

 Define viscosity and discuss
how intermolecular forces
influence the flow properties of
liquids.

13
 Viscosity

● Viscosity is the attraction between unlike molecules.

● Liquids that flow easily have low viscosity while liquids

that do not flow readily have high viscosity.

● Molecules with stronger intermolecular forces have

greater resistance to flow because it is difficult for the
molecules to move and slide past one another.

14
Viscosity

Honey is a viscous liquid.

15
 Enthalpy of Vaporization

● Vaporization is the process where a fraction of the

kinetic energy of a liquid escapes from the surface to
enter the vapor phase.

● The enthalpy of vaporization (∆Hvap), also called the

heat of vaporization, is the heat required to induce
this phase change.

16
Enthalpy of Vaporization

17
 Enthalpy of Vaporization

● Vaporization occurs more readily with:

○ increased temperature
○ increased surface area of the liquid
○ decreased strength of intermolecular forces

18
 Solubility

● Solubility refers to the ability of a substance to

dissolve in a given amount of solvent at a specified
temperature.

● Like dissolves like. When the solute and the solvent

both exhibit the same intermolecular forces of
attraction, they form a solution.
● Miscible and Immiscible liquids
19
 Solubility

● Substances or molecules that form interactions with

water are also described as hydrophilic.
● Dipole-dipole forces and hydrogen bonding are the
primary IMFA in hydrophilic substances.
● Substances or molecules that repel water are
described as hydrophobic.
● London dispersion forces are the primary IMFA in
these substances.
20
 Solubility

● Molecules with both hydrophilic and hydrophobic

regions are known as amphipathic molecules.

● Soap and detergent molecules have hydrophilic heads

and hydrophobic tails.

21
 Solubility

● Soap and detergent molecules have hydrophilic heads

and hydrophobic tails.

amphiphatic
molecules micelles layers

22
 Boiling Point

● Boiling point is the temperature at which a substance

changes from liquid to gas.

● Boiling happens when the molecules of a liquid gain

enough energy to overcome the intermolecular forces of
attraction that hold the molecules together.

● Stronger intermolecular forces mean a greater amount of

energy is needed to break the attractive forces between
molecules. 23
 Boiling Point

● Hydrogen bonding makes

the boiling point of water Compound Boiling
significantly higher than Point (°C)
similar binary compounds
of hydrogen and a Group H 2O 100
6A element. H 2S -61
H2Se -41
H2Te -2.2

24
The Structure and Unique
Properties of Water

General Chemistry 2
Science, Technology, Engineering, and Mathematics
26
What is the molecular shape of
water?

27
 The Structure of Water

The dipole moments do not cancel out. This means that

the water molecule is polar.

dipole moment of water

28
What intermolecular forces are
present in water?

29
What are the unique properties
of water?

30
 Properties of Water

Physical Properties of Water

Property Values
specific heat capacity (liquid water) 4.18 J/g · ºC
specific heat capacity (water vapor) 2.11 J/g · ºC
specific heat capacity (ice) 2.00 J/g · ºC
melting point 0 ºC
boiling point 100 ºC

31
Properties of Water

32
 Properties of Water

Water as a Universal Solvent

● Water is known to be the universal solvent due to its

capacity to dissolve a wide variety of substances.

● Its ability to dissolve ionic solids stems from the polarity

of the water molecule.

33
 Properties of Water

Water can dissolve ionic compounds through ion-dipole

interactions.

34
 Properties of Water

Water can dissolve polar compounds through dipole-dipole

interactions.

35
Amphiphatic molecules
contain a polar and
nonpolar regions. Fatty
acids, such as palmitic
acid, contain a long
hydrophobic tail and a
hydrophilic head.

36
When placed in water it
forms a micelle, which
is an assembly of fatty
acids, wherein the
hydrophobic tails are
hidden inside the
structure and the
hydrophilic heads are
exposed.
37
 Properties of Water

Heat Capacity

● Heat capacity is the amount of heat required to raise the

temperature of a substance by 1 ºC.
● Specific heat capacity refers to the amount of heat
required to raise the temperature of one gram of a
substance by 1 ºC.
● For liquid water, the specific heat capacity is 4.18 (J/g ✕ ºC).

38
 Properties of Water

Heat Capacity
Specific heat, c Specific heat, c
Substance Substance
(J/g ✕ ºC) (J/g ✕ ºC)
liquid water 4.18 magnesium 1.024
water vapor 2.11 aluminum 0.903
ice 2.00 iron 0.449
dry air 1.01 zinc 0.389
granite 0.79 copper 0.385

39
Remember

Heat capacity is an extensive property

while specific heat is an intensive
property.

40
 Properties of Water

Phases of Water

● Water, like all matter, can exist as a solid (ice), liquid

(water), or gas (steam).
● From the kinetic molecular theory, solids typically have a
more compact arrangement of particles than liquids and
gases.
● However, this is not true for water.

41
When liquid water
solidifies to ice, it
arranges itself based on
the hydrogen bonding
requirements.

42
Challenge Yourself

How can you relate the properties of

water to its important role in
maintaining life.

43
Lesson 1.5

Structural Features of
Solids

General Chemistry 2
Science, Technology, Engineering, and Mathematics
Solids, unlike liquid and gas, have definite shape and
volume.

2
Diamond is known as
the hardest natural
substance to exist in
the world.
3
But not all solids are
like diamonds. Some
are soft and can be
easily manipulated,
while others are
brittle and strong
resistance to shape.
4
How are the structural
features of a solid related to
its distinguishing
properties?

5
 Learning Objectives
At the end of the lesson, you should be able to do the
following:

● Describe the properties of solids.

● Explain the effect of the intermolecular forces of

attraction on these properties.

● Differentiate the characteristic properties of crystalline

and amorphous solids.
6
 Kinetic Molecular Theory of Solids

Kinetic Theory of Matter

● a simple microscopic model that explains the three
different phases (solid, liquid, and gas) of matter
● explains how they can change from one phase to
another

7
 Kinetic Molecular Theory of Solids

Kinetic Theory of Matter

● All matter is composed of particles that have a certain
amount of energy to make them move at different
speeds depending on temperature
○ temperature - average kinetic energy of particles in
a system

8
 Kinetic Molecular Theory of Solids

● In solids, the intermolecular forces

between neighboring particles are
held together by strong covalent
or ionic bonding
● noncompressible due to lack of
spaces between them
● definite shape and volume Particles in a solid state

9
 Structure of Solids

● Solids can appear in different forms.

● Two main categories:
○ crystalline solids
○ amorphous solids

10
What is the difference
between crystalline and
amorphous solids?

11
 Structure of Solids

Crystalline Solids
● solids in which the atoms,
ions, or molecules are
arranged in a definite
repeating pattern
● held together by uniform,
Silicon dioxide (SiO2), also known as
strong intermolecular forces quartz, is a crystalline solid.

12
 Structure of Solids

Crystalline Solids
● the orderly arrangement of atoms: highly regular
shapes
● examples: quartz and sodium chloride

13
 Structure of Solids

Unit Cell
● the small repeating pattern in crystalline solids
● made up of unique arrangement of atoms
● represent the structure of solid
● the crystal structure can be built by piling the unit
over and over

14
 Structure of Solids

Amorphous Solids
● from the Greek words for
“without form”
● lacks the order found in
crystalline solids
● structures at the atomic level
Obsidian (typically KAlSi3O8) is
similar to the structures of an amorphous solid.

liquids 15
 Structure of Solids

Amorphous Solids
● atoms, ions, or molecules have little freedom to move
unlike in liquids
● do not have well-defined shapes of a crystal
● examples: obsidian (volcanic glass) and rubber

16
Remember

Crystalline solids consist of particles in

an organized form, while the particles of
amorphous solids are not formed in an
orderly manner.

17
What are the characteristic
properties of solids?

18
 Properties of Solids

● Solids exhibit characteristic properties that are

distinguishable from other states of matter
● dependent on the composition of solids

19
 Properties of Solids

Melting Point
● the temperature at which a solid loses it definite
shape and is converted to a liquid

Freezing Point
● the temperature at which liquid changes to solid

20
 Properties of Solids

Melting/Freezing Point
● at this temperature, solid and liquid forms of a
substance are in equilibrium with each other
● crystalline solids have a precise melting point
● amorphous solids melt over a wide range of
temperature
● a stronger interaction has a higher melting point

21
 Properties of Solids

Cristobalite melts at 1713 oC, while soda-lime glass, the most prevalent type
of glass, can melt between 550 OC and 1450 oC. 22
 Properties of Solids

Heat of Fusion
● the quantity of heat necessary to melt a solid

● solids with stronger intermolecular forces: higher

values of heat of fusion
● crystalline: heat of fusion is fixed and definite
● amorphous: no precise value 23
 Properties of Solids

Sublimation
● the process of direct passage from solid to vapor
phase, bypassing the liquid state

Deposition
● the reverse process of sublimation (vapor to solid
phase)

24
 Properties of Solids

Enthalpy of Sublimation
● the quantity of heat to convert
solid to vapor
Example:
● ice - solid with significant
sublimation pressure
Ice exhibits an appreciable
● temperature may not rise above sublimation pressure (4.58
mm Hg).
0 OC, but snow may disappear. 25
 Properties of Solids

Anisotropy
● the property of substances where the physical and
mechanical properties vary with different orientation
and molecular axes
● exhibited by crystalline solids

26
 Properties of Solids

Isotropy
● the property of substances where the physical and
mechanical properties are equal in all direction
● exhibited by amorphous solids

27
 Properties of Solids

Malleability
● the ability of solid to undergo compressive stress
without breaking it
● metals are highly malleable
○ shaped through forging, rolling, extrusion, and
indenting

28
 Properties of Solids

Ductility
● the ability of solid to undergo tensile stress
● can be measured
● describes the extent to which the solid can be
stretched without fracture

29
 Properties of Solids

● Most of the time, malleability

and ductility properties coexist
● Some examples: silver and gold

Gold is known to be the most

malleable and ductile.

30
 Properties of Solids

Electrical Conductivity
● the measurement of the ability of atoms, molecules,
or ions to transfer electrons from one to another
● metallic bond - easy for electrons to move
● ionic or covalent bond - hard to conduct electricity
● electrical insulators - solids that do not conduct
electricity

31
 Properties of Solids

Thermal Conductivity
● the measurement of the ability of atoms, molecules,
or ions to move and collide with its neighboring
particles
● metallic bond - good heat conductor due to
nondirectional nature of bonds
● ionic/covalent bonds - low thermal conductivity
because of rigidity between atoms 32
 Properties of Solids

Anisotropy in Crystalline Solids

● properties (malleability, ductility, thermal and
electrical conductivity) vary on the direction from
which the force is applied
● solids with nondirectional bonds: high malleability
and ductility, good conductors
● solids with rigid bonds: brittle, heat insulators

33
Why is copper used to make
electrical wires?

34
 Check Your Understanding

State if the following accounts for a crystalline or an

amorphous solid

1. precise heat of fusion

2. wide range of melting temperature
3. cotton candy
4. anisotropic
5. dry ice

35
 Let’s Sum It Up!

● In solids, particles are being held by strong covalent

and ionic intermolecular forces between neighboring
particles to keep them in a fixed position and rotate and
vibrate in place.
● Crystalline solids are arranged in a definite repeating
pattern held together by uniform, strong intermolecular
forces.
36
 Let’s Sum It Up!

● Amorphous solids do not have a repeating pattern

arrangement and the exact opposite of crystalline solid.
● Solids exhibit characteristic properties:
○ The melting point is the temperature at which the
solid loses its definite shape and converts to liquid
when heated.
○ The quantity of heat necessary to melt a solid is the
enthalpy of fusion.
37
Let’s Sum It Up!

○ Sublimation is the process of direct passage of

molecules from solid to the vapor phase, bypassing
the liquid state.
○ Malleability describes the ability of the solid to
undergo compressive stress without breaking it.
○ Ductility is the ability of a solid to undergo tensile
stress.
38
Let’s Sum It Up!

○ Electrical conductivity is the measurement of the

ability of atoms, molecules, or ions to transfer
electrons from one to another.
○ Thermal conductivity is when the temperature is
used as a measurement of the movement of atoms,
molecules, or ions.

39
Challenge Yourself

Why do some solids are better in

conducting heat?

40
Bibliography

Brown T.L. et al. 2012. Chemistry: The Central Science. Pearson Prentice Hall.Brown. Chemistry: The
Central Science. Prentice-Hall, 2005.

Bettelheim, Frederick A., et al. 2015. Introduction to General, Organic and Biochemistry. Boston:
Cengage Learning.

Ebbing, Darrell and Steven Gammon. 2016. General Chemistry. Boston: Cengage Learning.

Moore, John W, and Conrad L. Stanitski. 2015. Chemistry: The Molecular Science, 5th ed. USA: Cengage
Learning.

Petrucci, Ralph H. General Chemistry: Principles and Modern Applications. Toronto, Ont.: Pearson
Canada, 2011. Print.
41
GENERAL CHEMISTRY 2
 POWER COMPETENCY

Apply the properties of liquids and

solids to the nature of forces in
designing a simple investigation to
determine the effect on boiling point
or freezing point when a solid is
dissolved in water.
 LEARNING TARGETS

➢I can interpret the phase diagram

of water.
➢I can determine and explaining the
heating and cooling curve of a
substance.
Lesson 2.1

Energy Changes Accom

panying
Phase Changes

General Chemistry 2
Science, Technology, Engineering, and Mathematics
 In this lesson, you are going to learn about phase
changes and how they occur on a molecular level.
You will also learn to calculate the change of energy
that occurs whenever there is a transition between
phases of matter.

5
How can phase change
affect the molecular order
in matter?

6
 Learning Objectives
At the end of the lesson, you should be able to do the following:

● Define phase changes.

● Describe changes in molecular order and energy

changes during solid-liquid, liquid-vapor, and
solid-vapor phase transitions.

● Calculate heat changes in phase and temperature

changes.
7
 Phases and Phase Changes

● Phase is defined as a homogeneous state in which the

substance has a uniform composition and governed by
the same intermolecular forces throughout the
material.

● There are three fundamental phases of matter:

solid, liquid, and gas.

8
 Phases and Phase Changes

In solids, molecules are

tightly packed and can be
compared to people inside a
train during rush hour.

9
 Phases and Phase Changes

In liquids, there is more

space. This can be analogous
to people walking in a busy
street.

10
 Phases and Phase Changes

The gaseous state has very

large spaces between them.
This figure is an attempted
representation of this phase,
but in reality gas molecules
are much farther part.

11
Phase Changes

12
 Molecular Order in Phase Changes

Phase Changes
Phase changes are
accompanied by a change
that occurs as a result of
energy interactions and the
intermolecular forces in the
substance itself.

13
 Molecular Order in Phase Changes

In solids, molecules are well

ordered. The particles are
not free to move around.
When a solid transitions to a
liquid or gas, there is a
decrease in the order in the
material.

14
 Molecular Order in Phase Changes

The same is true for a liquid

that would turn to a gaseous
state. There is a decrease in
the order in the material
because gas molecules are
free to move around.

15
 Molecular Order in Phase Changes

The gaseous state has the

most disorder because of the
large spaces these molecules
can move around. Gas
transitioning to liquid or gas
would result in decrease in
order.

16
 Molecular Order in Phase Changes

In summary, there is a decrease in molecular order if

the transition is

more condensed less condensed

state state

17
 Molecular Order in Phase Changes

How about if you reverse the process? What is the

expected change in molecular order?

more condensed less condensed

state state

18
 Energy in Phase Changes

● When molecular order decreases, this results in

absorption of energy. It is an endothermic process.

● When molecular order increases, this results in

release of energy. This is an exothermic process

19
 Energy in Phase Changes

The amount of heat energy

transferred from the
surroundings to the
substance is a change in
enthalpy, represented by
the symbol ΔH.

20
 Energy in Phase changes

● For an endothermic process, the change in enthalpy is

always positive.

● When ice melts, heat energy from the environment

causes the water molecules to break free from the
intermolecular forces that hold it in the solid state.

21
 Energy in Phase changes

● For an exothermic process. The change in enthalpy will

always be negative.

● When water condenses on a leaf on a cold morning,

intermolecular forces become stronger, and the spaces
between molecules become closer. Heat is released to
the environment.
22
 Energy in Phase Changes

Energy in phase changes can be likened to breaking sticks.

23
 Energy in Phase Changes

You need energy to break a stick. You produced more,

smaller sticks when you break a single one.

24
 Energy in Phase Changes

You release energy when you reattach the pieces of the stick.
You form a single stick from multiple, smaller pieces.

25
Tips

To remember if a process is
endothermic or exothermic:
○ When heat is absorbed, it is
endothermic; heat enters the
system.
○ When heat is released, it is
exothermic; heat exits the
system.
26
Tips

● In endothermic processes, heat is

added (+). Therefore, ΔH is positive
(+).

● In exothermic processes, heat is

removed (-). Therefore, ΔH is
negative (-).

27
 Specific Heat

● The temperature of a substance is independent of the

amount of material, since it is a measure of the
average kinetic energy of each particle in the
substance.

● However, the amount of heat will depend on the

quantity of material.

28
 Specific Heat

● The amount of heat needed to raise the temperature

of a specific substance by one degree Celsius is
defined as specific heat.

● In the case of water, the specific heat for ice is 2.00

J/(g∙°C); for liquid it is 4.186 J/(g∙°C); and for water
vapor it is 2.11 J/(g∙°C).

29
 Enthalpy and Phase Changes

In endothermic processes like melting, evaporation and

sublimation, energy is supplied to overcome the
intermolecular attractive forces that hold the particles in
their present state.

30
 Enthalpy and Phase changes

● The temperature at which a

substance melts is called the
melting point.
● The amount of heat needed
to melt a substance is called
the heat of fusion or the
enthalpy of fusion
symbolized by ΔHfus.

31
 Enthalpy and Phase changes

● The reverse process is

freezing and the temperature
where this occurs is the
freezing point.
● The melting point and the
freezing point of any
substance is the same.

32
 Enthalpy and Phase changes

● The temperature at which

liquid turns to gas is called
the boiling point.
● The amount of heat needed
to complete the evaporation
process is called the heat of
vaporization or the enthalpy
of vaporization, symbolized
by ΔHvap.
33
 Calculations involving Energy Change

In calculations where no phase changes are involved and

under constant pressure, use the equation

where q is the amount of heat energy in joule, m is the

mass, c is the specific heat, ΔT is the change in
temperature.

34
 Calculations involving Energy Change

In calculations involving purely phase changes under

constant pressure, use the equation

where q is the amount of heat, n is the number of moles,

and ΔH is the molar enthalpy of the process.

35
 Let’s Practice!

How much energy is required to raise the

temperature of 70.0 g of water in the liquid state by
12 ºC, if the specific heat of water is 4.186 J/(g∙°C) ?

36
 Let’s Practice!

How much energy is required to raise the

temperature of 70.0 g of water in the liquid state by
12 ºC, if the specific heat of water is 4.186 J/(g∙°C) ?

The amount of heat required is equal to 3516 J.

37
 Let’s Practice!

At a temperature of 0 ºC, how many grams of ice can

be melted by 500 joules of heat energy if the molar
enthalpy of fusion of water is 6.01 kJ/mol?

38
Let’s Practice!

39
 Let’s Practice!

At a temperature of 0 ºC, how many grams of ice can

be melted by 500 joules of heat energy if the molar
enthalpy of fusion of water is 6.01 kJ/mol?

The amount of ice that can be melted is 1.5 g.

40
 Let’s Practice!

The amount of energy required to heat 2.00 kg of

water from 25 ºC to 75 ºC, is the same amount of
energy needed to melt a certain amount of ice
completely. How many moles of ice would that be?
The molar enthalpy of fusion of water is 6.01 kJ/mol
and the specific heat of water is 4.186 J/(g∙°C).

41
Let’s Practice!

42
 Let’s Practice!

The amount of energy required to heat 2.00 kg of

water from 25 ºC to 75 ºC, is the same amount of
energy needed to melt a certain amount of ice
completely. How many moles of ice would that be?
The molar enthalpy of fusion of water is 6.01 kJ/mol
and the specific heat of water is 4.186 J/(g∙°C).

Melting 69.65 moles of ice would require the same

amount of energy as that of heating 2.00 kg of water.

43
Try It!

Calculate the amount of energy needed

to raise the temperature of 300 grams
of water in the liquid state from 20 ºC to
27 ºC.

44
Try It!

Calculate the mass of liquid water at

100 ºC that can be totally converted to
steam by 3000 J if the molar enthalpy
for vaporization is 40.67 kJ/mol.

45
Try It!

About 4.00 g of hydrogen gas undergoes

combustion and produces water in its
liquid state. Eventually, the
temperature of the liquid is lowered to
273 K. How much energy must be
released to the surroundings so that
water will be totally frozen?

46
Which phase changes absorb
heat from its surroundings?

47
 Check Your Understanding

Predict if there is an increase or decrease in the

property in the description.

1. the molecular order in the process of deposition

1. the change in enthalpy in an exothermic process

1. the molecular order when water turns to steam

48
 Check Your Understanding

Solve the following problems. Show the complete

solution in your notebook.

1. How much energy is required to melt 72.0 g of ice at

0 ºC?

49
 Check Your Understanding

Solve the following problems. Show the complete

solution in your notebook.

2. Ethylene glycol is a component of antifreeze

formulations. The specific heat of this compound is 2.43
J/g-°C. How much heat is required to raise the
temperature of 7.00 g ethylene glycol from 9 ºC to 18 ºC?

50
 Check Your Understanding

Solve the following problems. Show the complete

solution in your notebook.

3. One kilogram of water has maintained its liquid state as

its temperature increased from 273.15 K to 373.15 K.
How much energy is needed for this to be attained?

51
 Let’s Sum It Up!

● A phase is defined as a homogeneous state in

which the substance has a uniform composition
and governed by the same intermolecular forces
throughout the material.
● Phase changes are transitions that occur
between the fundamental phases of matter: solid,
liquid and gas.

52
 Let’s Sum It Up!

● Molecular order decreases in the following phase

changes: melting (solid to liquid), sublimation
(solid to gas), and evaporation (liquid to gas).
● Molecular order increases in the following phase
changes: freezing (liquid to solid), condensation
(gas to liquid), and deposition (gas to solid).

53
 Let’s Sum It Up!

● When molecular order decreases, this results in

absorption of energy. It is an endothermic
process.
● When molecular order increases, this results in
the release of energy. This is an exothermic
process.
● The amount of heat needed to raise the
temperature by 1 °C is defined as specific heat.
54
 Let’s Sum It Up!

● The temperature at which a substance melts is

called the melting point.
● The reverse process is freezing and the
temperature where this occurs is the freezing
point.
● The melting point and the freezing point of any
substance is the same.The temperature at which
liquid turns to gas is called the boiling point.
55
 Let’s Sum It Up!

● Under constant pressure, the amount of heat

needed to melt a substance is called the heat of
fusion or the enthalpy of fusion (ΔHfus) . The
amount of heat needed to complete the phase
change of a substance is called the heat of
vaporization or the enthalpy of vaporization,
(ΔHvap).

56
 Key Formulas

Concept Formula Description

Use this formula

Energy change
when the material
without phase
undergoes
changes where:
temperature
● m is mass (in g)
changes but not
● c is specific heat (in
phase changes.
J/(g✕0C))
● ΔT is the change in
temperature

57
 Key Formulas

Concept Formula Description

Use this formula

Energy change
when the material
during phase
undergoes phase
changes where:
changes but its
● q is the amount of heat,
temperature does
● n is the number of moles,
not change.
● ΔH is the molar enthalpy of
the specific process

58
Challenge Yourself

Assuming that steam behaves as an

ideal gas at 100 ºC, what is the
amount of heat that needs to be
released to the surroundings for the
gas to be totally converted to liquid
form? The gas occupies a volume of
16.0 L at 1.00 atm.
59
Bibliography
Bettelheim, Frederick A. ,William H. Brown, Mary K. Campbell, and Shawn O. Farrell. Introduction to
General, Organic, and Biochemistry. Boston, MA: Cengage Learning, 2016. Print.

Brown, Theodore E. Hill, James C.,, H. Eugene LeMay, Bruce Edward. Bursten, Catherine J. Murphy,
Patrick M. Woodward, and Matthew Stoltzfus. Chemistry: The Central Science, 13th Edition. NJ:
Pearson, 2015. Print.

Chang, Raymond, and Kenneth A. Goldsby. General Chemistry: The Essential Concepts. New York:
McGraw-Hill, 2014. Print.

Petrucci, Ralph H, F G. Herring, Jeffry Madura, and Carey Bissonnette. General Chemistry: Principles
and Modern Applications. 2016. Print.

Silberberg, Martin S. Principles of General Chemistry. Boston: McGraw-Hill Higher Education, 2007.
Print. 60
General Chemistry 2
Science, Technology, Engineering, and Mathematics
Try It!

Calculate the amount of energy needed

to raise the temperature of 300 grams
of water in the liquid state from 20 ºC to
27 ºC.

2
Lesson 2.3

Heating and Cooling

Curves

General Chemistry 2
Science, Technology, Engineering, and Mathematics
When you take the ice out of the refrigerator, it starts to
melt.

4
Water can be further
heated in a kettle to
boil, forming water
vapor rushing out
and mixing with the
air.
5
As water changes
phase, hydrogen
bonds are broken,
along with weaker
intermolecular forces
of attraction.
6
 But how much heat is needed to transform ice into water
vapor? Is there a way to represent these heat changes and
interpret the energetics of phase changes more
systematically?

In this lesson, you will learn how to construct and interpret

heating and cooling curves and calculate the associated
heat changes at each step.

7
How do you interpret
heating and cooling curves?

8
 Energetics of Phase Changes: A Recall

Changes in phases are

accompanied by energy
changes → molar
enthalpies

The phase changes and their energy

9
 Energetics of Phase Changes: A Recall

Phase changes only occur at specific temperatures and

pressures
○ freezing of water: at 0 OC and 1 atm
○ evaporation of water: at 100 OC and 1 atm

10
 Energetics of Phase Changes: A Recall

For phase changes at constant P,

where q = heat absorbed or released (J);

n = amount of substance
(moles), and
ΔH = molar heat associated with phase change
(J/mol).
11
 Energetics of Phase Changes: A Recall

For temperature changes,

where q = heat absorbed or released, J;

m = mass, g;
c = specific heat, J/(g x OC-1); and
ΔT = change in temperature.
12
 Heating and Cooling Curves

● Transformations of substances in real life involve

multiple phase changes and temperature changes.
● There are instances when a substance in its solid state
is converted into its gaseous state.
● In between phase changes, additional heat is needed
to satisfy temperature changes.

13
 Heating and Cooling Curves

The amount of heat in

complex phase
transformations can be
tracked in heating and
cooling curves.

14
 Heating and Cooling Curves

In these curves, the

temperature (y-axis) is
plotted against total heat
changes (x-axis).

15
 Heating and Cooling Curves

This heating curve shows

the heat associated with
the transformation of 1
mol of ice initially set at –
25 ºC and 1 atm to steam
at 125 ºC and 1 atm.

16
 Heating and Cooling Curves

How do we calculate the

amount of heat associated
with the entire process?

17
How can one calculate the
heat associated with
consecutive phase
transformations?

18
 Heating and Cooling Curves

The amount of heat can be

calculated by adding all
heat absorbed from each
phase and temperature
change.

19
 Heating and Cooling Curves

The process can be divided

as represented by the
segments.

20
 Heating and Cooling Curves

Blue segments represent

temperature changes
without phase change,
while red segments
represent phase changes
without temperature
change.

21
 Heating and Cooling Curves

Segment AB
Segment AB represents a
change in temperature of
ice from –25 ºC to 0 ºC.

22
 Heating and Cooling Curves

Segment BC
The ice cube starts to melt
from point B and ends with
point C.

23
 Heating and Cooling Curves

Segment CD
Segment CD represents a
change in temperature of
water from 0 ºC to 100 ºC.

24
 Heating and Cooling Curves

Segment DE
Water starts to boil from
point D and ends with
point E.

25
 Heating and Cooling Curves

Segment EF
Segment EF represents a
change in temperature of
steam from 100 ºC to 125
ºC.

26
 Heating and Cooling Curves

Total Heat
Total heat is additive, from
point A to F.

What is the total heat of the

phase change of ice to steam
from point A to F?

27
 Heating and Cooling Curves

Constructing Heating Curves

● To create a heating curve, one must plot temperature
(y-axis) against heat (x-axis).
● Processes, where heat is absorbed to increase
temperature, are represented by lines slanted
upright, while horizontal lines represent phase
changes at a constant temperature.

28
 Heating and Cooling Curves

Constructing Cooling Curves

● Cooling curves are the exact reverse of heating curves.
● Processes that require the release of heat are
represented by lines sloping downwards, while
horizontal lines still represent phase changes.
● A cooling curve can be produced from the heating
curve by reading the plot from the top right to bottom
left.

29
Remember

In heating curves, processes with

temperature changes without phase change
are represented by a line slanting upward,
while horizontal lines represent processes
with phase changes without temperature
change.

30
How can you describe the
cooling curve when steam at
150 ºC is transformed into
–15 ºC?

31
 Let’s Practice!

Calculate the heat required to transform 5 grams of

ice at -10 ºC to liquid water at 80 ºC. Use the following
specific heats: cice = 2.108 J/(g ✕ ºC), cwater = 4.186 J/(g ✕
ºC), cvapor = 1.97 J/(g ✕ ºC). Water has a heat of fusion
(ΔHfus) of 6 000 J/mol, and a heat of vaporization
(ΔHvap) of 40 700 J/mol.

32
 Let’s Practice!

Calculate the heat required to transform 5 grams of

ice at -10 ºC to liquid water at 80 ºC. Use the following
specific heats: cice = 2.108 J/(g ✕ ºC), cwater = 4.186 J/(g ✕
ºC), cvapor = 1.97 J/(g ✕ ºC). Water has a heat of fusion
(ΔHfus) of 6 000 J/mol, and a heat of vaporization
(ΔHvap) of 40 700 J/mol.

The total heat required to transform 5 grams of ice at

-10 ºC to liquid water at 80 ºC is 3446.5 J.
33
Try It!

Calculate the heat required to transform

15 grams of ice at -5 ºC to liquid water at
90 ºC. Use the following specific heats: cice
= 2.108 J/(g ✕ ºC), cwater = 4.186 J/(g ✕ ºC),
cvapor = 1.97 J/(g ✕ ºC). Water has a heat of
fusion (ΔHfus) of 6000 J/mol, and a heat of
vaporization (ΔHvap) of 40 700 J/mol.

34
 Let’s Practice!

Calculate the heat required to transform a gram of

ice at -50 ºC to steam at 120 ºC. Use the following
specific heats: cice = 2.108 J/(g ✕ ºC), cwater = 4.186 J/(g ✕
ºC), cvapor = 1.97 J/(g ✕ ºC). Water has a heat of fusion
(ΔHfus) of 6 000 J/mol, and a heat of vaporization
(ΔHvap) of 40 700 J/mol.

35
 Let’s Practice!

Calculate the heat required to transform a gram of

ice at -50 ºC to steam at 120 ºC. Use the following
specific heats: cice = 2.108 J/(g ✕ ºC), cwater = 4.186 J/(g ✕
ºC), cvapor = 1.97 J/(g ✕ ºC). Water has a heat of fusion
(ΔHfus) of 6 000 J/mol, and a heat of vaporization
(ΔHvap) of 40 700 J/mol.

The total heat required to transform a gram of ice at

-50 ºC to liquid water at 120 ºC is 3157.8 J.
36
Try It!

Calculate the heat required to transform

10 grams of ice at -200 ºC to steam at 200
ºC. Use the following specific heats: cice =
2.108 J/(g ✕ ºC), cwater = 4.186 J/(g ✕ ºC),
cvapor = 1.97 J/(g ✕ ºC). Water has a heat of
fusion (ΔHfus) of 6 000 J/mol, and a heat of
vaporization (ΔHvap) of 40 700 J/mol.

37
 Let’s Practice!

Calculate the heat required to transform a mole of

ice that underwent the change from A to F described
in the heating curve next slide. Use the following
specific heats: cice = 2.108 J/(g ✕ ºC), cwater = 4.186 J/(g ✕
ºC), cvapor = 1.97 J/(g ✕ ºC). Water has a heat of fusion
(ΔHfus) of 6 000 J/mol, and a heat of vaporization
(ΔHvap) of 40 700 J/mol.

38
Let’s Practice!

39
 Let’s Practice!

Calculate the heat

required to transform a
mole of ice that
underwent the change
from A to F described in
the heating curve.

The total heat

required is 56069.9 J.
40
 Check Your Understanding

Identify if the following statements are true or false.

1. When ice melts, the temperature remains at 0 ºC.

2. In a heating curve, the x-axis is temperature, in ºC, and
the y-axis is heat added, in J.
3. Above 100 ºC, the relevant specific heat to be used is
that of steam.

41
 Check Your Understanding

Use the cooling curve for a

hypothetical substance Z shown
to answer the questions that
follow.
1. What is the boiling point of the
hypothetical substance Z?
2. Which segment represents
condensation?
3. How long does it take for
gaseous Z to completely liquefy? 42
 Let’s Sum It Up!

● Heating and cooling curves are used to track heat

changes associated with complex phase
transformations.
● A heating curve is produced when the temperature
changes (y-axis) are plotted against heat changes (x-
axis).
● A cooling curve can be constructed from a heating curve
by reading the latter from top right to bottom left.

43
 Key Formulas

Concept Formula Description

Use this formula

Energy change
when the material
without phase
undergoes
changes where
temperature
● m is mass (in g);
changes but not
● c is specific heat (in
phase changes.
J/(g✕0C)), and
● ΔT is the change in
temperature.

44
 Key Formulas

Concept Formula Description

Use this formula

Energy change
when the material
during phase
undergoes phase
changes where
changes but its
● q is the amount of heat,
temperature does
● n is the number of moles,
not change.
and
● ΔH is the molar enthalpy of
the specific process.

45
Challenge Yourself

In a heating curve, temperature (in

ºC) is plotted against heat changes (in
J). Changes in temperatures for a
specific state of a substance are
represented by slanted lines. What
does the slope of these diagonal lines
represent? Briefly explain its
significance.
46
Bibliography

Chang, Raymond, and Kenneth A. Goldsby. General Chemistry: The Essential Concepts. New York:
McGraw-Hill, 2014.

Handwerker, Mark J. Science Essentials. San Francisco, CA.: Jossey-Bass, 2005.

Hawe, Alan, Dan Davies, Kendra McMahon, Lee Towler, Chris Collier, and Tonie Scott. Science 5–11: A
Guide for Teachers. 2nd ed. New York, NY: David Fulton Publishers, 2009.

Petrucci, Ralph H. General Chemistry: Principles and Modern Applications. Toronto, Ont.: Pearson
Canada, 2011.

Silberberg, Martin S. Principles of General Chemistry. New York: McGraw-Hill, 2013.

47
GENERAL CHEMISTRY 2
 POWER COMPETENCY

Apply the properties of solutions, solubility, and the

stoichiometry of reactions in solutions in calculating
stoichiometric calculations for reactions in solution and
to apply the energy changes in chemical reaction in
calculating enthalpy change.
 LEARNING TARGETS

➢I can use different ways of expressing

concentration of solutions: percent by mass,
mole fraction, molarity, molality, percent by
volume, percent by mass, ppm,
STEM_GC11PP-IIId-f-1.
 Different Ways Of
Expressing Concentration
Of Solutions
Percentage by Mass,
Volume, and
Mass/Volume

General Chemistry 2
Science, Technology, Engineering, and Mathematics
How much pizza can you eat in an hour? A standard pizza is
usually divided into eight parts. Most customers can eat two
out of eight pizzas, or one-fourth of the whole pizza.

6
How is the concentration of
solutions expressed?

7
 Concentration of Solution

● The amount of solute present in a solution is described

by its concentration.
● Concentration is an intensive property as it
represents the proportion of a substance with respect
to the whole solution.
● Concentration often expresses the ratio of the amount
of the solute to the amount of the solution, but
sometimes it is also expressed with respect to the
amount of the solvent.
8
 Concentration of Solution

Concentration Term Unit

Percentage by mass, % m/m

Percentage by volume, % v/v

Percentage by mass/volume,
% m/v

9
Concentration of Solution

Concentration Term Unit

Molarity, M

Molality, m

10
 Concentration of Solution

Concentration Term Unit

Mole Fraction, 𝛘

Parts per million, ppm

Parts per billion, ppb

11
 Percentage by Mass, % m/m

● Percentage by mass (% m/m), also known as mass

percent or weight percent (% w/w), is defined as the
mass of solute per mass of the solution.
● Mathematically, it is expressed as the mass fraction
times 100.
● The mass of solute and solvent should be expressed in
the same units.

12
 Percentage by Volume, % v/v

● Percentage by volume (% v/v), also known as volume

percent is defined as the volume of solute per volume
of the solution.
● Mathematically, it is expressed as the volume fraction
times 100.
● The volume of solute and solvent should be expressed
in the same units.

13
 Percentage by Mass per Volume, % m/v

● Percentage by mass per volume (% m/v), also known

as the percentage by weight per volume (% w/v), is
defined as the mass of solute per volume of the
solution.
● The mass of the solute is usually expressed in grams
while the volume of the solution is usually expressed in
milliliters.

14
Remember

Heat capacity is an extensive

property while specific heat is an
intensive property.

15
In what instances does the unit
% m/m more appropriate to use
than % v/v?

16
 Let’s Practice!

Calculate the concentration of a solution in

percentage by mass prepared by dissolving 12 g of
NaCl to produce 50 g of the solution.

17
Let’s Practice!

18
 Let’s Practice!

Calculate the concentration of ethanol, in percentage

per volume, in a solution prepared by dissolving 11
mL of ethanol with 10 mL of water.

19
Let’s Practice!

20
 Let’s Practice!

Calculate the mass of sodium acetate (NaCH3COO), in

grams, needed to be dissolved in water to produce a
150 mL solution with a concentration of 20 % m/v.

21
Let’s Practice!

22
Try It!

Calculate the concentration of a

solution in percentage by volume
prepared by dissolving 10 mL of acetone
to produce 100 mL of the solution.

23
Try It!

Calculate the concentration of sodium

hydroxide (NaOH), in percentage per
mass, in a solution prepared by
dissolving 22 g of NaOH with 53 g of
water.

24
Try It!

Calculate the mass of potassium nitrate

(KNO3), in grams, needed to be dissolved
in water to produce a 1000 mL solution
with a concentration of 15 % m/v.

25
Try It!

26
 Molarity, M

● The concentration of aqueous solutions are usually

expressed in molarity (M), defined as the number of
moles of the solute dissolved in one liter of the
solution.
● Molarity has a unit of M (read as molar), which can be
expressed as mol/L.

27
 Molality, m

● Molality, m, does not contain any volume term and is

defined as the number of moles of solute dissolved in
one kilogram of the solvent.
● Molality has a unit of m (read as molal), which can be
expressed as mol/kg.
● Mass does not change with temperature; hence,
molality is independent of temperature changes.

28
Why is it better to use molality
in laboratory experiments than
molarity?

29
Tips

Molarity (M) and molality (m) have

the same term in the numerator.
Remember that the denominator
term in molarity is the volume of
solution in liters, while that in
molality is the mass of solvent in
kilograms.
30
 Let’s Practice!

Calculate the molarity of a 2 L solution prepared by

dissolving 15 moles of glucose in enough water.

31
Let’s Practice!

32
 Let’s Practice!

Calculate the amount of sodium phosphate (Na3PO4),

in moles, to be added to 30 kg water to form a 2.5
molal solution of aqueous Na3PO4.

33
Let’s Practice!

34
Try It!

Calculate the molality of a solution

prepared by dissolving 8 moles of rock
salt in 20 kg of water.

35
Try It!
Calculate the molality of a solution
prepared by dissolving 8 moles of rock
salt in 20 kg of water.

36
Try It!

Calculate the amount of potassium

acetate (KCH3COO), in moles, to be
added to enough water to produce 0.250
L of a 1.5 molar solution of aqueous
KCH3COO.

37
Lesson 4.4

Mole Fraction, PPM, and

PPB

General Chemistry 2
Science, Technology, Engineering, and Mathematics
 Mole Fraction

Mole Fraction
● denoted by the Greek letter chi (𝜒)
● ratio between the moles of a substance and the total
moles present in the solution
● unitless

39
 Let’s Practice!

What is the mole fraction of glucose in a solution

that contains 0.50 moles of glucose and 2.50 moles of
water?

40
Let’s Practice!

41
 Let’s Practice!

What is the mole fraction of ethanol in a solution

that contains 25 g ethanol and 75 g water?

42
Let’s Practice!

43
Try It!

What is the mole fraction of NaCl in a

solution that contains 1.25 moles of
NaCl and 3.75 moles of water?

44
Try It!
What is the mole fraction of NaCl in a
solution that contains 1.25 moles of
NaCl and 3.75 moles of water?

45
Try It!

What is the mole fraction of KNO3 in a

solution that contains 10 g KNO3 and 50
g water?

46
Try It!
What is the mole fraction of KNO3 in a solution that
contains 10 g KNO3 and 50 g water?

47
 Parts per Million

Parts Per Million

● refers to “a portion out of one million”

48
Tips
A ppm may also be expressed in other ways.
For instance, 1 mg/kg is the same as 1 ppm.
This is because a milligram is equivalent to
0.001 g while a kilogram is equivalent to
1000 g.

Other ways of expressing 1 ppm are as

follows: 1 𝜇g/g, 1 𝜇g/mL, 1 mg/L
49
 Let’s Practice!

What is the concentration of a solution, in parts per

million, if 0.01 gram of NaCl is dissolved in 1000
grams of solution?

50
Let’s Practice!

51
 Let’s Practice!

What is the concentration of a solution, in parts per

million, if 25 micrograms of RbOH is dissolved to
make 750 milligrams of solution?

52
Let’s Practice!

53
Try It!

What is the concentration of a solution,

in parts per million, if 0.00700 grams of
Na2SO4 is dissolved in 325 grams of
solution?

54
Try It!

What is the concentration of a solution,

in parts per million, if 18 milligrams of
sucrose is dissolved to produce half a
kilogram of solution?

55
 Parts per Billion

Parts Per Billion

● refers to “a portion out of one billion”

56
Tips

Another way of expressing 1 ppb is 1 𝜇g/kg

or 1 𝜇g/L.

57
 Let’s Practice!

What is the concentration of a solution, in parts per

billion, if 2.5 ✕ 10-4 g of KCl is dissolved to make 1000
grams of solution?

58
Let’s Practice!

59
 Let’s Practice!

What is the concentration of a solution, in parts per

billion, if 75 𝜇g of AgNO3 is dissolved in 12 kg of
solution?

60
Let’s Practice!

61
Try It!

What is the concentration of a solution,

in parts per billion, if 0.00012 g of
Ba(C2H3O2)2 is dissolved to make 850
grams of solution?

62
Try It!

What is the concentration of a solution,

in parts per billion, if 218 𝜇g of Na2CO3 is
dissolved in 25.0 kg of solution?

63