Calculations
Involving Colligative
Properties
Freezing Point Depression and Boiling Point Elevation
Calculations
Objectives
• When you complete this presentation, you will be able to
• calculate the freezing point depression or boiling point
elevation of a solution when you know its molality
• calculate the molality of a solution when you know its
freezing point depression or boiling point elevation
Introduction
• We now understand colligative properties.
• To use this knowledge, we need to be able to predict
these colligative properties.
• Freezing Point Depression
• Boiling Point Elevation
Introduction
• We also need to use a different kind of concentration
determination.
• Instead of molarity, we will use
• molality, m
• mole fraction, X
Molality
• Molarity - we measure the number of mols of solute in
the volume of the solution.
•
• Msolution = nsolute/Vsolution
•
• Molality - we measure the number of mols of solute in
the mass of solvent.
•
• msolution = nsolute/msolvent
Molality
• msolution = nsolute/msolvent
• The mass of the solvent is measured in kilograms, kg.
• 1 mole of solute in
1,000 g of solvent gives
a 1 m solution.
Molality
Example 1:
Find the molality of 87.66 g of NaCl dissolved in 2.500 kg of
H2O.
First,
we can
write
down the
known
To
calculate
molality,
m,
we
need to
Therefore,
we
calculate
the
number
Now,
we
calculate
the
molality,
mNaCl = 87.66 g
values.
know
the
number
of
mols
ofsolution.
NaCl. of mols of NaCl.
m
of
the
mH2O = 2.500 kg
We
mass
NaCl.
This only
valueknow
comesthe
from
the of
sum
of the
MNaCl = 58.45 g/mol
atomic masses of Na, 22.99 g/mol, and
We
substitute
in known
values
We
put
ing/mol.
thecalculation.
known
values
… …
…
do
the
Cl,and
35.45
nNaCl = mNaCl/MNaCl = (87.66 g)/(58.44 g/mol) = 1.500 mol
m = nNaCl/mH2O = (1.500 mol)/(2.500 kg) = 0.600 mol/kg
Molality
Example Problems:
1. Find the molality of 100.0 g of NaOH in 1.500 kg of
H2O.
1.667
m
2. Find the molality of 32.00 g of KCl in 0.5000 kg of H2O.
0.8585 m
3. Find the molality of 142.5 g of CuCl2 in 2.000 kg of H2O.
1.667 m
4. Find the molality of 180.0 g of H2O in 1.500 kg of ethyl
alcohol.
6.667 m
Molality
Example 2:
How many grams of potassium iodide, KI, M = 166.0 g/mol,
must be dissolved in 500.0 g of water to produce a 0.060
First,
we
write
the
To
calculate
, the
mass
KI,mass
we
molal solution?
We
can
rearrange
equation
to
we
substitute
inthe
known
values
…
…
and
do
themcalculate
calculation.
Now,
we
can
mKIknown
,of
the
KIdown
values.
need
to
the
number
ofsolution.
mols of
solve
forknow
nKI …to
of KI needed
prepare
the
msoln = 0.060 mol/kg
KI, nKI. We only know the molality of
mH2O = 0.500 kg
the
solution,
Weand
putdo
inthe
them
known
values …
…
calculation.
soln.
MKI = 166.0 g/mol
msoln = nKI/mH2O
nKI = msolnmH20
nKI = (0.060 mol/kg)(0.500 kg) = 0.030 mol
mKI = MKInKI = (166.0 g/mol)(0.030 mol) = 5.0 g
Molality
Example Problems:
1. Find the mass of NaCl (58.5 g/mol) needed to prepare
a 0.600 m solution with a mass of 2.50 kg of water.
87.8 g NaCl
2. Find the mass of NaOH (40.0 g/mol) needed to prepare
a 1.20 m solutionwith a mass of 0.250 kg of water.
12.0 g NaOH
3. Find the mass of CrF2 (90.0 g/mol) needed to prepare a
1.50 m solution with a mass of 0.400 kg of water.
54.0 g CrF2
4. Find the mass of KNO3 (101 g/mol) needed to prepare
a 1.01 m solution with a mass of 0.100 kg of water.
10.2 g KNO3
Mole Fraction
• Mole Fraction is the ratio of number of mols of the solute
to the total number of mols of the solute plus the
solvent.
• We use the symbol Χ (the Greek letter chi) to represent
the mole fraction.
nsolute
• Χsolute = n
solute + nsolvent
Mole Fraction
Example 3
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 mols of EG and 4.00 mols of water?
nEG = 1.25 mol
nH2O = 4.00 mol
nEG
1.25 mol
=
ΧEG =
= 0.238
nEG + nH2O
(1.25 + 4.00) mol
4.00 mol
nEG
=
ΧH2O = n + n
= 0.762
(1.25
+
4.00)
mol
EG
H2O
Mole Fraction
Example Problems:
1. Find the mole fraction of 1.00 mol of NaCl in 55.6 mols
of water.
Χ = 0.177
2. Find the mole fraction of 5.00 mol of BaCl2 in 10.0 mols
of water.
Χ = 0.333
3. Find the mole fraction of 0.240 mol of C6H12O6 in 4.76
mols of ethanol.
Χ = 0.0480
4. Find the mole fraction of 0.0320 mol of KNO3 in 10.2
mols of water.
Χ = 0.00319
Colligative Calculations
• The magnitudes of freezing point depression (∆Tf) and
boiling point elevation (∆Tb) are
• directly proportional to the molal concentration of the
solute,
• if the solute is molecular and not ionic.
Colligative Calculations
• The magnitudes of freezing point depression (∆Tf) and
boiling point elevation (∆Tb) are
• directly proportional to the molal concentration of all
solute ions in the solution,
• if the solute is ionic.
Colligative Calculations
• ∆Tf = −Kf x m
• where
• ∆Tf is the freezing point depression of the solution.
• Kf is the molal freezing point constant for the solvent.
• m is the molal concentration of the solution.
Colligative Calculations
• ∆Tb = Kb x m
• where
• ∆Tb is the boiling point elevation of the solution.
• Kb is the molal boiling point constant for the solvent.
• m is the molal concentration of the solution.
Colligative Calculations
The number of particles in solution is equal to the number of
This
is a non-ionic compound
mols of the
compound.
Example 4
What is the freezing point depression, ∆Tf, of a benzene
(C6H6, 78.1 g/mol, Bz) solution containing 400 g of Bz and
200 g of the compound acetone (C3H6O, 58.0 g/mol, Ac)? Kf
for Bz is 5.12°C/m.
mBz = 400 g = 0.400 kg
mAc = 200 g
MAc = 58.0 g/mol
Kf = 5.12°C/m
nAc =
mAc
200 g
=
= 3.45 mol
MAc
58.0 g/mol
3.45 mol
nAc
msoln = m =
= 8.62 m
0.400
kg
Bz
∆Tf = −Kf × m = (5.12°C/m)(8.62 m) = −44.1°C
Colligative Calculations
For
each mol
of NaCl ininsolution,
is 1tomol
Na+ and
The
number
of particles
solution there
is equal
theof
number
of1
− ions. We This
n by 2.
istoanmultiply
ionic compound
of the need
compound.
Example 5 mol of Clions
What is the freezing point depression, ∆Tf, of a sodium
chloride (NaCl, 58.4 g/mol) solution containing 87.6 g of
NaCl in 1.20 kg water? Kf for water is 1.86°C/m.
mH2O = 1.20 kg
mNaCl = 87.6 g
MNaCl = 58.4 g/mol
Kf = 1.86°C/m
mNaCl
87.6 g
=
nNaCl = M
= 1.50 mol NaCl = 3.00 mol
58.4
g/mol
NaCl
3.00 mol
nNaCl
=
msoln = m
= 2.50 m
1.20
kg
H2O
∆Tf = −Kf × m = (1.86°C/m)(2.50 m) = −4.65°C
Colligative Calculations
Example Problems: Find the freezing point depression of:
1. 58. 44 g of NaCl (M = 58.44 g/mol) in 1.00 kg of H2O.
(Kf = 1.86°C/m)
∆T = −3.72°C
f
2. 228 g of octane, C8H18 (M = 114 g/mol), in 0.500 kg of
benzene. (Kf = 5.12°C/m)
∆T = −20.5°C
f
3. 33.5 g of NaC2H3OH (M = 67.0 g/mol) in 1.25 kg of
acetic acid. (Kf = 3.90°C/m)
∆T = −3.12°C
f
4. 217 g of C5H12 (M = 72.2 g/mol) in 0.600 kg of
cyclohexane. (Kf = 20.2°C/m)
∆T = −101°C
f
Colligative Calculations
Example 6
What is the boiling point, T, of a benzene (Bz) solution containing 0.600
kg of Bz and 200 g of the compound phenol (C6H5OH, 94.1 g/mol, Ph)?
Kb for Bz is 2.53°C/m. Normal boiling point for Bz is Tb = 80.1°C.
mBz = 0.600 kg
mPh = 200 g
Tb = 80.1°C
MPh = 94.1 g/mol
Kf = 5.12°C/m
nPh =
msoln =
mPh
MPh
=
nPh
=
mBz
200 g
94.1 g/mol
2.13mol
0.600kg
= 2.13 mol
= 3.54m
∆Tb = Kb × m = (2.53°C/m)(3.54 m) = 8.96°C
T = Tb + ∆Tb = 80.1°C + 8.96°C = 89.1°C
Colligative Calculations
Example 7
What is the boiling point, T, of a sodium chloride (NaCl, 58.4 g/mol)
solution containing 87.6 g of NaCl in 0.800 kg of acetic acid (AcOH)?
Kb for AcOH is 3.07°C/m. The boiling point of AcOH is Tb = 118.5°C.
mH2O = 1.20 kg
mNaCl = 87.6 g
Tb = 118.5°C
MNaCl = 58.4 g/mol
Kb = 3.07°C/m
nNaCl =
mNaCl
MNaCl
=
nNaCl
=
msoln = m
AcOH
87.6 g
58.4 g/mol
3.00 mol
0.800 kg
= 1.50 mol NaCl = 3.00 mol particles
= 3.75 m
∆Tb = Kb × m = (3.07°C/m)(3.75m) = 11.5°C
T = Tb + ∆Tb = 118.5°C + 11.5°C = 130.0°C
Colligative Calculations
Example Problems: Find the boiling point elevation of:
1. 58. 44 g of NaCl (M = 58.44 g/mol) in 1.00 kg of H2O.
(Kb = 0.512°C/m)
∆T = 1.02°C
b
2. 228 g of octane, C8H18 (M = 114 g/mol), in 0.500 kg of
benzene. (Kb = 2.53°C/m)
∆T = 10.1°C
b
3. 33.5 g of NaC2H3OH (M = 67.0 g/mol) in 1.25 kg of
acetic acid. (Kb = 3.07°C/m)
∆T = 2.46°C
b
4. 217 g of C5H12 (M = 72.2 g/mol) in 0.600 kg of
cyclohexane. (Kb = 2.79°C/m)
∆T = 14.0°C
b
Summary
• Molality - we measure the number of mols of solute in
the mass of solvent.
• msolution = nsolute/msolvent
• The mass of the solvent is measured in kilograms, kg.
• Mole Fraction is the ratio of number of mols of the solute
to the total number of mols of the solute plus the
solvent.
• We use the symbol X to represent the mole fraction.
• Xsolute = (nsolute)/(nsolute + nsolvent)
Summary
• ∆Tf = Kf x m
• ∆Tf is the freezing point depression of the solution
• Kf is the molal freezing point constant for the solute
• m is the molal concentration of the solution.
• ∆Tb = Kb x m
• ∆Tb is the boiling point elevation of the solution
• Kb is the molal boiling point constant for the solute
• m is the molal concentration of the solution.