Course Specification
Term 3 (AY 21-22)
Course Name:
Chemistry
Course Code:
CHM51
Grade:
10
Aim:
The Grade 10 Chemistry course provides students with a college-level foundation to
support future advanced coursework in Chemistry. The course will give students the
opportunity to cultivate their understanding of Chemistry through inquiry-based
investigations as they explore topics as Structure & Properties of Matter, Chemical
Bonding & Reaction, Matter, Energy & Equilibrium, Organic & Nuclear Chemistry.
Course Outline:
Topic
Number of Periods
Term 3
Topic 3: Matter, Energy, and Equilibrium
• Subtopic 3.1 – States of Matter
(Unit 3 – Module 11: Lesson 1 and 2)
• Subtopic 3.2 – Gases
(Unit 3 – Module 12: Lesson 1, 2 and 3)
• Subtopic 3.3 – Mixtures & Solutions
(Unit 3 – Module 13: Lesson 1, 2, and 3)
Color Code
Exam Coverage for Term 3 Material
Content Area
WHITE
Core – Short*
GREEN
Core – Long**
32
8
12
12
Percentage in Exam
85%
YELLOW
Extended
15%
*Core – Short: Content that takes less time and has lesser depth. Example: terms, facts, ……
**Core – Long: Content that takes more time and has greater depth. Example: main ideas,
procedures, ………
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Topic 3: States of Matter
3.1 – Kinetic Molecular Theory & The Three States of Matter
Inspire Chemistry – Module 11 – Lesson 1: Gases (4 Periods)
3.1.1 Describe the kinetic molecular theory of particles
The kinetic molecular theory describes the behavior of matter in terms of particles in motion
3.1.2
List the five assumptions of the Kinetic Molecular Theory
The five assumptions are:
1) Gases consist of large numbers of tiny particles that are far apart relative to their size
2) Collisions between gas particles and between particles and container walls are elastic collisions
3) Gas particles are in continuous, rapid, random motion
4) There are no forces of attraction between gas particles
5) The temperature of a gas depends on the average kinetic energy of the particles of the gas
3.1.3
Differentiate, using particulate diagram, among the different states of matter (solid, liquid, gas,
and plasma)
o
-
Solid State
Shape and volume: Definite shape and volume
Movement of particles: Vibrate in their fixed positions
Arrangement of particles: Regular repeating patterns
Strength of forces between the particles: Strong
Density and compressibility: Solids are highly dense and cannot be compressed
o
-
Liquid State
Shape and volume: Indefinite shape (take the shape of the container) but definite volume
Movement of particles: Can move freely, particles slide past each other
Arrangement of particles: Irregular arrangement
Strength of forces between the particles: Intermediate
Density and compressibility: Slightly compressible with definite density
o
-
Gaseous State
Shape and volume: Indefinite shape (take the shape of the container) and indefinite volume
Movement of particles: Can move freely in all directions
Arrangement of particles: Irregular arrangement
Strength of forces between the particles: Very weak with large spaces between the particles
Density and compressibility: Highly compressible with low density
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o Plasma
- It has some of the properties of gas but is a good conductor of electricity and is affected by
magnetic fields.
- Plasma exists inside stars where the temperature reaches several thousand degrees Celsius.
- Particles of plasma are charged
- No definite shape or volume
- It is formed when a gas is heated to extremely high temperatures where the atoms ionize, and
electrons and ions are formed.
3.1.4
Use the kinetic molecular theory to explain the physical properties of gases with respect to
expansion, fluidity, low density, compressibility, and diffusion
-
3.1.5
Expansion (Assumptions 3 and 4) - Gas particles move rapidly in all directions without
significant attraction between them
Fluidity (Assumption 4) - Attractive forces between gas particles are too weak to hold the
particles in a rigid structure
Low density and high compressibility (Assumption 1) - Gas particles are farther apart than
they are in any other state
Diffusion (Assumption 3) - Gas particles are in continuous and random motion
Compare between diffusion and effusion
Both involve the movement of gas particles.
Diffusion is the movement of one substance through another
Effusion is when a substance under pressure escapes through a tiny opening
Example of Diffusion
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3.1.6
Identify the factor that determines rates at which different molecules undergo diffusion and
effusion at a given temperature
Effusion and diffusion rates are inversely proportional to the molecular mass of a gas
3.1.7
State Graham's law of effusion and its mathematical relation
It states that rate of effusion of a gas is inversely proportional to the square root of its molar mass
𝑹𝒂𝒕𝒆𝟏
𝑴𝟐
= √
𝑹𝒂𝒕𝒆𝟐
𝑴𝟏
3.1.8
Use the mathematical formula of Graham’s law of effusion to compare the relative rates of
effusion of different gases and their molar masses
Example:
Which of the following gas particles will effuse the fastest at the same conditions of temperature
and pressure?
a) Hg
b) Cl2
c) SO2
d) N2O
N2O because it has the smallest molar mass
3.1.9
Use the mathematical formula of Graham’s law of effusion to calculate the relative rates of
effusion of different gases and their molar masses
Example:
How much faster does helium, He, effuse than nitrogen, N2, at the same temperature?
MN
RateHe
28
= √ 2 = √ = 2.7
RateN2
MHe
4
He travels 2.7 times faster than N2
3.1.10 Identify the most commonly used units for pressure and convert among units of pressure
atm, mmHg, cmHg, torr and kPa
1 atm = 760 mmHg = 76 cmHg = 760 torr = 101.3 kPa
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3.1.11 Describe the importance of barometer
An instrument used to measure the atmospheric pressure
In a barometer, when the atmospheric pressure rises, the height of the mercury column rises as
well. Similarly, when atmospheric pressure falls, the height of the column falls.
3.1.12 Use open-end and closed-end manometer to calculate the pressure of a gas
Closed-end
manometer
Open-end manometer
used to measure the
pressure of the gas
directly
Used to measure the pressure of a gas indirectly
Case 1
Case 2
Case 3
Pgas = h
Pgas < Patm
Pgas > Patm
Pgas = Patm
Pgas= Patm – h
Pgas= Patm + h
h = zero
Example:
Example:
Example:
Example:
If h = 76 cm
If h = 45 and
Patm = 760 mmHg
If h = 445 and
Patm = 760 mmHg
If Patm = 760 mmHg
Pgas = 76 cm Hg
Pgas = 760 – 45
= 715 mmHg
Pgas = 760 + 65
= 825 mmHg
Pgas = Patm
= 760 mmHg
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3.1.13 Define partial pressure of a gas and identify the factors that affect it and those that do not
Partial pressure is the portion of the total pressure contributed by a single gas
3.1.14 Identify the factors that affect the partial pressure and those that do not
Partial pressure depends on number of moles of gas, size of the container and temperature of the
mixture
Partial pressure does not depend on the identity of the gas
3.1.15 State Dalton’s law of partial pressures and its mathematical relation
The total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the
mixture
PTotal = P1 + P2 + P3 + ……..
3.1.16 Use the mathematical formula of Dalton’s law of partial pressures to calculate partial pressures
and total pressure of a mixture of gases
Example:
A gas mixture contains oxygen, O2, nitrogen, N2, and helium, He.
If the partial pressures of the gases are 0.20 atm, 0.46 atm and 0.27 atm respectively, what is the
total pressure of the gas mixture?
PTotal = PO2 + PN2 + PHe = 0.20 + 0.46 + 0.27 = 0.93 atm
Example:
A gas mixture containing oxygen, O2, nitrogen, N2, and carbon dioxide, CO2, has a total pressure
of 3.29 atm. If the partial pressure of O2 is 0.66 atm and that of N2 is 2.30 atm, what is the partial
pressure of CO2?
PTotal = PO2 + PN2 + PCO2
3.29 = 0.66 + 2.30 + PCO2
PCO2 = 0.33 atm
Example:
The diagram below shows a mixture of Argon, Ar, and ammonia, NH3, gases at 25℃. If the total
pressure is 759.0 mmHg, calculate the partial pressure of Ar.
PAr = XAr × Ptotal =
10
× 759.0 = 506.0 mmHg
15
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Inspire Chemistry – Module 11 – Lesson 2: Forces of Attraction (4 Periods)
3.1.17 Compare between intramolecular and intermolecular forces
Intramolecular forces of attraction
Intermolecular forces of attraction
Are the forces that occur within a molecule
and hold atoms together
Are those that occur between the molecules
Always represented by a solid line " − "
Always represented by a dashed line “-----”
3.1.18 Compare and contrast intramolecular forces (Ionic, covalent, and metallic bonding) with respect
to structure, type of atoms involved, strength and physical properties of the compound formed
o Ionic bonds
- Structure consists of giant lattice of cations and anions held together by strong electrostatic
forces of attraction
- Occurs between metal and non-metal atoms. Electrons are transferred between atoms
- Properties: Solid at room temperature, high melting point and boiling point, does not
conduct electricity when solid but conducts when molten or aqueous
o Metallic Bonding
- Structure consisting of giant metallic lattice of regularly arranged cations surrounded by a
sea of delocalized electrons
- Number of delocalized electrons and size of cations determine the strength of metallic
bond (bigger cations and more electrons causes metallic bond to be stronger)
- Properties: Solid at room temperature, high melting point and boiling point, good
conductor of electricity in all states, malleable and ductile
o Covalent Bonding
- A chemical bond formed by the sharing of electrons between non-metal atoms
- Properties: Generally soft solids, liquids, or gases. Insoluble in water, does not conduct
electricity at all states, low melting and boiling points
Ionic bonding is stronger than metallic bonding that are stronger than covalent bonding
Ionic bonding > Metallic bonding > Covalent bonding
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3.1.19 Compare and contrast the intermolecular forces (dispersion forces, dipole-dipole forces, and
hydrogen bond) with respect to type of molecules involved and strength
o London dispersion forces (Simply Dispersion forces)
- It results from the constant motion of electrons and the creation of instantaneous dipoles
- They are relatively weak forces of attraction that exist between non-polar molecules and noble gas
atoms.
- Larger non-polar molecules tend to have stronger London dispersion forces
For non-polar molecules, the further you go down the group, the stronger the London dispersion
forces
Ex. Going down the group of halogens, fluorine and chlorine are gases, bromine is a liquid and
iodine is a solid
o Dipole-dipole forces
- It is the force between oppositely charged ends of two polar molecules
- The more polar the molecules, the stronger the dipole-dipole force
o Hydrogen bond
- It is a strong dipole-dipole force of attraction
- It occurs between hydrogen atom that is covalently bonded to nitrogen, oxygen or fluorine and a
pair of non-bonding electrons (lone pair) on the nitrogen, oxygen or fluorine atom of a neighboring
molecule
- Strength of bonds H−F > H−O > H−N
(Fluorine is the most electronegative)
Hydrogen bond is stronger than dipole-dipole forces that is stronger than dispersion forces
Hydrogen bond > dipole-dipole forces > dispersion forces
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3.1.20 Explain why hydrogen bonds are stronger than most dipole-dipole forces
A hydrogen bond involves a large difference in electronegativity between the hydrogen atom and
the atom it is attached to (F, N or O), making the bond extremely polar
3.1.21 Compare between a temporary dipole and a permanent dipole
A temporary dipole forms when one molecule is close to another molecule, and the electrons repel
each other creating a greater electron density in one part of the molecule. Permanent dipoles are
found in polar molecules in which some regions of the molecule are always partially negative, and
others are partially positive
3.1.22 Explain why dispersion forces are weaker than dipole-dipole forces
Dispersion forces are between temporary dipoles while dipole-dipole forces are between
permanent dipoles
3.1.23 Explain, in terms of intermolecular forces, why different substances exist as solid, liquid or gas at
the same conditions of temperature and pressure
Example:
Explain why oxygen is a gas at room temperature while water is a liquid at the same conditions
Oxygen contains London dispersion forces between its molecules while water contains London
dispersion forces, dipole-dipole forces, and hydrogen bonds between its molecules. Hydrogen
bonds are stronger than dispersion forces thus water is a liquid while oxygen is a gas
3.2
– The Nature of Gases
Inspire Chemistry – Module 12 – Lesson 1: The Gas Laws (4 Periods)
3.2.1
Describe the relationship among temperature of gas, average kinetic energy (KE) of gas particles
and speed of gas particles
As temperature increases, average KE energy increases and speed of particles increases, but at
constant temperature average KE stays the same
3.2.2
Identify the most commonly used units for temperature while converting among units of
temperature
The most common temperature units are Kelvin and Celsius
𝐊 = ℃ + 𝟐𝟕𝟑
3.2.3
Identify the most commonly used units for volume while converting among units of volume
L and mL where 1 L = 1000 mL
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3.2.4
State Boyle's Law with its mathematical relation and graphical representation
Boyle’s Law states that the volume of a fixed amount of a gas held at a constant temperature
varies inversely with pressure
𝐏𝟏 𝐕𝟏 = 𝐏𝟐 𝐕𝟐 (𝐀𝐭 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐧 𝐚𝐧𝐝 𝐓)
3.2.5
Use the mathematical formula of Boyle’s law to calculate volume-pressure changes at constant
temperature
Example:
A balloon contains 30.0 L of helium gas at 1.00 atm. Calculate the volume of the helium when the
balloon rises to an altitude where the pressure is only 0.250 atm? (Assume that the temperature
remains constant)
P1 V1 = P2 V2
1.00 × 30.0 = 0.250 × V2
V2 = 120. L
Rationalize your result:
A decrease in pressure at constant temperature must correspond to a proportional increase in
volume.
3.2.6
Use pressure-volume graph to determine the volume when the pressure is given and vice versa
Lab 16 – Conduct an experiment to investigate the relationship between the pressure and volume of a gas
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3.2.7
State Charles’s Law, write its mathematical relation and sketch its graphical representation
Charles’s Law states that the volume of a given amount of a gases directly proportional to its
Kelvin Temperature at constant pressure
𝐕𝟏
𝐕𝟐
=
𝐓𝟏
𝐓𝟐
3.2.8
(𝐀𝐭 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐧 𝐚𝐧𝐝 𝐏)
Use the mathematical formula of Charles’s law to calculate volume-temperature changes at
constant pressure
Example:
A balloon inflated in a room at 24oC has a volume of 4.00 L. The balloon is then heated to a
temperature of 58oC. What is the new volume if the pressure remains constant?
V1
T1
=
V2
𝟒.𝟎𝟎
T2
𝟐𝟒+𝟐𝟕𝟑
=
𝐕𝟐
𝟓𝟖+𝟐𝟕𝟑
V2 = 4.46 L
Rationalize your result: The volume increases as the temperature increases.
3.2.9
Use volume-absolute temperature graph to determine the volume when the temperature is given
and vice versa
3.2.10 Define absolute zero
Absolute zero is the lowest possible theoretical temperature corresponds to 0 K or −273 C, where
the atoms are in the lowest possible energy state
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3.2.11 State Gay-Lussac’s Law with its mathematical relation graphical representation
Gay-Lussac’s Law states that the pressure of a fixed amount of gas varies directly with its Kelvin
temperature when the volume remains constant
𝐏𝟏
𝐏𝟐
=
𝐓𝟏
𝐓𝟐
(𝐀𝐭 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐧 𝐚𝐧𝐝 𝐕)
3.2.12 Use the mathematical formula of Gay-Lussac’s law to calculate pressure-temperature changes at
constant volume
Example:
The gas in a container is at a pressure of 1.00 atm at 25oC. If the gas was heated to a temperature
of 928oC, what is the new pressure?
P1
T1
=
P2
1.00
T2
25+273
=
P2
928+273
P2 = 4.03 atm
Rationalize your result:
From the kinetic theory, one will expect the increase in temperature of a gas to produce an
increase in pressure if the volume remains constant.
3.2.13 Identify the restrictions on the use of Gay-Lussac's law of combining volumes
It applies only to gases measured under the same temperature and pressure
3.2.14 Use pressure-absolute temperature graph to determine the pressure when the temperature is given
and vice versa
Lab 17 − Conduct an experiment to investigate the relationship between the pressure and temperature of a
confined gas
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3.2.15 Explain the importance of combined gas law while writing its mathematical relation
It states the relationships between pressure, temperature, and volume of a fixed amount of gas.
It allows to perform calculations when the constant is the amount of gas.
𝐏𝟏 𝐕𝟏 𝐏𝟐 𝐕𝟐
=
𝐓𝟏
𝐓𝟐
(𝐀𝐭 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐧)
3.2.16 Use the combined gas law to calculate volume-pressure-temperature changes
Example:
A scuba diver takes a ball of 3.20 L from the surface where the pressure is 1.0 atm and the
temperature is 24℃, to a depth of 28.0 m, where the pressure is 4.0 atm and the temperature is
15℃. Calculate the volume of the ball at the new depth.
P1 V1
T1
=
1.0 ×3.20
(24+273)
P2 V2
T2
=
4.0 × V2
(15+273)
V2 = 0.78 L
Inspire Chemistry – Module 12 – Lesson 2: The Ideal Gas Law (4 Periods)
3.2.17 Define ideal gas
An ideal gas is an imaginary gas whose particles are infinitely small and do not interact with each
other
3.2.18 Describe the relationship between an ideal gas and the kinetic molecular theory (KMT)
An ideal gas follows all the assumptions of KMT where: the particles take up no space, the
particles experience no intermolecular attractive forces, and the collisions of particles are perfectly
elastic
3.2.19 Explain why there are no true ideal gases in nature
Because all gas particles have some volume and are subject to intermolecular attractions
3.2.20 List the conditions for a gas to behave ideally
High temperature and low pressure
3.2.21 Describe a real gas
A gas that does not behave completely according to the assumptions of the KMT
3.2.22 List examples of gases following ideal behavior over wide range of temperature and pressure
Helium and neon
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3.2.23 List conditions of a gas at STP and RTP
o STP
o RTP
At 0oC and 1 atm, 1 mole of a gas occupies 22.4 L
At 25oC and 1 atm, 1 mole of a gas occupies 24 L
3.2.24 Analyze how the ideal gas law applies to real gases using the kinetic molecular theory
A real gas behaves most like an ideal gas under conditions that increase the distance and reduce
the attractions among gas particles. The best conditions for that are high temperature and low
pressure
3.2.25 Predict and explain the conditions under which a real gas might deviate from ideal behavior
A real gas might deviate from ideal behavior under conditions that decrease the distance and
increase the attractions among gas particles, such as low temperature and high pressure
3.2.26 State Avogadro’s law with its mathematical relation and graphical representation
It states that equal volumes of gases at the same temperature and pressure contain equal numbers
of particles
𝐕𝟏
𝐕𝟐
=
(𝐀𝐭 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐏 𝐚𝐧𝐝 𝐓)
𝐧𝟏
𝐧𝟐
3.2.27 Explain why Avogadro’s' law holds true for ideal gases that have small particles and for ideal
gases that have large particles
The size of any gas particle is so small compared to the volume of the gas, it is assumed that no
particle has any volume of its own
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3.2.28 Use the mathematical formula of Avogadro’s law to calculate volume-mole changes at constant
pressure
Example
12.0 moles of nitrogen gas have a volume of 315 L at STP condition. If 6.00 moles of nitrogen gas
is removed, what is the new volume of the gas at the same conditions?
V1
n1
=
V2
315
n2
12.0
=
V2
6.00
V2 = 157.5 L = 158 L (To be reported to the correct number of significant figures)
3.2.29 Describe the ideal gas law and write the equation that represents it
It is the mathematical relationship among pressure, volume, temperature, and the number of moles
of a gas
PV = nRT
3.2.30 List common units for each variable in the ideal gas law
P: atm, mmHg, torr
V: L, mL
T: K
n: mol
3.2.31 Identify the units and values of the ideal gas constant, R
0.0821 L.atm/mol.K
62.4 L.mmHg/mol.K
3.2.32 Use the ideal gas law to calculate pressure, volume, temperature, mass of a gas, when three
quantities are given
Example:
An 8.58 L rigid vessel contains 91.3 g of oxygen gas, O2, at 21℃. Calculate the pressure of the gas
at the same temperature.
P=
nRT
V
=
(
91.3
)×0.0821 ×(21+273)
32.0
8.58
= 8.03 atm
3.2.33 Use the ideal gas law to calculate density and molar mass of a gas
PM = DRT
Using the relation
Example:
An element X exists as a diatomic gas at room temperature. This element has a density of 1.553
g/L at 25℃ and 1.00 atm. Calculate the molar mass of element X.
PM = DRT
M=
DRT
P
=
1.553×0.0821×(25+273)
1.00
= 38.0 g/mol
Lab 18 – Conduct an experiment to find the molar mass of a volatile liquid
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Inspire Chemistry – Module 12 – Lesson 3: Gas Stoichiometry (4 Periods)
3.2.34 Define the molar volume of a gas at STP
It is the volume that 1 mole of a gas occupies at 0.00oC and 1.00 atm
3.2.35 Identify what the coefficients in a balanced chemical equation specify
Molar ratios and volume ratios for gaseous reactants and products that are at the same temperature
and pressure
Example:
Hydrogen and chlorine gases react together under the same conditions of temperature and pressure
to form hydrogen chloride gas.
H2 (g) + Cl2 (g) → 2 HCl (g)
1 mole 1 mole 2 moles
1L
1L
2L
Example:
Ammonia, NH3, burns in oxygen according to the equation:
4 NH3 (g) + 3 O2 (g) → 2 N2 (g) + 6 H2O (g)
If 24 mL of NH3 is burnt with an excess of oxygen, how many mL of H2O are produced at the
same conditions of temperature and pressure?
4 NH3 :
4 mol
4 mL
24 mL
6 H2 O
6 mol
6 mL
?
V of H2O = 36 mL
3.2.36 Identify the restrictions that applies to the use of volume ratios in solving stoichiometric
calculations
Volume ratios are valid only when the volumes of all gases, being compared, are measured under
the same temperature and pressure
Volume ratios apply only to gases (it cannot apply to solids and liquids)
3.2.37 Explain why it is not necessary to consider temperature and pressure when using a balanced
chemical equation to determine relative gas volume
Because temperature and pressure are the same for each gas involved in a reaction. These
conditions affect each gas in the same way
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3.2.38 Perform different gas stoichiometric calculations
Example: Volume-Volume and Volume-mass problems
In an effort to reduce pollution, car exhaust systems have been fitted with catalytic convertors to
ensure conversion of toxic gases into gases that already exist in the air. If 42.7g of carbon
monoxide, CO, reacts completely with nitrgen monoxide, NO, at STP to form nitrogen gas, N2, and
carbon dioxide gas, CO2, according to the equation below.
2 CO (g) + 2 NO (g) ⟶ N2 (g) + 2 CO2 (g)
a) Calculate the volume of NO needed to completely react with carbon monoxide.
2 CO
2 moles
2 × 28.01 g
42.7 g
2 NO
2 moles
2 × 22.4 L
????
Volume of NO = 34.1 L
b) Calculate the mass of N2 produced at STP conditions.
2 CO
2 moles
2 × 28.01 g
42.7 g
N2
1 mole
1 × 28.02 g
????
Mass of N2 = 21.4 g
Example: Using ideal gas law to calculate amount of gaseous reactant or product
Ammonium sulfate, (NH4)2SO4 , an important fertilizer, is prepared by the reaction of ammonia
gas, NH3, with sulfuric acid, H2SO4 according to the following reaction:
2NH3(g) + H2 SO4(aq) → (NH4 )2 SO4(aq)
Calculate the volume of ammonia, at 25℃ and 25.0 atm, needed to react with 196 kg of sulfuric
acid.
moles of sulfuric acid =
m 196 × 103
=
= 2.00 × 103 moles
M
98.0
2.00 × 103 moles × 2
moles of ammonia needed =
= 4.00 × 103 moles
1
Ideal gas law: PV = nRT
25.0 V = (4.00 × 103 )(0.0821)(298)
V = 3.92 × 103 L
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3.3 – Mixtures and Solutions
Inspire Chemistry – Module 13 – Lesson 1: Types of Mixtures (2 Periods)
3.3.1
Distinguish, using examples, between heterogeneous and homogenous mixtures
o Heterogeneous mixture: A mixture that does not have a uniform composition and in which the
individual substances remain distinct
Examples: Milk, blood, smoke, and dust in air
o Homogenous mixture: A mixture that has a uniform composition throughout and always has a
single phase.
Examples: Salt water, and sugar solution
3.3.2
Define solution
A solution is a homogenous mixture or a uniform mixture that contains solids, liquids, or gases
3.3.3
Define solute and solvent
Solute: One or more substances dissolved in a solution
3.3.4
Solvent: The substance that dissolves in a solute to form a solution, the most plentiful substance in
the solution
Identify, using examples, the various types of solutions (liquid , solid or gas) and identify the
solute and solvent in the six types of solutions
Type of Solution
Example
Solute
Solvent
Gas in gas
Oxygen in nitrogen
Oxygen
Nitrogen
Gas in liquid
Carbon dioxide in water
Carbon dioxide
Water
Liquid in liquid
Alcohol in water
Alcohol
Water
Sugar in water
Sugar
Water
Salt in water
Salt
Water
Liquid in solid
Mercury in silver
Mercury
Silver
Solid in solid
An alloy as copper in nickel
Copper
Nickel
Solid in liquid
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3.3.5
Differentiate among different types of mixtures; solution, colloid, and suspension in terms of type
of mixture, separation upon standing, separation by filtration and Tyndall effect or scattering of
light
o Type of mixture:
Solution is homogenous, while colloid and suspension are heterogeneous
o Separation upon standing:
Solution and colloid do not separate upon standing while suspension does
o Separation by filtration:
Solution and colloid cannot be separated by filtration while suspension can
o Tyndall effect or scattering of light:
Solution does not scatter light, colloid scatters light while suspension may scatter light but is not
transparent
Properties of Solutions, Colloids and Suspensions
System
Property
Solution
Colloid
Ions, atoms, and small
Large molecules or
Particle Type
molecules
particles
Suspension
Large particles or
aggregates
Particle Size
0.1 – 1 nm
1 – 1000 nm
1000 nm and larger
Effect of Light
No scattering
Exhibits Tyndall effect
Exhibits Tyndall effect
Effect of Gravity
Stable
Does not separate
Stable
Does not separate
Unstable
Sediment forms
Filtration
Particles not retained on
filter
Particles not retained on
filter
Particles retained on
filter
Uniformity
Homogenous
Heterogeneous
Heterogeneous
3.3.6
Distinguish, using examples, between miscible and immiscible liquids
o Miscible liquids: Two liquids that are soluble in each other in any
proportion
o Immiscible liquids: Two liquids that can be mixed together but separate
shortly after
Polar liquids include ethanol, water
Nonpolar liquids include gasoline, toluene, carbon tetrachloride
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Inspire Chemistry – Module 13 – Lesson 2: Solution Concentration (6 Periods)
3.3.7
Define concentration
Concentration is measure of how much solute is dissolved in a specific amount of solvent or
solution
3.3.8
List units of concentration
mol/L
3.3.9
mol/kg
g/100 mL
Define molarity
Molarity is the number of moles of solute dissolved per liter of solution; also known as molar
concentration
𝐌𝐨𝐥𝐚𝐫𝐢𝐭𝐲 =
𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧 (𝐢𝐧 𝐋)
𝐂=
𝐧
𝐕
3.3.10 Calculate molarity when the moles or the mass of solute and volume of solution are given and vice
versa
𝐌𝐨𝐥𝐚𝐫𝐢𝐭𝐲 =
𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧 (𝐢𝐧 𝐋)
𝐂=
𝐧
𝐕
Example:
Calculate the molarity of 12 mL of a gas dissolved in 100. mL of water at RTP.
AT RTP
1 mole of a gas
???
occupies
24 L
12 /1000 L
Number of moles of gas = 5.0 × 10−4 moles
C=
n
5.0 × 10−4
=
= 5.0 × 10−3 mol/L
V
100./1000
3.3.11 Define molality
Molality is the ratio of number of moles of solute dissolved in one kilogram of solvent; also
known as molal concentration
𝐌𝐨𝐥𝐚𝐥𝐢𝐭𝐲 =
𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐞
𝐌𝐚𝐬𝐬 𝐨𝐟 𝐒𝐨𝐥𝐯𝐞𝐧𝐭 (𝐢𝐧 𝐤𝐠)
3.3.12 Calculate molality when the moles or the mass of solute and mass of solvent are given and vice
versa
𝐌𝐨𝐥𝐚𝐥𝐢𝐭𝐲 =
𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐞
𝐌𝐚𝐬𝐬 𝐨𝐟 𝐒𝐨𝐥𝐯𝐞𝐧𝐭 (𝐢𝐧 𝐤𝐠)
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Example:
Calculate the molality of 1.80 g of glucose, C6H12O6, dissolved in 200. mL of water (density is 1.0
g/mL).
Mass of solvent = density × volume = 1 × 200. = 200 g = 0.200 kg
Molality =
Number of moles of Solute
Mass of Solvent (in kg)
=
1.80
)
180.156
(
0.200
= 0.0500 molal
3.3.13 Explain the effect of temperature on molarity and molality
Molality does not depend on temperature of solution while molarity depends on temperature of
solution because volume of solution is proportional to the temperature
3.3.14 Define percent by volume of a solution
Percent by volume is the comparison between the volume of the solute and the total volume of the
solution
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐞
𝐏𝐞𝐫𝐜𝐞𝐧𝐭 𝐛𝐲 𝐕𝐨𝐥𝐮𝐦𝐞 =
× 𝟏𝟎𝟎
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧
3.3.15 Calculate percent by volume of a solution
Example:
Calculate the percent by volume f ethanol in a solution that contains 35 mL of ethanol dissolved in
155 mL of water.
Percent by Volume =
Volume of Solute
35
× 100 =
× 100 = 18 %
Volume of Solution
190
3.3.16 Define percent by mass of a solution
Percent by mass is the comparison between the mass of solute and the total mass of solution
𝐏𝐞𝐫𝐜𝐞𝐧𝐭 𝐛𝐲 𝐌𝐚𝐬𝐬 =
𝐌𝐚𝐬𝐬 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐞
× 𝟏𝟎𝟎
𝐌𝐚𝐬𝐬 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧
3.3.17 Calculate percent by mass of a solution
Example:
A solution contains 3.6 g of NaCl per 100.0 g of water. Calculate the percent by mass of NaCl in
the solution.
Percent by Mass =
Mass of Solute
3.6
× 100 =
× 100 = 3.5 %
Mass of Solution
3.6 + 100.0
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Example:
A 12.0 mL of ethanol, C2H5OH, (density = 0.790 g/ml) were dissolved in 100. g of water (density
= 1 g/ml). What is the percent by mass of this solution?
Mass of C2H5OH = d × V = 0.790 × 12.0 = 9.48 g
Mass of H2O = d × V = 100. × 1 = 100. g
Mass of Solute
Percent by Mass = Mass of Solution × 100 =
9.48
9.48+100.
× 100 = 8.66 %
3.3.18 Define mole fraction of a solute or solvent
Mole fraction is the ratio of the number of moles of solute or solvent in solution to the total
number of moles of solute and solvent
𝐌𝐨𝐥𝐞 𝐟𝐫𝐚𝐜𝐭𝐢𝐨𝐧 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 =
𝐌𝐨𝐥𝐞𝐬 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐞
𝐌𝐨𝐥𝐞𝐬 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧
𝐌𝐨𝐥𝐞 𝐟𝐫𝐚𝐜𝐭𝐢𝐨𝐧 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭 =
𝐌𝐨𝐥𝐞𝐬 𝐨𝐟 𝐒𝐨𝐥𝐯𝐞𝐧𝐭
𝐌𝐨𝐥𝐞𝐬 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧
3.3.19 Calculate mole fraction of a solute or solvent
𝐌𝐨𝐥𝐞 𝐟𝐫𝐚𝐜𝐭𝐢𝐨𝐧 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 =
𝐌𝐨𝐥𝐞𝐬 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐞
𝐌𝐨𝐥𝐞𝐬 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧
𝐌𝐨𝐥𝐞 𝐟𝐫𝐚𝐜𝐭𝐢𝐨𝐧 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭 =
𝐌𝐨𝐥𝐞𝐬 𝐨𝐟 𝐒𝐨𝐥𝐯𝐞𝐧𝐭
𝐌𝐨𝐥𝐞𝐬 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧
Example:
Ethylene glycol (EG), C2H6O2, is added to automobile cooling systems to protect against cold
weather. What is the mole fraction of each component in a solution containing 1.25 mol of
ethylene glycol and 4.00 mol of water?
Moles of water
Xwater = Moles of Solution =
Moles of EG
XEG = Moles of Solution =
4.00
1.25+4.00
1.25
1.25+4.00
= 0.762
= 0.238
Rationalize your result:
The sum of the mole fractions of all the components in the solution equals 1
(XEG + Xwater = 1.000)
The component with the smaller number of moles has the smaller mole fraction
3.3.20 Differentiate between dilute solution and concentrated solution
Dilute solution is a solution that contains a small amount of
solute
Concentrated solution is a solution containing a large amount
of solute
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3.3.21 Describe how to prepare a dilute solution given a concentrated solution and list possible lab safety
measures that need to be followed
1) Measure out the specific volume of stock solution.
1
2) Pour into the appropriate volumetric flask which already contains 2 or
to be added.
3) Add water to the mark.
1
2
3
of the needed solvent
2
If you are diluting an acid solution, pour 2 or 3of the water needed in the appropriate flask. Then
SLOWLY add the acid to water while swirling it. Finish by adding the last amount of water to the
mark. This is considered a safety issue.
When diluting acids, always add the concentrated acid to the water.
Never add water to concentrated acid solutions, as the heat generated may cause the
concentrated acid to splatter and burn your skin.
Example: How to make a 0.5 M solution
a) Add 0.5 moles of a solute to a 1 L volumetric flask that is half-filled with distilled water
b) Swirl the flask carefully to dissolve the solute
c) Fill the flask with water exactly to the meniscus of the 1 L mark
3.3.22 Perform calculation involving diluting a concentrated solution
Use the formula M1V1 = M2V2
Example:
How many milliliters of aqueous 3.00 M Na2SO4 solution must be diluted with water to prepare
200.0 mL of aqueous 0.300 M Na2SO4?
Number of moles of concentrated solution = Number of moles of dilute solution
M1V1 = M2V2
3.00 × V1= 0.300 × 200.0
V1 = 20.0 mL
Thus, 20.0 mL of the initial solution must be diluted by adding enough water to increase the
volume to 200.0 mL.
The volume of water needed to be added is 200.0 – 20.0 = 180.0 mL
Lab 19 – Conduct an experiment to prepare solution of given concentration and list laboratory steps
needed
Lab 20 – Perform and conduct an experiment to dilute a solution
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Inspire Chemistry – Module 13 – Lesson 3: Factors Affecting Solvation (4 Periods)
3.3.23 Define solubility
Solubility is the maximum amount of a solute that will dissolve in a given amount of solvent at a
specific temperature and pressure
3.3.24 Identify the conditions that must be specified when expressing the solubility of a substance
Temperature and pressure
3.3.25 Explain three factors that affect the rate at which a solid solute dissolves in a liquid
o Solute surface area
The smaller the particle size, the greater the surface area, the more collisions occur and the higher
the rate of solvation
o Agitation
Stirring or shaking moves dissolved solute particles away from the contact surfaces more quickly
and thus allows new collisions between solute and solvent particles to occur and thus rate of
solvation increases
o Temperature
As temperature of solvent increases, solvent molecules move faster and their average KE increases
and more collisions between solvent and solute particles, and thus rate of solvation increases
3.3.26 Distinguish among saturated, unsaturated, and supersaturated solutions
o Saturated solution
A solution that contains the maximum amount of dissolved solute. The visual evidence that
indicates a solution is saturated is that a residual quantity of undissolved solid remains in contact
with the saturated solution
o Unsaturated solution
A solution that contains less dissolved solute than a saturated solution under the existing
conditions. In other words, more solute can be dissolved in an unsaturated solution
o Supersaturated solution
A solution that contains more dissolved solute than a saturated solution contains under the existing
conditions. To make a supersaturated solution, a saturated solution is formed at a high temperature
and then cooled slowly. The slow cooling allows the excess solute to remain dissolved in solution
at the lower temperature
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3.3.27 Interpret temperature − solubility graph for solids
Example:
Use the following solubility curve to answer the questions below.
-
At 10℃, the least soluble salt is KClO3 and the most soluble is NaNO3.
-
What mass of potassium dichromate, K2Cr2O7, forms if a saturated solution in 100g of water is
cooled from 90ºC to 20ºC?
At 90C, the solubility is 70 g / 100 g of water
At 20C, the solubility is 10 g / 100 g of water
70 – 10 = 60 g of solid will form in 100g of water
-
By how much will the solubility of potassium nitrate, KNO3, increase if the temperature is
increased from 20°C to 50°C?
At 20C, the solubility is 30 g / 100 g of water
At 50C, the solubility is 90 g / 100 g of water
The solubility increases by 60 g/ 100 g of water
-
What mass of water is needed to dissolve 40 g of sodium nitrate, NaNO3, at 10ºC?
At 10C, the solubility is 80 g / 100 g of water
80 g of solid dissolve in 100g of water
40g of solid dissolve in ?g of water
(40)(100)
? = 80 = 50 g of water
-
What mass of potassium nitrate, KNO3, will dissolve in 2 kg of water at 20ºC?
At 20C, the solubility is 30 g / 100 g of water
30 g of solid dissolve in 100 g of water
? g of solid dissolve in 2000g of water
(30)(2000)
? = 100 = 600 g
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Lab 21 − Conduct an experiment to study the effect of changing temperature on the amount of solute that
will dissolve in a given amount of water
3.3.28 List two factors that affect the rate at which a gaseous solute dissolve in a liquid
Pressure
As pressure increases, solubility of gas increases
Temperature
As temperature increases, solubility of gas decreases
3.3.29 Interpret temperature − solubility graph for gases
Example:
Use the following solubility curve to answer the questions below.
a) In general, how does the solubility of gases change (increase, decrease, does not change) as
temperature increases?
Solubility of gases decreases
b) At the same temperature, which has a higher solubility: CH4 or CO?
CH4
3.3.30 State Henry's Law and its mathematical relationship
The solubility of a gas in a liquid is directly proportional to the partial pressure of that gas on the
surface of the liquid
S = kP
𝑺𝟏
𝑷𝟏
=
𝑺𝟐
𝑷𝟐
Where:
S = Solubility
P = Pressure
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3.3.31 Apply Henry's Law to calculate the solubility of a gas given its pressure and vice versa
Example:
A gas has a solubility in water of 0.77 g/L at 3.5 atm of pressure. What is its solubility (in g/L) at
1.0 atm of pressure at the same temperature?
𝑆1
𝑃1
=
𝑆2
0.77
𝑃2
3.5
=
S2
1.0
S2 = 0.22 g/L
Rationalize your result:
The new pressure is approximately one-third of the original pressure. So, the new solubility
should be approximately one-third of the original.
This document was created for educational purposes. The document is not meant for publication or mass
printing or production. It is made to support Applied Technology High School students for AY 2021 –
2022.
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