Chapter 13
The Properties of Solutions
If you are doing this lecture “online” then print the
lecture notes available as a word document, go
through this ppt lecture, and do all the example and
practice assignments for discussion time.
TYPES OF SOLUTIONS
Most of us think of a solid dissolved in a liquid as a
solution, but any type of solute dissolved in any type
of solvent is a solution
S/L = L phase (sugar water)
G/L = L phase (O2 in water for fish)
L/L = L phase (beer)
G/G = G phase (air)
G/S = S phase (H2 in Pt)
S/S = S alloy (like stainless steel)
If 1-phase = homogeneous mixture
If 2-phase = heterogeneous (like cold tea with ice cubes)
Figure 13.1
The major types of intermolecular forces in solutions.
Forces are listed in decreasing order of strength
(with values in kJ/mol), and an example of each is
shown with space-filling models
DEFINITIONS :
Two fluids that mix in all proportions are miscible.
Solubility = maximum amount of solute that can be dissolved in a
given amount of solvent at a given fixed T, at equilibrium.
Saturated = solution containing maximum amount of solute.
Any additional solute appears as a precipitate or a gas, or a
separate liquid phase.
Unsaturated = more solute can be added.
Supersaturated = temporary condition where more solute has
dissolved, but add just 1 crystal to this, many will precipitate.
More types of solutions and
definitions
Types of Solutions: Solubility and Saturation
Part 2 - YouTube
Figure 13.19
from 4th ed.
Equilibrium in a saturated solution.
solute (undissolved)
solute (dissolved)
Figure 13.8
Sodium acetate crystallizing from a supersaturated solution.
Supersaturated
solution: add just
one little “seed”
crystal.
From one “seed”
excess solute will
crystallize from the
solution.
Now it’s just a
saturated solution
with some solute
present.
Factors affecting solubility:
Natural inclination of the universe towards
disorder so substances do mix
Strength of Force of Attraction between
molecules and ions and the solvent
IP Force of attraction affects & limits solubility
1. Gases - small forces, mix freely, completely
miscible
Factors affecting solubility:
2. Liquids:
Similar liquid molecules - heptane in octane - both
have only London forces involved, about the same
strength, both nonpolar
Different IP forces - octane and water – will NOT
mix
The less dense liquid will rise and stay in separate
phase on top of water.
LEADS TO GENERAL RULE: LIKE DISSOLVES LIKE.
Polar to polar, nonpolar to nonpolar.
Factors affecting solubility:
Using a drawing, explain which solvent is better
for dissolving ethanol, water or octane?
Figure 13.3
Like dissolves like: solubility of methanol in water.
The H-bonding force of attraction in water
and in methanol are similar in type and
strength, so they can substitute for one
another. Thus, methanol is soluble in
water; in fact, the two substances are
100% miscible in each other.
water
methanol
A solution of
methanol in water
Factors affecting solubility:
3.
Solids - must also be "like" the solvent
Glucose dissolves in water because of
extensive Hydrogen-bonding
Energy from attraction approx = energy to
break water to water H-bonding
I2 is not as soluble - London forces involved
CCl4 also London forces. I2 dissolves in
carbon tetrachloride, but glucose does not
SAMPLE PROBLEM 13.1
PROBLEM:
Predicting Relative Solubilities of Substances
Predict which solvent will dissolve more of the given solute:
(a) Sodium chloride in methanol (CH3OH) or in propanol (CH3CH2CH2OH)
(b) Ethylene glycol (HOCH2CH2OH) in hexane (CH3CH2CH2CH2CH2CH3)
or in water.
(c) Diethyl ether (CH3CH2OCH2CH3) in water or in ethanol (CH3CH2OH)
SOLUTION:
(a) Methanol. NaCl is ionic and will form ion-dipoles with the -OH groups of
both methanol and propanol. However, propanol is subject to London
dispersion forces to a greater extent.
(b) Water. Hexane has no dipoles to interact with the -OH groups in ethylene
glycol. Water can experience Hydrogen bonding attraction with ethylene glycol.
(c) Ethanol. Diethyl ether can interact through a dipole and London dispersion
forces. Ethanol can provide both while water would like to Hydrogen bond.
Practice

Work on problems 7 and 11 in chapter 13
Figure 13.14 from 4th ed.
The cyclic structure of b-D-glucopyranose
in aqueous solution.
Solid glucose is
named just Dglucose and is a
linear arrangement.
In water it reacts with
itself, C1 and C5, to
form a hemiacetal.
There are three processes involved in
dissolving a solute in a solvent
1. Have to separate solvent molecules: an
endothermic process
2. Have to separate solute molecules or ions:
another endothermic process
3. Surround solute particles with solvent
particles, an exothermic process. If solvent is
water, this is called Hydration. This step may
give back a lot of the energy that was taken in
by the first two processes.
Figure 13.2
Hydration shells around an aqueous ion.
When an ionic compound
dissolves in water, iondipole forces orient water
molecules around the
separated ions to form
hydration shells. The
cation shown here is
octahedrally surrounded
by six water molecules,
which form H-bonding
attractions with water
molecules in the next
hydration shell, and those
form H-bonding attractions
with others farther away.
Enthalpy of solution
For NaCl: NaCl(s)
lattice
energy
DHsoln
Na+(aq) + Cl-(aq)
enthalpy of
hydration (DHhyd)
Na+(g) + Cl-(g)
Hess’s Law tells us that DHsoln = sum of LE and DHhyd
Given that Lattice Energy is 786 kJ/mol and DHhyd is –783
kJ/mol, then DHsoln will be +3 kJ/mol or slightly endothermic.
If heat of hydration is about equal to Lattice Energy, the ionic
compound will dissolve.
Look at AgCl: LE = 916 kJ/mol, DHhyd = -851 kJ/mol, sum =
DHsoln = +65 kJ/mol to dissolve it, so AgCl is not soluble
Figure 13.5
Dissolving ionic compounds in water.
Endothermic for
NH4NO3
NaCl –
slightly
endothe
rmic
The enthalpy diagram for an
ionic compound in water
includes DHlattice (DHsolute;
always positive) and the
combined ionic heats of
hydration (DHhydr; always
negative).
NH4NO3
EXOTHERMIC!
NaOH
Figure 13.6
Enthalpy diagrams for dissolving NaCl and octane in hexane.
A, Because attractions between Na
(or Cl) ions and hexane molecules
are weak, DHmix is much smaller than
DHsolute. Thus, DHsoln is so positive
that NaCl does not dissolve in
hexane. B, Intermolecular forces in
octane and in hexane are so similar
that DHsoln is very small. Octane
dissolves in hexane because the
solution has greater entropy than the
pure components.
Very endothermic – NaCl will
not dissolve in hexane.
Temperature does cause solubility to vary for
all substances
Temperature increase ALWAYS causes gas
solubility to decrease
Sometimes increase in T can increase solubility
of a solid or liquid in liquid
Dissolving solids can be endothermic, so adding
heat causes solubility to increase
- If it was exothermic, adding heat may cause
solubility to decrease
Figure 13.9
The relation between solubility and temperature for
several ionic compounds.
Most ionic compounds
have higher solubilities at
higher temperatures. Cerium
sulfate is one of several
exceptions.
EFFECT OF P & T ON SOLUBILITY:
Le Chatelier's Principle - a change in any of the factors
determining an Equilibrium will cause the system to
adjust in order to counteract the effect of the change
as much as possible. (You will see this again soon!)
Pressure has little effect of liquid or solid in water, but
gases are different.
Picture two cylinders with pistons, same volume of
water, same moles of CO2 gas total.
ALL (nonreactive nondissociative) gases become more
soluble at higher Pressure.
Figure 13.10
The effect of pressure on gas solubility.
A, A saturated solution of a gas is in equilibrium at pressure P1. B, If the pressure is
increased to P2, the volume of the gas decreases. Therefore, the frequency of
collisions with the surface increases. C, As a result, more gas is in solution when
equilibrium is re-established.
Henry’s Law
Sgas = kH X Pgas
The solubility of a gas (Sgas) is
directly proportional to the
partial pressure of the gas
(Pgas) above the solution.
SAMPLE PROBLEM 13.2
PROBLEM:
SOLUTION:
Using Henry’s Law to Calculate Gas Solubility
The partial pressure of carbon dioxide gas inside a bottle of
cola is 4.0 atm at 250C. What is the solubility of CO2? The
Henry’s law constant for CO2 dissolved in water is 3.3 x10-2
mol/L*atm at 250C.
S
CO2
= (3.3 x10-2 mol/L*atm)(4 atm) = 0.13 mol/L
(Yes, it’s really this easy!)
Practice

Work on problems 15, 16, 25, and 34 in
chapter 13.
CONCENTRATION: AMOUNT OF SOLUTE IN
GIVEN QUANTITY OF SOLVENT OR
SOLUTION
Units of concentration:
M = Molarity = moles of solute per liter of solution
mass-percent = (mass of solute per mass of solution)x100
vol-percent = (vol of solute per volume of sol'n)x100
m = molality = moles of solute per kilogram of solvent
mole fraction = moles of solute per total moles
***Solubility = grams of solute per 100 grams of solvent
moles/L
%-wt(or mass)
%-vol
moles/kg
fraction,c
g/100g
***Normality: old concept, skip text, use N = M * (#H) if acid or
N = M * (#OH) if base
Units: N (normal)
***LEARN SOLUBILITY IN g/100g and NORMALITY:
NOT IN TEXTBOOK
Examples:
Mass percent: 3.5% NaCl solution means 3.5 g NaCl in
100 g solution
Preparation: put in 3.5 g, add water to make 100 grams.
(96.5g, NOT 100 g of water!)
How would you prepare 425 g of a 2.40%-wt aqueous
solution of sodium acetate? Molar mass NOT
needed!!!
EXAMPLES
Molality:
molal = mol solute/kg solvent
What is molality of a solution where 0.20 mol ethylene glycol is
dissolved in 2.0 x 103 g of H2O?
molality = 0.20 mol/2.0 kg = 0.10 molal
(little m can mean meter or milli, so write out molal)
TRY TO KEEP M AND m STRAIGHT!
Calculate the molality of a solution which has 4.57 g glucose in
25.2 g of H2O. (Turn the answer in to the instructor.)
EXAMPLES
Mole fraction: ci = molessubstance A/total moles
What is the mole fraction of ethylene glycol if a
solution has 1 mole ethylene glycol in 9 mol
H2O?
Total moles = 10, ci = 1/10 = 0.10
What is the mole fraction of glucose if 4.57 g
glucose is dissolved in 25.2 g H2O? (Turn the
answer in to the instructor.)
SAMPLE PROBLEM 13.4
Expressing Concentration in Parts by Mass,
Parts by Volume, and Mole Fraction
PROBLEM: (a) Find the concentration of calcium (in ppm) in a 3.50-g pill that
contains 40.5 mg of Ca.
(b) The label on a 0.750-L bottle of Italian Chianti indicates
“11.5% alcohol by volume”. How many liters of alcohol does the
wine contain?
(c) A sample of rubbing alcohol contains 142 g of isopropyl
alcohol (C3H7OH) and 58.0 g of water. What are the mole
fractions of alcohol and water?
SAMPLE PROBLEM 13.4
Expressing Concentrations in Parts by Mass,
Parts by Volume, and Mole Fraction
continued
g
SOLUTION:
(a)
40.5 mg Ca x
103 mg
x 106
= 1.16x104 ppm Ca
3.5 g
(b)
(c)
11.5 L alcohol
0.750 L Chianti x
100 L Chianti
moles ethylene glycol = 142 g
moles water = 38.0g
2.39 mol C2H8O2
2.39 mol C2H8O2 + 3.22 mol H2O
= 0.423
cC2H6O2
= 0.0862 L alcohol
mole
= 2.36 mol C2H6O2
60.09 g
mole
= 3.22 mol H2O
18.02 g
3.22 mol H2O
2.39 mol C2H8O2 + 3.22 mol H2O
= 0.577
cH2O
CONVERSIONS BETWEEN CONCENTRATION
UNITS:
Given that you have 0.100 L of ethanol/H2O
solution made with 10.00 mL of ethanol and
the Deth = 0.789g/mL, Dsoln= 0.982g/mL. Find
a) vol %, b) mass %, c) molarity, d) molality,
and e) mole fraction.
Other conversions:
1. Converting molality to mole fraction: given a 0.120 molal
glucose solution (0.120 mol glucose/1.00 kg H2O), find ci for
glucose and water.
mol H2O = 1000 g/18.015 g/mol = 55.51 mol
Total moles = 0.120 mol + 55.51 mol = 55.63
cglucose = 0.120 mol glucose/55.63 mol = 0.00216
cwater = 55.51 mol/55.63 mol = 0.998
2. Converting mole fraction to molal: Given a solution that is
0.150 mole fraction glucose and 0.850 mole fraction water,
find molality. (So assume you have a total of 1.000 moles.)
Convert water to kg: 0.850 mol x 18.015 g/mol = 15.3 g
molality = 0.150 mol/0.0153 kg = 9.80 molal
Other conversions:
3. Convert molality to Molarity: need D of solution,
Given a 0.273 molal KCl sol'n, D = 1.011 g/mL,
find Molarity.
molality = 0.273 mol KCl/1.00 kg water
Find mass KCl: 0.273 mol x 74.553 g/mol = 20.35 g
Total mass = 1000.0 g + 20.35 g = 1020.35 g
Volume = 1020.35g/1.011 g/mL = 1009.25 mL
Molarity = 0.273 mol/1.00925 L = 0.2705 M
Other conversions:
4. Convert Molarity to molality:
Given a 0.907 M Pb(NO3)2 solution with D =
1.252 g/mL, find molality.
Mass of solution: 1000 mL x 1.252 g/mL =
1252 g
0.907 mol x 331.2 g/mol = 300.4 g
Mass water = 1252 – 300.4 = 951.6 g
molality = .907 mol/0.9516 kg = 0.953 molal
Practice

Do problem 56 in chapter 13.
COLLIGATIVE PROPERTIES
COLLIGATIVE PROPERTIES: the effects of solutes
on four physical properties of the solvents: VP, BP,
FP (MP), and osmotic pressure
VP, vapor pressure: rate of evaporation decreases and
VP at a given T decreases
***Raoult's Law: VP lowering is proportional to
mole fraction of solvent, which is always <1.0
Psoln = csolv * Posolv for a solid nonvolatile solute
(Or you can use two steps: DP = csolute*Posolv
and then Psoln = Posolv - DP
Figure 13.25
from 4th ed.
The three types of electrolytes.
STRONG
Review: what makes a strong vs.
a weak electrolyte?
nonelectrolyte
weak
Figure 13.11
The effect of a solute on the vapor pressure of a solution.
Solute particles are interfering with solvent particles trying
to go into vapor phase: affects VP, BP, etc.
A, Equilibrium is established between a
pure liquid and its vapor when the
numbers of molecules vaporizing and
condensing in a given time are equal.
B, The presence of a dissolved solute decreases the number of
solvent molecules at the surface so fewer solvent molecules
vaporize in a given time. Therefore, fewer molecules need to
condense to balance them, and equilibrium is established at a
lower vapor pressure.
SAMPLE PROBLEM 13.6
PROBLEM:
Calculate the vapor pressure lowering, DP, when 10.0 mL of
glycerol (C3H8O3) is added to 500. mL of water at 50.0C. At this
temperature, the vapor pressure of pure water is 92.5 torr and
its density is 0.988 g/mL. The density of glycerol is 1.26 g/mL.
SOLUTION:
DP = csolute*Posolv
10.0 mL C3H8O3
500.0 mL H2O
DP =
Using Raoult’s Law to Find the Vapor Pressure
Lowering
1.26 g C3H8O3
x
mL C3H8O3
0.988 g H2O
mL H2O
x
mol C3H8O3
92.09 g C3H8O3
mol H2O
18.02 g H2O
0.137 mol C3H8O3
x
92.5 torr
= 0.137 mol C3H8O3
= 27.4 mol H2O
= 0.461 torr
0.137 mol C3H8O3 + 27.4 mol H2O
cglycerol = 0.00498
Psoln = 92.5 - 0.461 = 92.039 torr
Raoult’s Law for two liquids
If 2 or more liquids are involved, say A & B, then
PA=cA*PoA and PB=cB*PoB
THIS HOLDS TRUE FOR IDEAL SOLUTIONS
WHICH OBEY THESE LAWS:
- Intermolecular forces between solute particles and
solvent are same as between two solvent particles,
leading to NO volume change and no DHsoln
HOWEVER, NOT ALL LIQ/LIQ SOLNS ARE
IDEAL!
If there's an enhanced attraction like Hydrogen-bonding
between solute and solvent, the observed vapor
pressure is even lower than expected.
Raoult’s Law for two liquids
EXAMPLE: A solution of 0.500 mol benzene and 0.500 mol toluene is in a
distillation apparatus. At 25°C (before heating), what will be the
equilibrium VP's and concentrations of each? Given PoB = 95.1 torr,
PoToluene = 28.4 torr; PB+PToluene=Ptot
0.500 mol * 95.1 torr + 0.500 mol * 28.4 torr = 61.8 torr
Mole fraction in vapor using partial pressure of gas:
cToluene = PTolene/Ptot = 14.2/61.8 = 0.230 toluene
cB = PB/Ptot = 47.6/61.8 = 0.770 benzene
If we heated the solution, the mole fraction of benzene would be even higher!
Then if we cooled the vapor through a condenser, collected it and repeated the
process, we could approach making pure benzene. This is called fractional
distillation.
Practice Raoult’s Law
A solution contains 20.00 grams of methanol in
100.0 g of ethanol. The vapor pressures are:
MeOH = 94 torr, EtOH = 44 torr. Find the
vapor pressure of each alcohol and the total
pressure over the solution. Also find the mole
fraction of each alcohol in the vapor phase.
(PMeOH = 21 torr, PEtOH = 34 torr, Pt = 55 torr
XMeOH = 21/55 = 0.38, XEtOH = 34/55 = 0.62
Figure 13.12
Phase diagrams of solvent and solution.
Phase diagrams of an
aqueous solution
(dashed lines) and of
pure water (solid lines)
show that, by lowering
the vapor pressure (P), a
dissolved solute elevates
the boiling point (Tb) and
depresses the freezing
point (Tf).
More Colligative Properties:
FP & BP
The VP effect causes the FP and BP of solutions to change also.
FP depression: solvent begins to crystallize before solute does,
solute interferes and gets in the way of the crystal lattice. Two
step calculation:
Step 1. DTf = Kf*molality
and Step 2. new FP = FPo - DTf for
nonvolatile covalent solutes
BP elevation: because VP is lower at any given T, must have
higher T at 1 atm to boil. Similar two-step calculation:
Step 1. DTb = Kb*molality and Step 2. new BP = BPo + DTb
Table 13.5
Molal Boiling Point Elevation and Freezing Point
Depression Constants of Several Solvents
Typo in text
Boiling
Point (0C)*
Kb (0C/m)
Melting
Point (0C)
117.9
3.07
16.6
3.90
Benzene
80.1
2.53
5.5
4.90
Carbon disulfide
46.2
2.34
-111.5
3.83
Carbon tetrachloride
76.5
5.03
-23
Chloroform
61.7
3.63
-63.5
4.70
Diethyl ether
34.5
2.02
-116.2
1.79
Ethanol
78.5
1.22
-117.3
1.99
100.0
0.512
0.0
1.86
Solvent
Acetic acid
Water
*at 1 atm.
Kf (0C/m)
30.
Practice with FP & BP
Calculate both the new FP & BP for a solution that has
100.0 g of glucose in 500.0 mL of water, given the
density of water is 0.998 g/mL at room temperature
kg water = 500.0 mL (0.998 g/mL)(1 kg/1000g) = 0.499 kg
molality = 100.0 g (1mol/180.16 g)/0.499 kg = 1.112 molal
DTf = 1.86oC/molal (1.112 molal) = 2.068oC
new FP = 0.00 – 2.068oC = -2.068oC
DTb = 0.512oC/molal (1.112 molal) = 0.569oC
new BP = 100.000 + 0.569 = 100.569oC
You practice again!
A 24.0 g sample of an organic compound, molar
mass 58.0 g/mol, is added to 600.0 g of water.
Since barometric pressure is low that day, the
boiling point of pure water is 99.725oC. What
is the boiling point of the solution?
(m = 0.690 mol/kg; DTb is 0.354, so new Tb is
100.079oC)
Another Colligative Property:
Osmotic Pressure
Osmosis is the movement of solvent molecules thru a
semi permeable membrane from a region of lower
solute concentration to a region of higher solute conc.
Semi perm membrane is a thin sheet of material thru
which only the solvent molecules in a solution can
pass in either direction.
Osmotic Partial Pressure is the force exerted by solvent
molecules passing thru a semi perm membrane in a
solution system at equilibrium. Symbol is p .
If a nonvolatile covalent solute is involved, p = MRT.
This can be used to find the molar mass of a solute!
Figure 13.13
The development of osmotic pressure.
osmotic pressure
pure
solvent
solution
semi permeable
membrane
net movement of solvent
solute
molecules
solvent
molecules
See notes below or in textbook.
Applied pressure
needed to prevent
volume increase
SAMPLE PROBLEM 13.8
Determining Molar Mass from Osmotic Pressure
PROBLEM: A physician studying a mutated variety of hemoglobin
associated with a fatal disease first finds its molar mass (M).
She dissolves 21.5 mg of the protein in water at 5.00C to make
1.50 mL of solution and measures an osmotic pressure of 3.61
torr. What is the molar mass of this variety of hemoglobin?
SOLUTION:
p
M=
3.61 torr
=
RT
atm
760 torr
= 2.08 x10-4 M
(0.082057 L*atm/mol*K)(278.1 K)
L
2.08 x10-4 mol (1.50 mL)
= 3.12x10-8 mol
L
103 mL
21.5 mg
g
103 mg
1
= 6.89 x104 g/mol
3.12 x10-8 mol
Hemoglobin is a protein – a biological polymer of high M.
Practice with Osmosis
1. Find the osmotic pressure at 17.0oC for a
solution containing 1.78 g of sucrose in a
150.0 mL solution. (Turn the work into the
instructor, showing how to get the answer
given below.)
(0.8254 atm or 627 torr)
2. Find the molar mass of a polymer called PIB
(polyisobutene) given that a solution
containing 0.200 g PIB per 100.0 mL of
benzene has a height is an osmosis tube of
2.40 mm when compared to pure benzene.
The density of the solution is 0.880 g/mL.
(Turn the work into the instructor, showing
how to get the answer given below.)
(2.4 x 105 g/mol)
Ionic Compounds and
Colligative Properties
We have to define what happens when an ionic solute is
dissolved - it contributes more particles than a
covalent solute does (glucose vs. NaCl).
We define colligative molality and Molarity to be the
given molality or Molarity times the number of solute
particles or the van’t Hoff factor.
Colligative molality or Molarity:
molal * i or M * I
where i = van't Hoff factor
Colligative Properties of
Electrolytic Solutions
For electrolyte solutions, the compound formula tells us
how many particles are in the solution.
The van’t Hoff factor, i, tells us what the “effective”
number of ions are in the solution (see table).
For vapor pressure lowering: DP = i(csolute*P0solv)
For boiling point elevation:
DTb = i(b*m)
For freezing point depression:DTf = i(f*m)
For osmotic pressure :
p = i(MRT)
Figure 13.14
Nonideal behavior of
electrolyte solutions.
(Expected factors vs.
measured factors for
effect of ionic
compounds on
colligative properties.)
Figure 13.15
An ionic atmosphere model for nonideal behavior of
electrolyte solutions.
Hydrated anions cluster
near cations, and vice
versa, to form ionic
atmospheres of net
opposite charge.
Because the ions do not
act independently,
their concentrations are
effectively less than
expected. Such
interactions cause
deviations from ideal
behavior.
SAMPLE PROBLEM 13.9
Depicting a Solution to Find Its
Colligative Properties
PROBLEM: A 0.952-g sample of magnesium chloride is dissolved in 100. g
of water in a flask.
(a) Which scene depicts the solution best?
(b) What is the amount (mol) represented
by each green sphere?
(c) Assuming the solution is ideal, what is its freezing point (at 1 atm)?
SAMPLE PROBLEM 13.9
Depicting a Solution to Find Its
Colligative Properties
continued
(a) The formula for magnesium chloride
is MgCl2; therefore the correct depiction
must be A with a ratio of 2 Cl-/ 1 Mg2+.
0.952 g MgCl2
(b)
mols MgCl2 =
95.21 g MgCl2
= 0.0100 mol MgCl2
mol MgCl2
mols
Cl-
=
0.0100 mol MgCl2 x
mols/sphere =
0.0200 mols Cl8 spheres
2 mols Cl1 mol MgCl2
= 0.0200 mols Cl-
= 2.50 x 10-3 mols/sphere
SAMPLE PROBLEM 13.9
Depicting a Solution to Find Its
Colligative Properties
continued
(c)
0.0100 mol MgCl2
molality (m) =
100. g x
= 0.100 m MgCl2
1 kg
103 g
Assuming this is an IDEAL solution, the van’t Hoff factor, i, should be 3.
Using just the number of particles expected: 1 Mg2+ and 2 Cl-:
DTf = i (Kf*m) = 3(1.86 0C/m x 0.100 m) = 0.558 0C
Tf = 0.000 0C - 0.558 0C = - 0.558 0C
Using the van’t Hoff factor:
DTf = i (Kf*m)= 2.7(1.86 0C/m x 0.100 m) = 0.502 0C
Tf = 0.000 0C - 0.5o2 0C = - 0.502 0C
More Practice
Given a solution that has 20.0 g of MgCl2 in 500.0 mL
of water with Dwater = 0.998 g/mL. Find molarity and
use Van’t Hoff factor to find osmotic pressure of
solution at 25.0oC.
M = (20.0 g/95.216 g/mol)/.500 L = 0.4201 M
Look up Van’t Hoff factor in table.
Now find the other three properties as well: FP, BP and
VP of the solution.
FP & BP PROBLEMS: Turn in Answers
1. Given 864 g of NaCl in 5.50 kg of water. Find new
MP & BP.
(m = 2.69 molal)
2. Given 25.0 g of CaCl2 in 500 g of water. Find new
FP & BP.
(m = 0.450 molal)