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1 CHAPTER 4 Solutions B By Dr. Hisham Ezzat 2011- 2012 First year.
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Transcript of 1 CHAPTER 4 Solutions B By Dr. Hisham Ezzat 2011- 2012 First year.
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1
CHAPTER 4
Solutions
BBy
Dr. Hisham Ezzat
2011- 2012
First year
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2
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 C H O
kg H O
g C H O
90.0 g H O 6 12 6
2
6 12 6
2
10 0.
OH kg 1
OH g 1000
OH g 90.0
OHC g 0.10
OH kg
OHC mol ?
2
2
2
6126
2
6126
.molalityin ion concentrat theis This
OHC 617.0OHC g 180
OHC mol 1
OH kg 1
OH g 1000
OH g 90.0
OHC g 0.10
OH kg
OHC mol ?
6126
6126
6126
2
2
2
6126
2
6126
m
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3
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!
61266126
6126
6126
2
6126
OHC 578.0OHC g 180
OHC mol 1
L 1
mL 1000
nsol' mL
nsol' g 04.1
n sol' g 100.0
OHC g 0.10
OH L
OHC mol ?
M
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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!You do it!
? mol C H COOH
kg C H
g C H COOH
200.0 mL C H
mL C H
0.879 g C H
g C H
1 kg C H
mol C H COOH
122 g C H COOH C H COOH
6 5
6 6
6 5
6 6
6 6
6 6
6 6
6 6
6 5
6 56 5
7 25 1
1000 10 338
.
. m
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Example 8 What are the mole fractions of glucose
and water in a 10.0% glucose solution (Example 6)?
You do it!You do it!
6126
6126
6126
61266126
2
OHC mol 0556.0OHC g 180
OHC mol 1
OHC g 0.10OHC mol ?
water.of g 90.0 and glucose of g 10.0
are heresolution t thisof g 101.00In
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6
OH mol 00.5OH g 18
OH mol 1OH g 0.90OH mol ? 2
2
222
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Now we can calculate the mole fractions.
011.0989.000.1
011.0
OHC mol 0.0556 +OH mol 00.5
OHC mol 0556.0
989.0
OHC mol 0.0556 +OH mol 00.5
OH mol 00.5
61262
6126OHC
61262
2OH
6126
2
X
X
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Chapter 13 8
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).
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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.
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(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.
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(Miscibility
For example, methanol, CH3OH, is very soluble in water
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(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.
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(Miscibility)
Nonpolar molecules essentially “slide” in between each other. تنزلق For example, carbon tetrachloride and benzene
are very miscible.
C Cl
Cl
Cl
Cl
C
CC
C
CC
H
H
H
H
H
H
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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.
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Effect of Temperature on Solubility
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SolventSolvent
SolutionSolution
HHsolventsolvent HHsolutesolute
solutionsolution
SoluteSolute
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exothermicSolute + solvent → solution + heat
or
endothermic Solute + solvent + heat → solution
ΔH = H solution - (H solute + H solvent).
ΔH solution negative = exothermic
positive. = endothermic,
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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. Heating or cooling the system.
2. Changing the pressure of the system.
3. Changing the concentrations of reactants or products.
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Example of endothermic dissolution
21 kJ + KI(s) K+ + I-
18
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.
Le Chatelier's principle
the solubility of KI increases with increasing temperature.
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Temperature EffectsTemperature Effects
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Example of exothermic dissolution
solubility of lithium iodide decreases with an increase in temperature
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.)
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Factors Affecting SolubilityFactors Affecting SolubilityTemperature EffectsTemperature Effects
the solubility of gas decreases with temperature.
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Effect of Pressure on Solubility
The effect of pressure on the solubility of gases in liquids is described by Henry’s Law.
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.
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Factors Affecting SolubilityFactors Affecting Solubility
Henry’s Law – The solubility of a gas increases in direct proportion to its partial pressure above the solution.
Cg - solubility of gas Pg - the partial pressure of the gas k - Henry’s law constant.
ggkPC
Pressure Effects
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Carbonated beverages are bottled under PCO2> 1 atm. As
the bottle is opened, PCO2 decreases and the solubility of
CO2 decreases. Therefore,
bubbles of CO2 escape from
solution.
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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?
Problem:
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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.
Example 5:
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P total = PH2O + PO2
PO2 = P total - PH2O = 101-3 = 98 kPa
Solution: