CHAPTER 12: SOLUTIONS AND
THEIR PROPERTIES
Vanessa N. Prasad-Permaul
CHM 1046
Valencia Community College
1
Introduction
1. A mixture is any intimate combination of two or more pure
substances
2. Can be classified as heterogeneous or homogeneous
Heterogeneous
-The mixing of components is visually
nonuniform and have regions of different
composition
Homogenous
-Mixing is uniform, same composition
throughout
-Can be classified according to the size of their
particles as either solutions or colloids
2
Solutions
Solution
1. Homogeneous mixtures
2. Contain particles with diameters in the range of
0.1–2 nm
3. Transparent but may be colored
4. Do not separate on standing
Colloids
1. Milk & fog
2. Diameters 2-500 nm
3. Do not separate on standing
3
Types of Solutions
4
Solution Formation
Solute
 Dissolved substance, or smaller quantity
substance
Solvent
 Liquid dissolved in, larger quantity substance
Saturated solution
 Contains the maximum amount of solute that
will dissolve in a given solvent.
5
Solution Formation
Unsaturated
Contains less solute than a solvent has the capacity
to dissolve.
Supersaturated
Contains more solute than would be present in a
saturated solution.
Crystallization
The process in which dissolved solute comes out of
the solution and forms crystals.
6
EXAMPLE 12.1:
A: GIVE AN EXAMPLE OF A SOLID SOLUTION
PREPARED FROM A LIQUID AND A SOLID
A dental filling made up of liquid mercury and solid
silver is a solid solution
B: GIVE AN EXAMPLE OF A LIQUID SOLUTION
PREPARED BY DISSOLVING A GAS IN A LIQUID
Aqueous ammonia
7
EXERCISE 12.1: IDENTIFY THE SOLUTE & SOLVENT IN
THE FOLLOWING SOLUTIONS.
a) 80g of chromium & 5g of molybdenum
b) 5g of MgCl2 dissolved in 1000g of H2O
c) 39% N2, 41% Ar, and the rest O2
8
Energy Changes and the Solution Process
Three Types of interactions
1. Solvent-solvent
2. Solvent-solute
3. Solute-solute
“Like dissolves like”
solutions will form when three types of
interactions are similar in kind and magnitude
9
Energy Changes and the Solution Process
Example NaCl and water:
Ionic solid NaCl dissolve in polar solvents like water
because the strong ion-dipole attractions between
Na+ and Cl- ions and polar water molecules are
similar in magnitude to the strong dipole-dipole
attractions between water molecules and to the
strong ion-ion attractions between Na+ and Cl- ions
Example Oil and water
Oil does not dissolve in water because the two
liquids have different kinds of intermolecular
forces. Oil is not polar or an ionic solvent
10
Energy Changes and the Solution Process
NaCl in H2O
1. Ions that are less tightly held because of their position at a corner or
an edge of the crystal are exposed to water molecules
2. Water molecules will collide with the NaCl until an ion breaks free
3. More water molecules then cluster around the ion, stabilizing it by
ion-dipole attractions
4. The water molecules attack the weak part of the crystal until it is
dissolved
5. Ions in solution are said to be solvated they are surrounded and
stabilized by an ordered shell of solvent molecules
11
EXAMPLE 12.2: WOULD NAPTHALENE (C10H8) BE
MORE SOLUBLE IN ETHANOL OR IN BENZENE?
EXPLAIN.
Naphthalene is more soluble in benzene because nonpolar
naphthalene must break the strong hydrogen bonds
between ethanol molecules and replace them with weaker
London forces.
12
EXERCISE 12.2: WHICH OF THE FOLLOWING
COMPOUNDS IS LIKELY TO BE MORE SOLUBLE IN
WATER: C4H9OH or C2H9SH? EXPLAIN.
13
Energy Changes and the Solution Process
G, Free energy change
1. If G is negative the process is spontaneous, and
the substance is dissolved
2. If G is positive the process is non-spontaneous, the
substance is not dissolved
3. G = H -TS
H, enthalpy, heat flow in or out of the system,
Hsoln heat of solution
S, entropy, disorder, Ssoln entropy of solution
14
Energy Changes and the Solution Process
15
Energy Changes and the Solution Process
Ssoln Entropy of Solution
Usually a positive number because when you dissolve
something you are increasing disorder
Hsoln Heat of Solution
1. Harder to predict because it could be exothermic
(- Hsoln) or endothermic (+Hsoln)
2. The value of the heat of solution for a substance
results from an interplay of the three kinds of
interactions
16
Energy Changes and the Solution Process
1)
Solvent-solvent interactions: Energy is required
(endothermic) to overcome intermolecular forces
between solvent molecules because the molecules
must be separated and pushed apart to make room
for solute particles
2) Solute-solute interactions: Energy is required
(endothermic) to overcome interactions holding
solute particles together in a crystal. For an ionic
solid, this is the lattice energy. Substances with
higher lattice energies therefore tend to be less
soluble than substances with lower lattice energies.
17
Energy Changes and the Solution Process
3) Solvent-solute interactions: Energy is released
(exothermic) when solvent molecules cluster around
solute particles and solvate them. For ionic
substances in water, the amount of hydration energy
released is generally greater for smaller cations than
for larger ones because water molecules can
approach the positive nuclei of smaller ions more
closely and thus bind more tightly. Hydration energy
generally increases as the charge on the ion
increases.
18
Energy Changes and the Solution Process
Exothermic -Hsoln
The solute–solvent
interactions are stronger
than solute–solute or
solvent–solvent.
Favorable process
Exothermic rxn.
19
Energy Changes and the Solution Process
Endothermic +Hsoln
The solute–solvent
interactions are
weaker than
solute–solute or
solvent–solvent.
Unfavorable
process.
Endothermic rxn
20
Energy Changes and the Solution Process
 Hydration
 The attraction of ions for water molecules
 Hydration Energy
 The energy associated with the attraction
between ions and water molecules
 Lattice energy
 The energy holding ions together in a crystal
lattice
21
EXAMPLE 12.3: WHICH OF THE FOLLOWINGIONS
WOULD BE EXPECTED TO HAVE THE GREATER ENERGY
OF HYDRATION. Mg2+ OR Al3+?
Al3+ because water molecules can approach the positive
nuclei of smaller ions more closely and thus bind more
tightly. Hydration energy generally increases as the
charge on the ion increases.
22
EXERCISE 12.3: WHICH ION HAS THE LARGER
HYDRATION ENERGY, Na+ or K+ EXPLAIN.
23
Units of Concentration
Molarity (M)
M = mole of solute / Liter of solution
Molality (m)
m = moles of solute/mass of solvent (kg)
Mole Fraction (x)
X = mole of component / total moles
24
EXAMPLE 12.6: GLUCOSE, C6H12O6, IS A SUGAR THAT IS
IN FRUITS. IT IS ALSO FOUND IN BLOOD AND IS THE
BODY’S MAIN SOURCE OF ENERGY. WHAT IS THE
MOLALITY OF A SOLUTION CONTAINING 5.67g OF
GLUCOSE DISSOLVED IN 25.2g OF WATER?
5.67g C6H12O6 x 1 mol C6H12O6 = 0.0315 mol C6H12O6
180.2g C6H12O6
0.0315 mol C6H12O6 = 1.25 m C6H12O6
25.2g x 1kg
1000g
25
EXERCISE 12.6: TOLUENE, C6H5CH3, IS A STARTING
MATERIAL FOR TNT (TRINITROTLOUENE) . FIND THE
MOLALITY OF TOLUENE IN A SOLUTION THAT
CONTAINS 35.6g OF TOLUENE AND 125g OF BENZENE
26
EXAMPLE 12.7: WHAT ARE THE MOLE FRACTIONS OF
GLUCOSE AND WATER IN A SOLUTION CONTAINING 5.67g OF
GLUCOSE DISSOLVED IN 25.2g OF WATER?
5.67g C6H12O6 x 1 mol C6H12O6
180.2g C6H12O6
= 0.0315 mol C6H12O6
25.2g H2O x 1 mol H2O = 1.40 mol H2O
18.0g H2O
1.40mol + 0.0315mol = 1.432mol
MOLE FRACTION OF GLUCOSE 0.0315mol = 0.0220 x 100% = 2.20%
MOLE FRACTION OF WATER
1.432mol
1.40mol = 0.978 x 100% = 97.8%
1.432mol
27
EXERCISE 12.7: CALCULATE THE MOLE FRACTIONS OF
TOLUENE AND BENZENE IN THE SOLUTION FROM
EXERCISE 12.6
28
Units of Concentration
Mass Percent (mass %)
Mass % =
(mass of component / total mass of sol’n) x 100%
Parts per million, ppm =
(mass of component / total mass of solution) x 106
Parts per billion, ppb =
(mass of component / total mass of solution) x 109
29
EXAMPLE 12.5: HOW WOULD YOU PREPARE 425g OF AN
AQUEOUS SOLUTION CONTAINING 2.40% BY MASS OF
SODIUM ACETATE, NaC2H3O2?
MASS OF SOLUTE: 0.0240 x 425g = 10.2g
MASS OF WATER: 425g – 10.2g = 414.8 = 415g
MASS OF SOLUTION: 425g
YOU WOULD PREPARE THIS SOLUTION BY DISSOLVING 10.2g
OF NaC2H3O2 IN 415g OF WATER FOR A TOTAL MASS OF 425g.
30
EXERCISE 12.5: AN EXPERIMENT CALLS FOR 35.0g OF
HYDROCHLORIC ACID THAT IS 20.2% HCl BY MASS. HOW
MANY GRAMS OF HCl IS THIS? HOW MANY GRAMS OF
WATER IS THIS?
31
Some Factors Affecting Solubility
Solubility
The amount of solute per unit of solvent needed to form
a saturated solution
Miscible
Mutually soluble in all proportions
Effect of Temperature on Solubility
1. Most solid substances become more soluble as
temperature rises
2. Most gases become less soluble as temperature
rises
32
Some Factors Affecting Solubility
Effect of Pressure on Solubility
1. No effect on liquids or solids
2. The solubility of a gas in a liquid at a given
temperature is directly proportional to the partial
pressure of the gas over the solution, @ 25°C
Henry’s Law
solubility = k x P
k = constant characteristic of specific gas, mol/Latm
P = partial pressure of the gas over the sol’n
33
Some Factors Affecting Solubility
a) Equal numbers of gas molecules escaping liquid and
returning to liquid
b) Increase pressure, increase # of gas molecules
returning to liquid, solubility increases
c) A new equilibrium is reached, where the #’s of
escaping = # of returning
34
EXAMPLE 12.4: 27g OF ACETYLENE (C2H2) DISSOLVES
IN 1L OF ACETONE AT 1.0atm. IF THE PARTIAL PRESSURE
OF ACETYLENE IS INCREASED TO 12atm, WHAT IS THE
SOLUBILITY OF ACETONE?
S2 = P2
S1 P1
S2
27g C2H2/L acetone
=
12atm
1.0atm
S2 = 27g C2H2 x 12atm = 3.2x102g C2H2/L acetone
L acetone 1.0atm
 320g of acetylene will dissolve in 1L of acetone @ 12atm
35
EXERCISE 12.4: A LITER OF WATER @ 250C DISSOLVES
0.0404g OF O2 WHEN THE PARTIAL PRESSURE OF THE
OXYGEN IS 1.00atm. WHAT IS THE SOLUBILITY OF
OXYGEN FROM AIR, IN WHICH THE PARTIAL PRESSURE
OF O2 IS 159mmHg?
36
Physical Behavior of Solutions: Colligative Properties
Colligative properties
Properties that depend on the amount of a
dissolved solute but not its chemical identity
There are four main colligative properties:
1. Vapor pressure lowering
2. Freezing point depression
3. Boiling point elevation
4. Osmotic pressure
37
Physical Behavior of Solutions: Colligative Properties
In comparing the properties of a pure solvent with those
of a solution…
1. Vapor pressure of sol’n is lower
2. Boiling point of sol’n is higher
3. Freezing point of sol’n is lower
4. Osmosis, the migration of solvent molecules through
a semipermeable membrane, occurs when solvent
and solution are separated by the membrane
38
Vapor-pressure Lowering of Solutions: Raoult’s Law
1. A liquid in a closed container is in equilibrium with its vapor
and that the amount of pressure exerted by the vapor is called
the vapor pressure.
2. When you compare the vapor pressure of a pure solvent with
that of a solution at the same temperature the two values are
different.
3. If the solute is nonvolatile and has no appreciable vapor
pressure of its own (solid dissolved) the vapor pressure of the
solution is always lower that that of the pure solvent.
4. If the solute is volatile and has a significant vapor pressure
(2 liquids) the vapor pressure of the mixture is intermediate
between the vapor pressures of the two pure liquids.
39
Solutions with a Nonvolatile Solute
When solute molecules displace solvent molecules at
the surface, the vapor pressure drops since fewer gas
molecules are needed to equalize the escape rate and
capture rates at the liquid surface.
40
Solutions with a Nonvolatile Solute!!!
Raoult’s Law
Psoln = Psolv · Xsolv
Psoln = vapor pressure of the solution
Psolv = vapor pressure of the pure solvent
Xsolv = mole fraction of the solvent in the
solution
41
Raoult’s Law applies to only Ideal
solutions
1. Law works best when solute concentrations are low and
when solute and solvent particles have similar
intermolecular forces.
2. If intermolecular forces between solute particles and
solvent molecules are weaker than solvent molecules alone,
solvent molecules are less tightly held, vapor pressure is
higher than Raoult predicts
3. If intermolecular forces between solute and solvent are
stronger than solvent alone, solvent molecules are more
tightly held and the vapor pressure is lower than predicted
4.
No Van’t Hoff factor!!! is a measure of the effect of a
solute upon colligative properties
42
Solutions with a Nonvolatile Solute
Close-up view of part of the vapor pressure curve
for a pure solvent and a solution of a nonvolatile
solute. Which curve represents the pure solvent,
and which the solution?
Why?
43
 The lower vapor pressure of a sol’n relative to that of a
pure solvent is due to the difference in their entropies of
vaporization, Svap. Because the entropy of the solvent in
a sol’n is higher to begin with, Svap is smaller for the sol’n
than for the pure solvent. As a result vaporization of the
solvent from the sol’n is less favored (less negative Gvap),
and the vapor pressure of the solution is lower.
44
Solutions with a Volatile Solute!!
Ptotal = PA + PB
Ptotal = (P°A · XA) + (P°B · XB)
P°A = vapor pressure of pure A
XA = mole fraction of A
P°B = vapor pressure of pure B
XB = mole fraction of B
Ptotal should be intermediate to A & B
45
Close-up view of part of the vapor pressure curves for
two pure liquids and a mixture of the two. Which
curves represent the mixture?
1) Red
2) Green
3) Blue
46
EXAMPLE 12.12: CALCULATE THE VAPOR-PRESSURE
WHEN 5.67g OF GLUCOSE IS DISSOLVED 25.2g OF
WATER @ 25oC. THE VAPOR PRESSURE OF WATER @
25oC IS 23.8mmHg. WHAT IS THE VAPOR PRESSURE OF
THE SOLUTION?
P = PAOXB = 23.8 mmHg x 0.0220 = 0.524 mmHg
THE VAPOR PRESSURE OF THE SOLUTION IS:
PA = PAO - P
= (23.8mmHg – 0.524mmHg)
= 23.3mmHg
47
EXERCISE 12.12: NAPTHALENE IS USED TO MAKE
MOTHBALLS. A SOLUTION IS MADE BY DISSOLVING
0.515g OF NAPTHALENE IN 60.8g OF CHLOROFORM.
CALCULATE THE VAPOR PRESSURE LOWERING OF
CHLOROFORM @ 20oC FROM NAPTHALENE. THE
VAPOR PRESSURE OF CHLOROFORM @ 20oC IS
156mmHg. NAPTHALENE CAN BE ASSUMED TO BE
NON-VOLATILE COMPARED WITH CHLOROFORM.
WHAT IS THE VAPOR PRESSURE OF THE SOLUTION?
48
Boiling Point Elevation and Freezing
Point Depression of Solutions
49
Boiling Point Elevation and Freezing Point Depression of Solutions
1. Red line is pure solvent
2. Green line solution of nonvolatile solute
3. Vapor pressure of sol’n is lower
4. Temp at which vapor pressure = 1 atm for sol’n is higher
5. Boiling point of sol’n is higher by Tb
6. Liquid/vapor phase transition line is lower for sol’n
7. Triple point temp is lower for sol’n
8. Solid/liquid phase transition has shifted to a lower temp.
9. The freezing point of the sol’n is lower by Tf
50
Boiling Point Elevation and Freezing Point Depression of Solutions
 Tb = Kb · m
 Tf = Kf · m
Kb = molal boiling-point elevation constant
Kf = molal freezing-point depression constant
m = molality
NO Van’t Hoff Factor!!
51
Boiling Point Elevation and Freezing Point Depression of Solutions
The higher boiling point of a solution relative to that of a pure
solvent is due to a difference in their entropies of vaporization,
Svap. Because the solvent in a solution has a higher entropy to
begin with, Svap is smaller for the solution than for the pure
solvent. As a result, the boiling point of the solution Tb is higher
than that of the pure solvent.
52
Boiling Point Elevation and Freezing Point Depression of Solutions
The lower freezing point of a solution relative to that of a
pure solvent is due to a difference in their entropies of fusion,
Sfusion. Because the solvent in a solution has a higher
entropy level to begin with, Sfusion is larger for the solution
than for the pure solvent. As a result the freezing point of
the solution Tf is lower than that of the pure solvent.
53
EXAMPLE 12.13: AN AQUEOUS SOLUTION IS 0.0222
m GLUCOSE. WHAT ARE THE BOILING POINT AND
THE FREEZING POINT OF THIS SOLUTION?
Tb = Kbcm = 0.512oC/m x 0.0222 m = 0.0114oC
Tf = Kfcm = 1.86oC/m x 0.0222 m = 0.0413oC
THE BOILING POINT OF THE SOLUTION IS
100.000oC + 0.0114oC = 100.011oC
THE FREEZING POINT OF THE SOLUTION IS
0.000oC – 0.0413oC = -0.041oC
54
EXERCISE 12.13: HOW MANY GRAMS OF ETHYLENE
GLYCOL MUST BE ADDED TO 37.8g OF WATER TO GIVE
A FREEZING POINT OF -0.150oC?
55
Osmosis and Osmotic Pressure
1.
Semipermeable membranes allow water or
other small molecules to pass through, but they
block the passage of large solute molecules or
ions.
2. When a solution and a pure solvent are
separated by the right kind of semipermeable
membrane, solvent molecules pass through the
membrane in a process known as osmosis.
3. Passage of solvent through the membrane
takes place in both directions
56
Osmosis and Osmotic Pressure
4. Passage from the pure solvent side to the
solution side is more favored and faster.
5. The amount of liquid on the pure solvent side
decreases
6. The amount of liquid on the solution side
increases
7. The concentration of the solution decreases
57
Osmosis and Osmotic Pressure
58
Osmosis and Osmotic Pressure
Osmotic Pressure
1. The amount of pressure necessary to achieve
equilibrium
2.  = MRT
 = osmotic pressure
M = molarity
R = gas constant, .08206 L atm/K mol
T = temperature in kelvins
59
Some uses of Colligative Properties
1.
The most important use of colligative
properties in the laboratory is for determining
the molecular mass of an unknown substance.
2.
Any of the four colligative properties can be
used but using osmotic pressure gives the most
accurate results
60
If you have to calculate the molar mass of a compound
which of the following will give you the most accurate
results?
Osmotic pressure
2) Vapor pressure lowering
3) Boiling point elevation
4) Freezing point depression
1)
61
EXAMPLE 12.8: AN AQUEOUS SOLUTION IS 0.120 m
GLUCOSE. WHAT ARE THE MOLE FRACTIONS OF EACH
COMPONENT IN THE SOLUTION?
0.120 m = 0.120mol IN 1.00kg OF WATER
1.00kg H2O x 1000g x
1 mol
= 55.6mol H2O
1kg
18.0 g H2O
MOLE FRACTION OF GLUCOSE:
0.120 mol
(0.120mol + 55.6mol)
= 0.00215
MOLE FRACTION OF WATER:
55.6mol
(0.120mol + 55.6mol)
= 0.998
62
EXERCISE 12.8: A SOLUTION IS 0.120 m METHANOL
DISSOLVED IN ETHANOL. CALCULATE THE MOLE
FRACTIONS OF CH3OH AND CH3CH2OH IN THE
SOLUTION.
63
EXAMPLE 12.9: A SOLUTION IS 0.150 mol FRACTION
GLUCOSE, AND 0.850 mol FRACTION WATER. WHAT IS
THE MOLALITY OF GLUCOSE IN THE SOLUTION?
0.850 mol x 18.0 g H2O
1 mol H2O
= 15.3 g H2O (0.0153kg H2O)
0.150 mol C6H12O6 = 9.80 m C6H12O6
0.00153kg SOLVENT
64
EXERCISE 12.9: A SOLUTION IS 0.250 mol FRACTION
METHANOL AND 0.750 mol FRACTION ETHANOL. WHAT
IS THE MOLALITY OF METHANOL IN THE SOLUTION?
65
EXAMPLE 12.10: AN AQUEOUS SOLUTION IS 0.273 m KCl.
WHAT IS THE MOLAR CONCENTRATION OF KCl? THE
DENSITY OF THE SOLUTION IS 1.011 x 103g/L.
0.273 mol KCl x 74.6g KCl = 20.4g KCl
1 mol KCl
1.000 x 103g + 20.4g = 1.020 x 103g
1.020 x 103g x
1L
= 1.009L
1.011x103g
0.273mol KCl
1.009L solution
= 0.271 M KCl
66
EXERCISE 12.10: UREA (NH2)2CO IS USED AS
FERTILIZER. WHAT IS THE MOLAR CONCENTRATION OF
AN AQUEOUS SOLUTION THAT IS 3.42 m UREA? THE
DENSITY OF THE SOLUTION IS 1.045 g/mL.
67
EXAMPLE 12.11: AN AQUEOUS SOLUTION IS 0.907 M
Pb(NO3)2. WHAT IS THE MOLALITY? THE DENSITY OF
THE SOLUTION IS 1.252g/mL.
1.000 L x 1000mL x 1.252 g
1L
1mL
= 1.252 x 103g
0.907 mol Pb(NO3)2 x 331.2 g Pb(NO3)2 = 3.00 x 102 Pb(NO3)2
1 mol
1.252 x 103 g – 3.00 x 12 g = 9.52 x 102g (0.952kg)
0.907 mol Pb(NO3)2 = 0.953 m Pb(NO3)2
0.952kg solvent
68
EXERCISE 12.11: AN AQUEOUS SOLUTION IS 2.00 M
UREA. THE DENSITY IS 1.029 g/mL. WHAT IS THE MOLAL
CONCENTRATION OF UREA IN THE SOLUTION?
69
Example 1
Arrange the following in order of their expected
increasing solubility in water:
Br2, KBr, C7H8
70
Example 2
Which would you expect to have the larger (more
negative) hydration energy?
1. Na+
1. Mg2+
2. Cs+
2. Na+
3. Li+
3. Li+
4. Rb+
71
Example 3
What is the mass % concentration of a saline sol’n
prepared by dissolving 1.00 mol of NaCl in 1.00 L
of water? DensityH2O=1.00 g/mL MMNaCl = 58.443
g/mol
72
Example 4
Assuming that seawater is an aqueous solution of
NaCl what is its molarity? The density of
seawater is 1.025 g/mL at 20C and the NaCl
concentration is 3.50 mass %
 Assume 1 L to make easier, 1000 mL
73
Example 5
What is the molality of a solution prepared by
dissolving 0.385 g of cholesterol, C27H46O in 40.0
g of chloroform, CHCl3? What is the mole
fraction of cholesterol in the solution?
74
Example 6
What mass in grams of a 0.500 m solution of sodium
acetate, CH3CO2Na, in water would you use to obtain
0.150 mol of sodium acetate?
75
Example 7
The density at 20°C of a 0.258 m solution of glucose
in water is 1.0173 g/mL and the molar mass of
glucose is 180.2 g/mol. What is the molarity of the
solution?
 Assume 1 kg
76
Example 8
The density at 20°C of a 0.500 M solution of acetic
acid in water is 1.0042 g/mL. What is the
concentration of this solution in molality? The
molar mass of acetic acid, CH3CO2H, is 60.05
g/mol.
 Assume 1 L
77
Example 9
Which of the following will become less soluble
in water as the temperature is increased?
1) NaOH(s)
2) CO2(g)
78
Example 10
The solubility of CO2 in water is 3.2 x 10-2 M @ 25°C
and 1 atm pressure. What is the Henry’s-Law constant
for CO2 in mol/L atm?
solubility = k x P
79
Example 11
What is the vapor pressure (in mm Hg) of a solution
prepared by dissolving 5.00 g of benzoic acid
(C7H6O2) in 100.00 g of ethyl alcohol (C2H6O) at
35°C? The vapor pressure of the pure ethyl alcohol
at 35°C is 100.5 mm Hg
80
Example 12
What is the vapor pressure ( in mm Hg) of a sol’n
prepared by dissolving 25.0 g of ethyl alcohol
(C2H5OH) in 100.0 g of water at 25°C? The vapor
pressure of pure water is 23.8 mm Hg and the vapor
pressure of ethyl alcohol is 61.2 mm Hg at 25°C
81
Example 13
The following phase
diagram shows part of the
vapor pressure curves for
a pure liquid (green
curve) and a solution (red
curve) of the first liquid
with a second volatile
liquid (not shown)
a)
Is the boiling point of the second pure liquid higher or lower
than that of the first liquid?
b)
On the diagram where is the approximate position of the
second pure liquid?
82
Example 14
What is the normal boiling point in °C of a solution
prepared by dissolving 1.50 g of aspirin (C9H8O4) in
75.00 g of chloroform (CHCl3)? The normal boiling
point of chloroform is 61.7 °C and Kb of chloroform
is 3.63 °C kg/mol
83
Example 15
What osmotic pressure in atm would you expect for
a solution of 0.125 M C6H12O6 that is separated
from pure water by a semipermeable membrane
at 310 K?
 = MRT
84
Example 16
A solution of unknown substance in water at
300 K gives rise to an osmotic pressure of 3.85
atm. What is the molarity of the solution?
 = MRT
M = /RT
85
Example 17
 What is the molar mass of sucrose if a
solution prepared by dissolving 0.822 g of
sucrose in water and diluting to a volume of
300.0 mL has an osmotic pressure of 149 mm
Hg at 298 K?
86