Download The Properties of Mixtures: Solutions and Colloids Dr.ssa Rossana

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The Properties of Mixtures:
Solutions and Colloids
Dr.ssa Rossana Galassi
320 4381420
[email protected]
13-1
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The Properties of Mixtures: Solutions and Colloids
1 Types of Solutions: Intermolecular Forces and Predicting Solubility
2 Intermolecular Forces and Biological Macromolecules
3 Why Substances Dissolve: Understanding the Solution Process
4 Solubility as an Equilibrium Process
5 Quantitative Ways of Expressing Concentration
6 Colligative Properties of Solutions
7 The Structure and Properties of Colloids
13-2
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Solutions
• Solutions are homogeneous mixtures obtained
by mixing two or more substances in a single
phase.
• By convention the component in the largest
amount is identified as the solvent and the
other component as the solute.
13-3
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Types of solutions
•
•
•
•
•
•
13-4
Solid solid
Gas solid
Gas gas
Gas liquid
Liquid liquid
Solid liquid
alloy
clathrate
air
river/sea water
alcoholic drinks
salty water
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The arrangement of atoms in two types of alloys.
brass - a substitutional alloy
13-5
carbon steel -an interstitial alloy
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Clathrate hydrates constitute a class
of solids in which the guest molecules
occupy, fully or partially, cages in host
structures made up of H-bonded
water molecules. The usually unstable
empty clathrate is stabilised by
inclusion of the guest species. In case
of guest molecules which are gaseous
at ambient conditions the resulting
clathrate hydrate is often called a
gas hydrate
13-6
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Preparing a solution : dissolving a solid in a liquid
Solvent = water
Solute = CuCl2
13-7
Interactions between
water molecules and
Cu2+ and Cl- ions
allow the solid to
dissolve
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Preparing a solution : diluting a more concentrated
solution
13-8
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A solution does not scatter light
A colloidal dispersion does.
13-9
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Units of concentration
The concentration of a solution
represents the amount of solute
dissolved in a certain amount of
solvent or solution at a fixed
temperature.
Molarity (M, mol/L)
Molality (m, mol/Kg)
Percent (w/w; w/V; V/V)
Normality (N, eq/L)
Mole fraction (, n/n+N)
Parts per milion (ppm, mg/L)
13-10
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Molarity
M = number of mole of solute / 1 liter of solution
Normality
N = number of equivalent / 1 liter of solution
molality
m = number of mole of solute / 1Kg of solvent
Mole fraction
 =n/n+N
Weight percent
%W/W = mass of A / mass of A + mass of B + etc..
Volume percent
%V/V = volume of A / volume of A + volume of B +
Weight/volume
% W/V = weight of A / volume of A + volume of B
13-11
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Normality
number of equivalent /1 liter of solutions
This unit of concentration is particularly usefull in titrations.
The number of equivalent takes into account the valence of
the chemical system in the chemical reaction occurring in the
titration. In example :
Zn0 + 2H+ => H2 + Zn2+
In this redox reaction occurs an exchange of 2 electrons, so
the equivalent mass is given by MM/2 and the number of
equivalent are equal to eq = grams /EM
2NaOH + H2SO4 => Na2SO4 + H2O
The equivalent number of H2SO4 is equal to eq = grams /
MM/2 taking into account the two H+ exchanged during the
acid base reaction.
13-12
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The solution process
LIKE DISSOLVES LIKE
Substances giving similar types of
intermolecular forces dissolve in each
other.
When a solute dissolves in a solvent,
solute-solute interactions and solventsolvent interactions are being replaced
with solute-solvent interactions. The
forces must be comparable in strength
in order to have a solution.
13-13
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The major types of intermolecular forces in
solutions.
13-14
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Hydration shells around an aqueous ion. Salt in water…..
13-15
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13-16
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Like dissolves like: solubility of methanol in water.
water
methanol
A solution of
methanol in water
13-17
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Liquids dissolving in liquids
If two liquids mix to an appreciable extent to
form a solution, they are said to be miscibile.
In constrast, immiscibile liquids do not mix to
form a solution, they exist in contact with
each other as separated layers.
13-18
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I2 is a non polar molecule.
In the left tube it dissolves
in water (superior layer)
after mixing it prefers the
CCl4 (it is non polar and
non miscibile with water)
Less dense octane
CuSO4 in
water
CuSO4 in water
More dense CCl4
13-19
CCl4 and
octane
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The solution process
When the solid is added to the water it starts to
dissolve till a point when no additional solid seems
to dissolve. The solution is saturated and the
concentration is the maximum reachable in those
conditions
(Solubility of CuCl2, 70.6 g in 100 mL of H2O at 20°C)
13-20
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Equilibrium in a saturated solution.
solute (undissolved)
13-21
solute (dissolved)
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Heats of solution and solution cycles
1. Solute particles separate from each other - endothermic
solute (aggregated) + heat
solute (separated)
Hsolute > 0
2. Solvent particles separate from each other - endothermic
solvent (aggregated) + heat
solvent (separated) Hsolvent > 0
3. Solute and solvent particles mix - exothermic
solute (separated) + solvent (separated)
<0
solution + heat
Hsoln = Hsolute + Hsolvent + Hmix
13-22
Hmix
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Solution cycles and the enthalpy components of the heat of solution.
Exothermic solution process
Endothermic solution process
13-23
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Heats of solution
so
The enthalpy of solution is the algebric sum
of all the enthalpies involved in the solution
process.
When a solid dissolves in a liquid the solid
separation is actually: Hsolute = Hlattice.
Then, high Hlattice generally corresponds to
low solubility
high Hlattice are for highly charged and small
ions
HlatticeMgS > HlatticeKCl
13-24
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Heats of Hydration
The solvation of ions by water is always exothermic
M+ (g) [or X-(g)]
H2O
M+(aq) [or X-(aq)] Hhydr of the ion < 0
Hhydr is related to the charge density of the ion, that is,
coulombic charge and size matter (ion radius)
Lattice energy is the H involved in the formation of an ionic solid
from its gaseous ions.
M+ (g) + X-(g)
13-25
MX(s) Hlattce is always (-)
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Trends in Ionic Heats of Hydration
Ion
Ionic Radius (pm)
Hhydr (kJ/mol)
Group 1A(1)
Li+
Na+
K+
Rb+
Cs+
76
102
138
152
167
-510
-410
-336
-315
-282
72
100
118
-1903
-1591
-1424
133
181
196
220
-431
-313
-284
-247
Group 2A(2)
Mg2+
Ca2+
Sr2+
Ba2+
13-26
Group 7A(17)
FClBrI-
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Dissolving ionic compounds in water.
NaCl
13-27
NH4NO3
NaOH
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Enthalpy diagrams for dissolving NaCl and octane in hexane.
13-28
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The relation between solubility and temperature for
several ionic compounds.
13-29
• For many solids dissolved in
liquid water, the solubility
increases with temperature.
• The increase in kinetic energy
that comes with higher
temperatures allows the solvent
molecules to more effectively
break apart the solute
molecules that are held
together by intermolecular
attractions.
• The increased vibration (kinetic
energy) of the solute
molecules causes them to
dissolve more readily because
they are less able to hold
together
• Na2SO4 and Ce2(SO4)3
solubilities decreases with the
increase of temperature
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The structure and function of a soap.
13-30
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Factors affecting the solubility of
gases in liquid: pressure and
temperature
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.
13-31
Some values for kH for gases
dissolved in water at 298 K include:
oxygen (O2) : 769.2 L·atm/mol
carbon dioxide (CO2) : 29.41
L·atm/mol
hydrogen (H2) : 1282.1 L·atm/mol
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The effect of pressure on gas solubility.
13-32
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SAMPLE PROBLEM
PROBLEM:
PLAN:
The partial pressure of carbon dioxide gas inside a bottle of
cola is 4 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.
Knowing kH and Pgas, we can substitute into the Henry’s law
equation.
SOLUTION:
S
CO2
13-33
Using Henry’s Law to Calculate Gas Solubility
= (3.3 x10-2 mol/L*atm)(4 atm) = 0.1 mol/L
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Table Concentration Definitions
Concentration Term
Molarity (M)
Molality (m)
Ratio
amount (mol) of solute
volume (L) of solution
amount (mol) of solute
mass (kg) of solvent
Parts by mass
Parts by volume
mass of solute
mass of solution
volume of solute
volume of solution
Mole fraction 
amount (mol) of solute
amount (mol) of solute + amount (mol) of solvent
13-34
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SAMPLE PROBLEM
PROBLEM:
Calculating Molality
What is the molality of a solution prepared by dissolving 32.0 g
of CaCl2 in 271 g of water?
PLAN: We have to convert the grams of CaCl2 to moles and the grams of
water to kg. Then substitute into the equation for molality.
SOLUTION:
32.0 g CaCl2 x
mole CaCl2
110.98 g CaCl2
0.288 mole CaCl2
molality =
kg
271 g H2O x
13-35
103 g
= 0.288 mole CaCl2
= 1.06 m CaCl2
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SAMPLE PROBLEM
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?
PLAN:
13-36
(a) Convert mg to g of Ca, find the ratio of g Ca to g pill and multiply
by 106.
(b) Knowing the % alcohol and total volume, we can find volume of
alcohol.
(c) Convert g of solute and solvent to moles; find the ratios of parts
to the total.
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SAMPLE PROBLEM
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
13-37
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
H2O
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Interconverting Concentration Terms
To convert a term based on amount (mol) to one based on
mass, you need the molar mass. These conversions are
similar to mass-mole conversions.
To convert a term based on mass to one based on
volume, you need the solution density. Working with the
mass of a solution and the density (mass/volume), you can
obtain volume from mass and mass from volume.
Molality involves quantity of solvent, whereas the other
concentration terms involve quantity of solution.
13-38
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SAMPLE PROBLEM
Converting Concentration Units
PROBLEM: Hydrogen peroxide is a powerful oxidizing agent used in
concentrated solution in rocket fuels and in dilute solution a a
hair bleach. An aqueous solution H2O2 is 30.0% by mass and
has a density of 1.11 g/mL. Calculate its
(a) Molality
(b) Mole fraction of H2O2
(c) Molarity
PLAN: (a) To find the mass of solvent we assume the % is per 100 g of
solution. Take the difference in the mass of the solute and solution for
the mass of peroxide.
(b) Convert g of solute and solvent to moles before finding .
(c) Use the density to find the volume of the solution.
SOLUTION:
(a)
g of H2O = 100. g solution - 30.0 g H2O2 = 70.0 g H2O
30.0 g H2O2
34.02 g H2O2
molality =
70.0 g H2O
13-39
mol H2O2
kg H2O
103 g
0.882 mol H2O2
= 12.6 m H2O2
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SAMPLE PROBLEM
Converting Concentration Units
continued
(b)
70.0 g H2O
mol H2O
= 3.88 mol H2O
18.02 g H2O
0.882 mol H2O2
= 0.185  of H2O2
0.882 mol H2O2 + 3.88 mol H2O
(c)
mL
100.0 g solution
= 90.1 mL solution
1.11 g
0.882 mol H2O2
90.1 mL solution
L
103 mL
13-40
= 9.79 M H2O2
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Dilution law
When a solution is diluted (or concentrated) the amount of
solute does not change (just the solvent is added or
removed).
If C is the molar concentration term of a solution and V its
volume, the product of CV is the number of mole
CV = n (number of mole of solute). Upon dilution or
concentration C and V change their values but n is equal.
Accordingly to that :
C1 V1= C2V2= n
Es: If 200 mL of a 0.1 M solution, 200 mL of water are added, how much
will be the final concentration?
V1 = 200 mL, V2 = 400 mL; C1 = 0.1 M, C2 = ?
C2 = C1V1/C2 = 200 x 0.1 /400 = 0.05 M
13-41
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Colligative Properties
By considering the properties of liquids, such as the vapour pressure, the
boiling and the freezing temperatures the colligative properties are
involved in the study of the change of these properties upon addition of a
solute to a solvent (pure liquid) to form a solution.
13-42
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Colligative Properties
Raoult’s Law (vapor pressure of a solvent above a solution, Psolvent)
Psolvent = solvent X P0solvent
where P0solvent is the vapor pressure of the pure solvent
P0solvent - Psolvent = P = solute x P0solvent
Boiling Point Elevation and Freezing Point Depression
Tb = Kbm
Tf = Kfm
Osmotic Pressure
 = M R T where M is the molarity, R is the ideal gas
law constant and T is the Kelvin temperature
13-43
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The three types of electrolytes.
STRONG
nonelectrolyte
weak
13-44
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The effect of a solute on the vapor pressure of a solution.
13-45
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SAMPLE PROBLEM
PROBLEM:
PLAN:
Using Raoult’s Law to Find the Vapor Pressure
Lowering
Calculate the vapor pressure lowering, P, 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.
Find the mol fraction, , of glycerol in solution and multiply by the
vapor pressure of water.
SOLUTION:
10.0 mL C3H8O3
500.0 mL H2O
1.26 g C3H8O3
x
mL C3H8O3
0.988 g H2O
mL H2O
x
mol C3H8O3
= 0.137 mol C3H8O3
92.09 g C3H8O3
mol H2O
18.02 g H2O
= 27.4 mol H2O
 = 0.00498
P =
0.137 mol C3H8O3
0.137 mol C3H8O3 + 27.4 mol H2O
13-46
x
92.5 torr
= 0.461 torr
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Phase diagrams of solvent and solution.
13-47
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Table Molal Boiling Point Elevation and Freezing Point
Depresssion Constants of Several Solvents
Kb (0C/m)
Melting
Point (0C)
Kb (0C/m)
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
Boiling
Point (0C)*
Acetic acid
Water
*at 1 atm.
13-48
30.
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SAMPLE PROBLEM
Determining the Boiling Point Elevation and
Freezing Point Depression of a Solution
PROBLEM: You add 1.00 kg of ethylene glycol (C2H6O2) antifreeze to your
car radiator, which contains 4450 g of water. What are the
boiling and freezing points of the solution?
Find the # mols of ethylene glycol; m of the solution; multiply by
the boiling or freezing point constant; add or subtract,
respectively, the changes from the boiling point and freezing point
of water.
SOLUTION:
mol C2H6O2
1.00x103 g C2H6O2
= 16.1 mol C2H6O2
62.07 g C2H6O2
PLAN:
16.1 mol C2H6O2
4.450 kg H2O
Tbp = 0.512 0C/m x 3.62m
BP = 101.85 0C
13-49
= 3.62 m C2H6O2
= 1.850C
Tfp =
1.86 0C/m x 3.62m
FP = -6.73 0C
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Osmotic Pressure
The symbol for osmotic pressure is .

nsolute
or
 M
Vsolution

13-50
nsolute
=
Vsolution
RT
= MRT
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The development of osmotic pressure.
osmotic pressure
pure
solvent
solution
semipermeable
membrane
net movement of solvent
solute
molecules
solvent
molecules
13-51
Applied pressure
needed to prevent
volume increase
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13-52
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13-53
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SAMPLE PROBLEM
Determining Molar Mass from Osmotic Pressure
PROBLEM: Biochemists have discovered more than 400 mutant varieties of
hemoglobin, the blood protein that carries oxygen throughout
the body. A physician studing a variety 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?
We know as well as R and T. Convert  to atm and T to degrees K.
Use the  equation to find M and then the amount and volume of the
sample to get to M.
atm
3.61 torr

SOLUTION:
M=
=
= 2.08 x10-4 M
760 torr
RT
(0.0821 L*atm/mol*K)(278.1 K)
L
2.08 x10-4 mol (1.50 mL)
= 3.12x10-8 mol
L
103 mL
PLAN:
21.5 mg
13-54
g
103 mg
1
= 6.89 x104 g/mol
3.12 x10-8 mol
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Colligative Properties of Electrolyte Solutions
For electrolyte solutions, the compound formula
tells us how many particles are in the solution.
The van’t Hoft factor, i, tells us what the “effective”
number of ions are in the solution.
van’t Hoff factor (i)
i
13-55
measured value for electrolyte solution
=
expected value for nonelectrolyte solution
For vapor pressure lowering:
P = i(solutex P0solvent)
For boiling point elevation:
Tb = i(bm)
For freezing point depression:
Tf = i(fm)
For osmotic pressure :
 = i(MRT)
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Figure 13.30
Nonideal behavior of
electrolyte solutions.
13-56
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An ionic atmosphere model for nonideal behavior of
electrolyte solutions.
13-57
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SAMPLE PROBLEM
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)?
PLAN:
13-58
(a) Consider the formula for magnesium chloride, an ionic compound.
(b) Use the answer to part (a), the mass given, and the mol mass.
(c) The total number of mols of cations and anions, mass of solvent,
and equation for freezing point depression can be used to find the
new freezing point of the solution.
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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 =
13-59
0.0200 mols Cl8 spheres
2 mols Cl1 mol MgCl2
= 0.0200 mols Cl-
= 2.50 x 10-3 mols/sphere
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SAMPLE PROBLEM
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.
Tf = i (Kfm) = 3(1.86 0C/m x 0.100 m) = 0.558 0C
Tf = 0.000 0C - 0.558 0C = - 0.558 0C
13-60
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Exercise: A 0.100 L solution is made by dissolving a sample of CaCl2(s) in water.
(a) Is CaCl2 an electrolyte or a nonelectrolyte? (b) The solution has an osmotic
pressure of 3.55 atm at 27°C. What is the approximate molarity of CaCl2 in the
solution? (c) The van't Hoff factor for CaCl2 is 2.6 in the concentration range
0.04 - 0.12 M. Using this value, calculate the molarity of the NON Ideal CaCl2
solution . (d) The enthalpy of solution for CaCl2 is H = -81.3 kJ/mol. If the final
temperature of the solution was 27.0°C, what was its initial temperature?
(Assume that the density of the solution is 1.0 g/mL, that its specific heat is 4.18
J/g-K, and that the solution loses no heat to its surroundings.)
SOLUTION
(a) Soluble ionic compounds are strong electrolytes. CaCl2 consists of metal
cations (Ca2+) and nonmetal anions (Cl–) and is hence a strong electrolyte.
(b) we have :
This concentration is the total effective concentration of particles in the solution.
Because each CaCl2 unit ionizes to form three ions (one Ca2+ and two Cl–), the
concentration of CaCl2 is approximately 0.144 M/3 = 0.048 M.
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(c) In the concentration range of 0.4-1.2 M CaCl2, the ion pairing reduces the
effective number of free ions per CaCl2 unit from the ideal or limiting value of 3
to an actual value of 2.6. This number is the van't Hoff factor, i. Using this value,
we calculate that the concentration of CaCl2 equals 0.144 M/2.6 = 0.055 M, just
a bit higher than what we estimated in part (b).
(d) If the solution is 0.055 M in CaCl2 and has a total volume of 0.100 L, the
number of moles of solute is (0.100 L)(0.055 mol/L) = 0.0055 mol. Hence the
quantity of heat generated in forming the solution is (0.0055 mol)(-81.3 kJ/mol) =
-0.45 kJ. The solution absorbs this heat, causing its temperature to increase.
The relationship between temperature change and heat is given by
The heat absorbed by the solution is q = +0.45 kJ = 450 J. The mass of the
0.100 L of solution is (100 mL)(1.0 g/mL) = 100 g (to 2 significant figures). Thus
the temperature change is:
A kelvin has the same size as a degree Celsius. Because the solution
temperature increases by 1.1°C, the initial temperature was 27.0°C - 1.1°C =
25.9°C.
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Table Types of Colloids
Colloid Type
Dispersed
Substance
Dispersing
Medium
Example
Aerosol
Liquid
Gas
Fog
Aerosol
Solid
Gas
Smoke
Foam
Gas
Liquid
Whipped cream
Solid foam
Gas
Solid
Marshmallow
Emulsion
Liquid
Liquid
Milk
Solid emulsion
Liquid
Solid
Butter
Sol
Solid
Liquid
Paint; cell fluid
Solid sol
Solid
Solid
Opal
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Light scattering and the Tyndall effect.
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Photo by C.A.Bailey, CalPoly SLO
(Inlay Lake, Myanmar)
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Colloids.
A colloid is one of the three primary types of mixtures, with the other two
being a solution and suspension. A colloid is a solution that has particles
ranging between 1 and 1000 nanometers in diameter, yet are still able to
remain evenly distributed throughout the solution. These are also known as
colloidal dispersions because the substances remain dispersed and do not
settle to the bottom of the container. In colloids, one substance is evenly
dispersed in another.
The types of colloids includes sol, emulsion, foam, and aerosol.
Sol is a colloidal suspension with solid particles in a liquid.
Emulsion is between two liquids.
Foam is formed when many gas particles are trapped in a liquid or solid.
Aerosol contains small particles of liquid or solid dispersed in a gas.
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Solutions toward suspensions….
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Comparison among the properties of solutions
toward suspensions
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The most important colloids are those in which the dispersing medium is water. Such
colloids are frequently referred to as hydrophilic (water loving) or hydrophobic (water
fearing).
In the human body the extremely large molecules that make up such important substances as enzymes and
antibodies are kept in suspension by interaction with surrounding water molecules. The molecules fold in such a
way that the hydrophobic groups are away from the water molecules, on the "inside" of the folded molecule,
while the hydrophilic, polar groups are found on the surface, interacting with the water molecules. These
hydrophilic groups generally contain oxygen or nitrogen
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A Cottrell precipitator for removing particulates from
industrial smokestack gases.
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The steps in a typical municipal water treatment plant.
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Ion exchange for
removal of hard-water
cations.
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Reverse osmosis for the removal of ions.
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