CHAPTER 4
First year
 Solutions
B
By
Dr. Hisham Ezzat
2011- 2012
1
Example 6

Calculate the molarity and the molality of an
aqueous solution that is 10.0% glucose, C6H12O6.
The density of the solution is 1.04 g/mL. 10.0%
glucose solution has several medical uses. 1 mol
C6H12O6 = 180 g
? mol
10
.0Hg OC6 H12
O6 g H
1000
g H 2O
6 H12O
6 .0 g
? mol
C 6 HC
O
10
C
1000
O
6
6 12 6
2






? mol12C 6 H
O
10
.
0
g
C
H
O
12 6
6 12 6
kg2 OH 2O 90.0
1 kg H 2O

1 kg H 2 O
kg H
g90.0
H 2 Og H 2 O
kg H 2 O
90.0 g H 2 O
1 mol C6 H12O 6
 0.617m C6 H12O 6
180 g C6 H12O 6
This is the concentration in molality .
2

Calculate the molality and the molarity of an
aqueous solution that is 10.0% glucose, C6H12O6.
The density of the solution is 1.04 g/mL. 10.0%
glucose solution has several medical uses. 1 mol
C6H12O6 = 180 g
You calculate the molarity!
? mol C 6 H12O 6 10.0 g C 6 H12O 6 1.04 g sol' n



L H 2O
100.0 g sol' n
mL sol' n
1000 mL 1 mol C 6 H12O 6

 0.578 M C 6 H12O 6
1L
180 g C 6 H12O 6
3
Example 7

Calculate the molality of a solution that contains
7.25 g of benzoic acid C6H5COOH, in 2.00 x 102
mL of benzene, C6H6. The density of benzene is
0.879 g/mL. 1 mol C6H5COOH = 122 g
You do it!
? mol C 6 H 5COOH 7.25g C 6 H 5COOH
1 mL C 6 H 6


kg C 6 H 6
200.0 mL C 6 H 6
0.879 g C 6 H 6
1000 g C 6 H 6 1 mol C 6 H 5COOH

 0.338 m C 6 H 5COOH
1 kg C 6 H 6
122 g C 6 H 5COOH
4
Example 8

What are the mole fractions of glucose
and water in a 10.0% glucose solution
(Example 6)?
You do it!
In 1.00  10 2 g of this solution t here are
10.0 g of glucose and 90.0 g of water.
? mol C6 H12O 6  10.0 g C6 H12O 6 
1 mol C6 H12O 6
 0.0556 mol C6 H12O 6
180 g C6 H12O 6
5
1 mol H 2 O
? mol H 2 O  90.0 g H 2 O 
 5.00 mol H 2 O
18 g H 2 O
6

Now we can calculate the mole fractions.
XH O
2
5.00 mol H 2O

5.00 mol H 2O + 0.0556 mol C6 H12O6
 0.989
XC H
6
1 2O 6
0.0556 mol C6 H12O6

5.00 mol H 2O + 0.0556 mol C6 H12O6
 0.011
1.00  0.989  0.011
7
The extent to which a solute dissolves in solvent depends
The nature of the solute.
The nature of the solvent.
The temperature.
The pressure (for gases).
Chapter 13
8

Few organic compounds that dissolve readily in water,
most contain - OH groups. methyl alcohol, ethyl alcohol,
and ethylene glycol, all of which are soluble in water in all
proportions.
9
(Miscibility



Pairs of liquids that mix in any proportions
are said to be miscible.
Example: Ethanol and water are miscible
liquids.
In contrast, immiscible liquids do not mix
significantly. Example: Gasoline and water
are immiscible.
10
(Miscibility

For example,
methanol,
CH3OH, is very
soluble in water
11
(Miscibility

Water and ethanol are miscible because the
broken hydrogen bonds

the more C atoms in the alcohol, the lower its
solubility in water.
Increasing the number of –OH groups within a
molecule increases its solubility in water.

12
(Miscibility)

Nonpolar molecules essentially “slide” in
between each other. ‫تنزلق‬

For example, carbon tetrachloride and benzene
are very miscible.
H
Cl
Cl
C
Cl
C
H
Cl
H
H
C
C
C
C
C
H
H
13
Effect of Temperature on Solubility
Experience tells us that sugar dissolves better
in warm water than in cold water.
As temperature increases, solubility of solids
generally increases.
Sometimes solubility decreases as
temperature increases (e.g., Ce2(SO4)3).
Gases are less soluble at higher temperatures.
An environmental application of this is thermal
pollution.
14
Solvent
DHsolvent
DHsolute
Solute
DHsolution
Solution
15
exothermic
Solute + solvent → solution + heat
or
endothermic
Solute + solvent + heat → solution
ΔH
=
ΔH solution
H solution - (H solute + H solvent).
negative = exothermic
positive. = endothermic,
16
Effect of Temperature on Solubility
According to LeChatelier’s Principle when stress is
applied to a system at equilibrium, the system
responds in a way that best relieves the stress.


Since saturated solutions are at equilibrium, LeChatelier’s
principle applies to them.
Possible stresses to chemical systems include:

1.
2.
3.
Heating or cooling the system.
Changing the pressure of the system.
Changing the concentrations of reactants or products.
17
Example of endothermic dissolution
21 kJ + KI(s)  K+ + ILe Chatelier's principle
equilibrium will shift to the right
 using up some of the added heat (and some of the
excess solid KI) and

increasing the concentration of K+ and I- ions in
solution.
the solubility of KI increases with increasing
temperature.
18
Temperature Effects
19

Example of exothermic dissolution
LiI (s)  Li+ + I- + 71 KJ
equilibrium shifts to the left
(1) using up some of the added heat (and Li and I- ions in solution) and
, (2)
forming more solid Lil. (We observe the precipitation of
some Lil out of solution.)

solubility of lithium iodide decreases with an increase in
temperature
20
Factors Affecting Solubility
Temperature Effects
the solubility of gas
decreases with
temperature.
21
Effect of Pressure on Solubility




Pressure changes have little or no effect on
solubility of liquids and solids in liquids.
Liquids and solids are not compressible.
Pressure changes have large effects on the
solubility of gases in liquids.
The effect of pressure on the solubility of gases in
liquids is described by Henry’s Law.
22
23
Henry’s Law – The solubility of a gas
increases in direct proportion to its partial
pressure above the solution.
24
Factors Affecting Solubility
Pressure Effects
Henry’s Law – The solubility of a gas increases in direct proportion to
its partial pressure above the solution.
C g  kPg
Cg - solubility of gas
Pg - the partial pressure of the gas
k - Henry’s law constant.
25
Carbonated beverages are
bottled under PCO > 1 atm. As
the bottle is opened, PCO2
decreases and the solubility of
CO2 decreases. Therefore,
bubbles of CO2 escape from
solution.
2
26
Problem:
At 740 torr and 20°C, nitrogen has
solubility in H2O of 0.018 g /I. At 620
torr and 20°C its solubility is 0.015
g/l.
Do these data show that nitrogen
obey Henry's law or not?
27
Example 5:
At 25°C oxygen gas collected over water
at a total pressure of 101 kPa is soluble
to the extent of 0.0393 g dm-3. What
would its solubility be if its partial
pressure over water were 107 kPa?
The vapor pressure of water is 3.0 kPa at
25°C.
28
Solution:
P total = PH2O + PO2
PO2 = P total - PH2O = 101-3 = 98 kPa
29