SHS
STEM
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General Chemistry 2
Quarter 2- Week 3 (Module 11)
(Boiling Point Elevation, Freezing Point
Depression, Molar Mass Calculation)
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General Chemistry 2 (SHS-STEM)
Quarter 2 – Week 3 (Module 11): Boiling Point Elevation, Freezing Point Depression,
Molar Mass Calculation
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Regional Director: Gilbert T. Sadsad
Asst. Regional Director: Jessie L. Amin
Development Team of the Module
Writer: Brian B. de Lima
Editors/Reviewers: Ronaldo C. Reyes
Noel V. Ibis
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• Calculate boiling point elevation and freezing
point depression from the concentration of a
solute in a solution.
• Calculate molar mass from colligative property
data
OBJECTIVES:
1. solve problems involving boiling point elevation and freezing
point depression from the concentration of a solute in a
solution.
2. indicate what happens to the boiling point and freezing point
of a solvent when solute is added to it.
3. calculate the molar mass from the colligative property data.
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Often times a solution is described in terms of concentration of one or more
solutes present in it. However, there are some important physical properties of solution
which are more directly dependent on the concentration of solute particles. Such
properties are called colligative (Latin, coligare – which means “tied together”)
properties which means, they depend on the collective effect of the concentration of
solute particles present in the solution. These properties include: (1) vapor pressure
lowering, (2) boiling point elevation, (3) freezing point depression, and (4) osmotic
pressure.
In this topic we will concentrate on how to calculate the boiling point elevation
and freezing point depression. And also to illustrate example on how to calculate molar
mass from colligative property data.
PRETEST
Answer the following questions.
1. Consider a solution made from a nonvolatile solute and a volatile solvent. Which
statement is TRUE?
a. The osmotic pressure is the same as vapor pressure of the solution.
b. The vapor pressure of the solution is always greater than the vapor pressure
of the pure solvent.
c. The boiling point of the solution is always greater than the boiling point of
the pure solvent.
d. The freezing point of the solution is always greater than the freezing point
of the pure solvent.
2. Dissolving a solute such as NaCl in a solvent such as water results in:
a. an increase in the melting point of the liquid
b. a decrease in the boiling point of the liquid
c. a decrease in the vapor pressure of the liquid
d. no change in the boiling point of the liquid
3. What is the freezing point of a solution that contains 10.0 g of glucose (C6H12O6)
in 100 g of H2O? Kf = 1.86 °C/m
a. –0.186°C
c. –0.10°C
b. 0.186°C
d. –1.03°C
4. A solution that contains 55.0 g of ascorbic acid (Vitamin C) in 250 g of water
freezes at –2.34°C. Calculate the molar mass (g/mol) of the solute. Kf = 1.86
°C/m
a. 1.26
c. 43.6
b. 10.9
d. 175
5. What mass of ethanol, C2H5OH, a nonelectrolyte, must be added to 10.0 L of
water to give a solution that freezes at -10.0°C? Assume the density of water is
1.0 g/mL.
a. 85.7 kg
c. 5.38 kg
b. 24. 8 kg
d. 2.48 kg
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PICTURE THIS OUT!
“It is a hot summer day and you have a picnic at the park or beach front with
your classmates, friends or relatives with watermelon and “dirty ice cream”.
Mmmmmm….. tastes good… refreshing…. The ice cream is an old-fashioned
homemade kind ice cream. The kind of where the maker has a tub full of mix of
ingredients immersed in a bigger tub filled with ice and salt. But wait a minute, why
salt? Why the ice cream vendor does add salt to the ice?
Effect of solute concentration on the colligative properties of solutions
The concentration or amount of nonvolatile solute (i.e., a solute that does not
have a vapor pressure of its own) in the solution has an effect on the colligative
properties of solutions. The effect would depend on the ratio of the number of particles
of solute and solvent in the solution and not on the identity of the solute. However, it
is necessary to take into account whether the solute is an electrolyte or a
nonelectrolyte.
Let us now describe the boiling point elevation and freezing point depression.
1. Boiling Point Elevation
What is the difference between the boiling point of a solution with that of a pure
solvent?
The addition of a nonvolatile solute lowers the vapor pressure of the solution;
consequently, the temperature must be raised to restore the vapor pressure of the
solution to the value conforming to the pure solvent. Specifically, the temperature at
which the vapor pressure is 1 atm will be higher than the normal boiling point by an
amount known as the boiling point elevation.
Figure 1 below shows the phase diagram of a solution and the effect that the
lowered vapor pressure has on the boiling point of the solution compared to the
solvent. In this case the sucrose solution has a higher boiling point than the pure
solvent. Since the vapor of the solution is lower, more heat must be supplied to the
solution to bring its vapor pressure up to the pressure of the external atmosphere. The
boiling point elevation is the difference in temperature between the boiling point of
the pure solvent and that of the solution.
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Figure 1: The lowering of the vapor pressure in a solution causes the boiling
point of the solution to be higher than pure solvent
Figure 2: Normal boiling point for water (solvent) as a function of molality in
several solutions containing sucrose (a non-volatile solute).
The boiling point elevation of a solution, ΔTb, is directly proportional to molality,
that is,
ΔTb = Kb m
Where Kb is the molal boiling point elevation constant of the solvent and m is
the molal concentration of the solution (mol/kg)
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Table 1: Boiling Point Elevation Constant (Kb) and Freezing Point Depression
Constant (Kf) for some solvents.
The Kb of water is 0.52°C/m, so 1 m aqueous solution of a nonvolatile solute
like sucrose will boil at 0.52°C higher than pure water. The boiling point elevation is
proportional to the number of solute particles in the solution. When 1 mole of NaCl
dissolves in water, 2 moles of solute are formed ( 1 mole of Na and 1 mole of Cl).
Therefore, a 1 m solution of NaCl in water causes a boiling point elevation twice
as large as a 1 m solution of sucrose.
2. Freezing Point Depression
The freezing point of a substance is the temperature at which the solid and
liquid forms can coexist indefinitely, at equilibrium. Under these conditions molecules
pass between the 2 phases at equal rates because their escaping tendencies from the
two phases are identical.
Figure 3 below shows the phase diagram for a pure solvent and how it changes
when a solute is added to it. The solute lowers the vapor pressure of the solvent
resulting in a lower freezing point for the solution compared to the pure solvent. The
freezing point depression is the difference in temperature between the freezing point
of a pure solvent and that of a solution. On the graph, Tf represents the freezing point
depression.
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Figure 3: The lowering of the vapor pressure in a solution causes the
boiling point of the solution to be higher than pure solvent (purple). As a
result, the freezing point of a solvent decreases when any solute is
dissolved into it.
Like the boiling point elevation, the freezing point of a solution is directly
proportional to the molal concentration of the solution, that is,
ΔTf = Kf m
where ΔTf refers to the freezing point lowering, Kf, the freezing point depression
constant, and m, the molality of the solution. Some of the Kf values are shown
in Table 1. For water, Kf is 1.86 °C/m, therefore, any 1 m aqueous solution of
nonvolatile solute or a 0.5 m aqueous solution of NaCl will freeze at 1.86 °C
lower than pure water. This explains the question on the riddle given from the
introduction of this module.
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Figure 4: The use of salt in homemade ice cream is an application of freezing
point depression.
Illustrative Example 1:
A solution is prepared by dissolving 2.40 g of biphenyl, C12H10 (molar mass =
154 g/mol), in 75.0 g benzene. Find the (a) Boiling point and (b) Freezing point of the
solution.
Solution:
From Table 1:
BPC6H6 = 80.1 °C
Kb = 2.53°C kg/mol
FP C6H6 = 5.5 °C
Kf = 5.12°C kg/mol
a. Boiling point of solution:
Step 1
Calculate the molality of the solution.
2.40 g C12H10
m
=
nsolute
kgsolvent
=
75 g
9
x
x
1 mol
154 g C12H10
1 kg
1000 g
=
0.208 m
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Step 2
Calculate the boiling point elevation.
ΔTb = Kb m = (2.53 °C/m) (0.208 m) = 0.526 °C
Step 3
Calculate the boiling point of the solution.
BPsolution = BPsolvent + ΔTb
= 80.1 °C + 0.526 °C
= 80.6 °C
b. Freezing point of solution:
Step 1
Step 2
Use the value of m from (a) for Step 2.
Calculate the freezing point depression.
ΔTf = Kf m = (5.12 °C/m) (0.208 m) = 1.06 °C
Step 3
Calculate the freezing point of the solution.
FPsolution = FPsolvent - ΔTf
= 5.5 °C + 1.06 °C
= 4.44 °C
Illustrative Example 2:
A solution prepared from 0.3 g of unknown nonvolatile solute and 30 g of CCl 4
has a boiling point of 0.392 °C higher than that of pure CCl4. What is the molar mass
of the solute?
Solution:
From Table 1: KbCCl4 = 5.03 °C/m or 5.03°C • kg/mol
Step 1
Substitute the given values in the boiling point elevation equation.
ΔTb = Kb m
0.3 g
molar mass
0.03 kg
0.392°C
=
(5.03°C •
kg
mol
)(
(0.392°C)( 0.03 kg)
=
(5.03°C •
kg
mol
0.3 g
) ( molar
mass )
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)
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Step 2
Simplify the terms and solve for the molar mass.
1.509 °C •
Molar mass
=
g
mol
0.0118 °C
= 128 g/mol
Let’s perform this experiment.
Effect of Particle Size in Freezing and Boiling Point
Materials:
Triple beam balance
Four 100-mL beakers
Two 50-mL beakers
Ice
Sucrose (table sugar), C12H22O11
thermometer
stirring rod
distilled water
ethylene glycol, C2H4(OH)2
sodium chloride(salt), NaCl
Safety
Mercury is extremely toxic. If a mercury thermometer gets broken accidentally,
immediately report to the teacher so that any spilled mercury can be cleaned up
thoroughly and disposed of properly.
Procedure
A. Freezing Point
1. Weigh a clean, dry, and empty 100-mL beaker. Record its mass in Table
2.1 under the heading, Sodium Chloride Solute.
2. Fill three-fourths of the beaker with ice. Wipe the outer surface of the beaker
to make it dry before weighing it using the triple beam balance.
3. Pour 25-mL of distilled water into the beaker.
4. Use a thermometer to measure the temperature that results from mixing of
ice and water. Estimate the temperature to the nearest 0.1°C. Record the
lowest temperature reading that you get as the freezing point of pure water.
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5. Weigh about 5.0 g of NaCl. Add the salt to the ice and water mixture. Stir
the mixture for at least 3 minutes.
6. Measure the temperature that results from adding salt to the ice-water
mixture. Record the lowest temperature reading that you get as the freezing
point of the solution.
7. Weigh an empty and dry 50-mL beaker on the balance. Record its mass in
Table 2.1 under the heading, Ethylene Glycol Solute.
8. Repeat steps 2 to 4. Record the data in Table 2.1 under the heading,
Ethylene Glycol Solute.
9. Put about 4.5-mL of ethylene glycol into the 50-mL beaker. Reweigh and
record the combined mass.
10. Add the ethylene glycol sample to the ice and water mixture from step 8.
Rinse the last traces of ethylene glycol using some water from the ice-water
mixture. Pour the water used in rinsing back into the ice-water mixture. Stir
the resulting mixture for 3 minutes.
11. Measure the temperature that results from adding ethylene glycol to the icewater mixture. Record the lowest temperature reading that you get as the
freezing point of the solution.
12. Determine experimentally the freezing point constant of water from the
formula,
ΔTf = Kf m
Table 2.1: Data and Observations
NaCl
Sodium Chloride
Solute
C2H4(OH)2
Ethylene Glycol
Solute
Mass of beaker + ice
Mass of empty beaker
Volume of water added
Freezing point of pure water
Mass of solute sample + container
Mass of empty container
Freezing point of solution
Mass of water
Mass of solute
Delta Tf (ΔTf)
Mass of water in kg
Molality of solution (m)
Freezing point constant (Kf)
B. Boiling Point
1. Pour 25-mL of distilled water into a 100-mL beaker. Heat until the water boils
steadily.
2. Submerge the thermometer bulb in the boiling water and get the
temperature reading. Record this as the boiling point of water in Table 2.2.
3. Empty and dry the beaker.
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4. Pour 25-mL of distilled water into the beaker.
5. Weigh12 g of sucrose (table sugar). Add the sucrose to the water in the
beaker and stir well.
6. While stirring carefully, heat the mixture until it boils. Measure and record
the boiling point as in step 2.
7. Empty, rinse, and dry the beaker.
8. Repeat steps 4 to 6, but instead of sucrose, add 10-mL of ethylene glycol.
Weigh the ethylene glycol using 50-mL beaker. Rinse the last traces of
ethylene glycol from the beaker with some water from the 100-Ml beaker.
Swirl, then pour back into the 100-mL beaker.
9. Determine experimentally the boiling point constant of water from the
formula,
ΔTb = Kb m
Table 2.2: Data and Observations
C12H22O11
Sucrose Solute
C2H4(OH)2
Ethylene Glycol
Solute
Volume of water used
Mass of solute + container
Mass of empty container
Boiling point of pure water
Boiling point of solution
Mass of water
Mass of solute
Moles of solute
Delta Tb (ΔTb)
Mass of water in kg
Molality of solution (m)
Boiling point constant (Kb)
Questions for Analysis
1. What happened to the freezing point of water when salt was added to it?
__________________________________________________________
__________________________________________________________
2. What happened to the freezing point of water when ethylene glycol was
added up to it?
___________________________________________________________
___________________________________________________________
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3. Compare the freezing point depression caused by sodium chloride to that
caused by ethylene glycol.
___________________________________________________________
___________________________________________________________
4. What happened to the boiling point of water when sucrose was added to it?
___________________________________________________________
___________________________________________________________
5. What happened to the boiling point of water when ethylene glycol was
added to it?
___________________________________________________________
___________________________________________________________
6. Compare the boiling point elevation caused by sucrose to that of ethylene
glycol.
___________________________________________________________
___________________________________________________________
7. What is the effect of nonvolatile solutes on the magnitude of the freezing
point depression and boiling point elevation as observed in the activity?
___________________________________________________________
___________________________________________________________
ENRICHMENT
1. People who live in colder climates have seen the trucks put or sprinkle salt
on the roads when snow or ice is forecast. Why do they do that?
______________________________________________________________
______________________________________________________________
2. When planes fly in cold weather, the planes need to be de-iced before liftoff.
Why is that done? _______________________________________________
______________________________________________________________
Problem Solving:
1. A solution is made of 2.00 g of sucrose, C12H22O11, in water, H2O. The molar
mass (MM) of sucrose is 342 g/mol. Calculate the (a) boiling point and (b)
freezing point of the solution. (Hint: See Table 1 for given constants.)
2. A certain nonvolatile hydrocarbon that weighed 1.00 g was dissolved in
50.00 g of benzene. The resulting solution was boiled at 0.285 °C higher
than the boiling point of pure benzene which boils at 80.1°C. Find the molar
mass of the hydrocarbon. (Hint: See Table 1 for given constant.)
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➢ The boiling point of an electrolyte solution is normally higher than the boiling
point of a nonelectrolyte solution because of a higher number of moles of solute
dissolved in the electrolyte solution.
➢ Since there is a relatively larger number of moles of solutes dissolved in the
solution in electrolyte solutions, the freezing point of electrolyte solutions are
lower than nonelectrolyte solutions.
➢ Colligative properties depend on the concentration of solutes and not on their
nature.
•
Colligative Property is a property of solution that does not depend on the kind of
matter (like intensive property) but more specifically on the amount of solute
present in the solution.
•
Boiling Point is the temperature at which the vapor pressure is equal to the
atmospheric pressure.
•
Freezing Point is the temperature at which the liquid starts to become solid.
•
Volatile are those substance that readily evaporates.
•
Nonvolatile are those substance that does not turn to vapor easily.
•
Electrolyte are substances that tend to form ions and can conduct electricity
when placed in water.
•
Nonelectrolyte are substances that do not form ions and does not conduct
electricity when placed in water.
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POSTTEST
Answer the following questions.
1. Consider a solution made from a nonvolatile solute and a volatile solvent. Which
statement is TRUE?
a. The osmotic pressure is the same as vapor pressure of the solution.
b. The vapor pressure of the solution is always greater than the vapor pressure of
the pure solvent.
c. The boiling point of the solution is always greater than the boiling point of the
pure solvent.
d. The freezing point of the solution is always greater than the freezing point of
the pure solvent.
2.Dissolving a solute such as NaCl in a solvent such as water results in:
a.
b.
c.
d.
an increase in the melting point of the liquid
a decrease in the boiling point of the liquid
a decrease in the vapor pressure of the liquid
no change in the boiling point of the liquid
3.What is the freezing point of a solution that contains 10.0 g of glucose (C 6H12O6) in
100 g of H2O? Kf = 1.86 °C/m
a. –0.186°C
b. 0.186°C
c. –0.10°C
d. –1.03°C
4.A solution that contains 55.0 g of ascorbic acid (Vitamin C) in 250 g of water freezes
at –2.34°C. Calculate the molar mass (g/mol) of the solute. Kf = 1.86 °C/m
a. 1.26
b. 10.9
c. 43.6
d. 175
5.What mass of ethanol, C2H5OH, a nonelectrolyte, must be added to 10.0 L of water
to give a solution that freezes at -10.0°C? Assume the density of water is 1.0 g/mL.
a. 85.7 kg
b. 24. 8 kg
c. 5.38 kg
d. 2.48 kg
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Magdalena C. Jauco, Violeta L. Jerusalem, Julito A. Agudong, Rosalie E. Solivio,
Emily K. Bernardo. General Chemistry2 (Second Semester). 1st ed. Mindshapers Co.
Inc.,2016.
Sol Saranay M. Baguio. Breaking Through General Chemistry 2 For Senior High
School. C & E Publishing, Inc.,2017.
Raymond Chang. Chemistry. 8th ed. The McGraw-Hill Companies Inc.,2005.
Retrieved from http://www.chem.purdue.edu/gchelp/solutions/eboil.html
https://chem.libretexts.org/Courses/College_of_Marin/CHEM_114%3A_Introductory_
Chemistry/13%3A_Solutions/13.09%3A_Freezing_Point_Depression_and_Boiling_P
oint_Elevation-_Making_Water_Freeze_Colder_and_Boil_Hotter
https://www.cliffsnotes.com/study-guides/chemistry/chemistry/solutions/freezing-andboiling-points
https://whatscookingamerica.net/Desserts/StrawberryIceCream.html
Retrieved from
https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.deped.gov.ph/
wp-content/uploads/2019/01/General-Chemistry-1-and2.pdf&ved=2ahUKEwjTq9zHlNqAhWszIsBHTWwBX0QFjAHegQIAxAB&usg=AOvVa
w3PeQKPndnZ3KqGjOuV6FRm
ANSWER KEY
Lets check what you have learned (Pretest & Posttest)
1.
2.
3.
4.
5.
c
c
d
d
d
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