1 CHAPTER 4 Solutions C By Dr. Hisham Ezzat 2011- 2012 First year
1
CHAPTER 4
Solutions
CBy
Dr. Hisham Ezzat
2011- 2012
First year
1. Completely miscible liquids– Ideal solution– Non - ideal solution
2. Completely immiscible liquids H2O and aniline, H2O and chlorobenzene
3. Partially immiscible liquids, H2O and phenol, H2O and ether
Ideal Solution
1. The force of attraction between all molecules are identical i.e. the attraction force is not affected by addition of other components A - A = B-B = A - B.
2. No heat is evolved or absorbed during mixing i.e. H soln. = Zero
3. The volume of solution is the sum of volumes of the two liquids.
4. The solution obeys Raoult's law.
Figure (1): Vapor pressure of ideal solutions
Example 6:
Heptane (C7H16) and octane (C8H18) form ideal solutions What is the vapor pressure at 40°C of a solution that contains 3.0 mol of heptane and 5 mol of octane? At 40°C, the vapor pressure of heptane is 0.121 atm and the vapor pressure of octane is 0.041 atm.
Solution:
The total number of moles is 8.0. therefore X heptane = 3.0/8.0 = 0.375 X octane = 5.0/8.0 = 0.625Total = X heptane . Po heptane + X
octane. Po octane= 0.375 x 0.12 +0.625 x 0.04 = 0.045 atm + 0.026 atm. = 0.071 atm.
Example 7:
Assuming ideality, calculate the vapor pressure of 1.0 m solution of a non - volatile, on dissociating solute in water at 50°C. The vapor pressure of water 50°C is 0.122 atm.
Solution :
From example 2 the mole fraction of water in 1.0m solution is 0.982.
PH2O = XH2O PH2O = 0.982 x 0.122 = 0.120 atm.
Problem:
At 140°C, the V.P of C6H5CI is 939.4 torr and that of C6H5Br is 495.8 torr. Assuming that these two liquids from an ideal solution. Find the composition of a mixture of two liquids which boils at 140°C under 1 atm pressure?
Non- ideal solutions
Negative deviation Positive deviation1- The force of attraction
increase by mixing A - A, B-B < A-B
The force of attraction decrease by mixing A-A , B-B
> A-B2- The vapor pressure will be lower
than that given by Roault's lawThe vapor pressure will be higher than that given by
Raoult's law.3- H solution :- Ve (exothermic) H solution: + Ve
(endothermic)4- Temperature change when solution is formed: increase
Temperature change when solution is formed: decrease.
5- Example: Acetone-water Ethanol-hexane
Fig.2: Vapour pressure of non-ideal solution (-ve deviation)
Fig.3: Vapour pressure of non-ideal solution (+ve deviation)
Fractional Distillation of Binary Miscible liquids
• The separation of mixture of volatile liquids into their components is called fractional distillation,
• the distillate containing the more volatile component and the residue the less volatile one
a) Ideal solutions
• If a mixture of 2 liquids (A and B) form a completely miscible ideal solution and PA > PB result in B.P. of A < B.P of B thus on boiling:-– 1) The Liquid A boils at lower B.P than that of liquid B.– 2) The liquid A which is more volatile will be passed
from the fractionating column and the liquid B which is less volatile returned again to the distallating flask.
• A solution of intermediate B.p. between 2 pure liquid -called azeotropic solution
b) Non - ideal solutions (solutions that exhibit deviations from Raoults law)
If a solution having any other compositions is distilled, the azeotropic mixture will distill first and the excess of (A) or (B) will remains in the flask e.g 95 % ethanol and 5 % H2O.
Non - ideal solutions with minimum boiling point:
• If a solution having any other composition is distilled, the execs of acetone or CHCI3 will distill first leaving the azeotropic mixture in the flask.
2) Non - ideal solutions with maximum boiling point:
Example 8:
A solution is prepared by mixing 5.81 g acetone C3H6O, (M. wt = 58.1 g/mole) 11.9 g chloroform (CHCI3 M.wt 119.4 g/mole). At 35°C this solution has a total vapor pressure of 260 torr. Is this an ideal solution? Comment? The vapor pressure of pure acetone and pure CHCI3 at 35°C are 345 and 293 torr, respectively.
20
Colligative Properties of Solutions
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Colligative Properties of Solutions There are four common types of colligative
properties:1. Vapor pressure lowering
2. Freezing point depression
3. Boiling point elevation
4. Osmotic pressure Vapor pressure lowering is the key to all
four of the colligative properties.
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Lowering of Vapor Pressure and Raoult’s Law Addition of a nonvolatile solute to a solution
lowers the vapor pressure of the solution. The effect is simply due to fewer solvent
molecules at the solution’s surface. The solute molecules occupy some of the spaces
that would normally be occupied by solvent. Raoult’s Law models this effect in ideal
solutions.
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Lowering of Vapor Pressure and Raoult’s Law Derivation of Raoult’s Law.
P P
where P vapor pressure of solvent
P vapor pressure of pure solvent
mole fraction of solvent
solvent solvent solvent0
solvent
solvent0
solvent
X
in solution
X in solution
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Lowering of Vapor Pressure and Raoult’s Law Lowering of vapor pressure, Psolvent, is defined as:
0solventsolvent
0solventsolvent
0solvent
solvent0solventsolvent
)P1(
)P)((- P
PP P
X
X
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Lowering of Vapor Pressure and Raoult’s Law Remember that the sum of the mole fractions
must equal 1. Thus Xsolvent + Xsolute = 1, which we can
substitute into our expression.
Law sRaoult' iswhich
P P
- 10solventsolutesolvent
solventsolute
X
XX
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Lowering of Vapor Pressure and Raoult’s Law This graph shows how the solution’s vapor pressure
is changed by the mole fraction of the solute, which is Raoult’s law.
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The vapor pressure of water is 17.5 torr at 20°C. Imagine holding the temperature constant while adding glucose, C6H12O6, to the water so that the
resulting solution has XH2O = 0.80 and XGlu = 0.20.
What is , the vapor pressure of water over the solution 0
AAA PXP
torrXPXP AAA 5.1780.00
= 14 torr
Examples
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Glycerin, C3H8O3, is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25°C. Calculate the vapor pressure at 25°C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25°C is 23.8 torr
The vapor pressure of pure water at 110°C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110°C. Assuming that Raoult's law is obeyed, what is the mole fraction of ethylene glycol in the solution? Answer: 0.290
P°H2O =1070 torrPH2O = 1 Atm = 760 torr
XH2O = ---------PH2O
P°H2O
= ---------760 torr1070 torr
= 0.71028XXH2OH2O + X + XEGEG = 1 = 10.7103 + XEG = 1
1- 0.7103 = XEG
XEG = 0.28972 = 0.290
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More ExamplesSucrose is a nonvolatile, nonionizing solute in water. Determine the vapor pressure lowering, at 27°C, of a solution of 75.0 grams of sucrose, C12H22O11, dissolved in 180. g of water. The vapor pressure of pure water at 27°C is 26.7 torr. Assume the solution is ideal.
molSucg
SucmolgSucnSuc 219.0
3.342
10.75
molWatyerg
WatermolgWaternWater 99.9
18
1180
978541.02191.0991.9
991.9
UcWater
waterWater nn
nX
13.2697854.07.260 XtorrXPP WaterWaterWater
Vapor Pressure Lowered = 26.7-26.1= 0.6
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solution is made by mixing 52.1 g of propyl chloride, C3H8Cl, and 38.4 g of propyl bromide, C3H8Br. What is the vapor pressure of propyl chloride in the solution at 25°C? The vapor pressure of pure propyl chloride is 347 torr at 25°C and that of pure propyl bromide is 133 torr at 25°C. Assume that the solution is an ideal solution.
6633.054.78
11.52
CPg
CPmolCPgnCP
312.099.122
14.38
CBg
CBmolCBgnCB
67996.03122.06633.0
6633.0
PBPC
PCrPC nn
nX
TorrXXPP PCPCPC 23695.235679964.03470
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. At 25°C a solution consists of 0.450 mole of pentane, C5H12, and 0.250 mole of cyclopentane, C5H10. What is the mole fraction of cyclopentane in the vapor that is in equilibrium with this solution? The vapor pressure of the pure liquids at 25°C are 451 torr for pentane and 321 torr for cyclopentane. Assume that the solution is an ideal solution.
95.202451450.0 XXPP PenPenPen
25.80321250.00 XXPP CPenCPenCPen
RT
VPn
RT
VPn
RT
PVn CPen
CPenPen
Pen ;;
PenCPen
CPen
PenCPen
CPen
PenCPen
CPenCPen PP
P
RT
VP
RT
VPRT
VP
nn
nX
283.095.20225.80
25.80
CPenP
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Boiling Point Elevation Addition of a nonvolatile solute to a solution
raises the boiling point of the solution above that of the pure solvent. This effect is because the solution’s vapor
pressure is lowered as described by Raoult’s law. The solution’s temperature must be raised to
make the solution’s vapor pressure equal to the atmospheric pressure.
The amount that the temperature is elevated is determined by the number of moles of solute dissolved in the solution.
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Boiling Point Elevation
Boiling point elevation relationship is:
solvent for the
constantelevation point boiling molal K
solution ofion concentrat molal
elevationpoint boiling T :where
KT
b
b
bb
m
m
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Boiling Point Elevation
Example 14-4: What is the normal boiling point of a 2.50 m glucose, C6H12O6, solution?
C101.28=C28.1+C100.0 =solution theofPoint Boiling
C28.1T
)50.2)(C/ 512.0(T
K T
000
0b
0b
bb
mm
m
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Boiling-Point Elevation Boiling-Point Elevation The addition of a nonvolatile solute lowers the vapor pressure of the solution. At any given temperature, the vapor pressure of the solution is lower than that of the pure liquid
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The increase in boiling point relative to that of the pure solvent, Tb, is directly proportional to the number of solute particles per mole of solvent molecules. Molality expresses the number of moles of solute per 1000 g of solvent, which represents a fixed number of moles of solvent mKT bb
Solvent B.Point (°C) Kb (°C/m)
Freezing P. (°C)
Kf (°C/m)
Water, H2O 100.0 0.52 0.00 1.86Benzen, C6H6 80.1 2.53 5.5 5.12Ethanol, C2H6O 78.4 1.22 -114.0 1.99Carbon tetrachloride, CCl4 76.8 5.02 -22 29.8Chloroform, CHCl3
61.2 3.63 -63.5 4.68
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Automotive antifreeze consists of ethylene glycol, C2H6O2, a nonvolatile nonelectrolyte. Calculate the boiling point of a 25.0 mass percent solution of ethylene glycol in water.
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Freezing Point Depression Addition of a nonvolatile solute to a solution
lowers the freezing point of the solution relative to the pure solvent.
See table for a compilation of boiling point and freezing point elevation constants.
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Freezing Point Depression
Relationship for freezing point depression is:
T K
where: T freezing point depression of solvent
molal concentration of soltuion
K freezing point depression constant for solvent
f f
f
f
m
m
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Freezing Point Depression Notice the similarity of the two relationships
for freezing point depression and boiling point elevation.
Fundamentally, freezing point depression and boiling point elevation are the same phenomenon. The only differences are the size of the effect which is
reflected in the sizes of the constants, Kf & Kb.
This is easily seen on a phase diagram for a solution.
mm bbff K T vs.KT
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Freezing Point Depression
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Freezing Point Depression
Example 14-5: Calculate the freezing point of a 2.50 m aqueous glucose solution.
C4.65 - = C4.65 - C0.00=solution ofPoint Freezing
C65.4T
)50.2)(C/(1.86T
KT
000
0f
0f
ff
mm
m
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Freezing Point Depression
Example : Calculate the freezing point of a solution that contains 8.50 g of benzoic acid (C6H5COOH, MW = 122) in 75.0 g of benzene, C6H6.
You do it!You do it!
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Freezing Point Depression
C0.72=C4.76-C5.48 =F.P.
C76.4)929.0)(C/12.5(T
KT
solution. for this depression theCalculate .2
929.0COOHHC g 122
COOHHC mol 1
HC kg 0.0750
COOHHC g 50.8
HC kg
COOHHC mol ?
molality! Calculate .1
000
00f
ff
56
56
66
56
66
56
mm
m
m
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Determination of Molecular Weight by Freezing Point Depression The size of the freezing point depression
depends on two things:1. The size of the Kf for a given solvent, which are well
known.
2. And the molal concentration of the solution which depends on the number of moles of solute and the kg of solvent.
If Kf and kg of solvent are known, as is often the case in an experiment, then we can determine # of moles of solute and use it to determine the molecular weight.
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Determination of Molecular Weight by Freezing Point Depression Example : A 37.0 g sample of a new covalent
compound, a nonelectrolyte, was dissolved in 2.00 x 102 g of water. The resulting solution froze at -5.58oC. What is the molecular weight of the compound?
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Determination of Molecular Weight by Freezing Point Depression
g/mol 7.61mol 0.600
g 37 is massmolar theThus
compound mol 600.0
kg 0.200 3.00=OH kg 0.200in compound mol ?
water.of kg 0.200 mL 200
are thereproblem In this
00.3C1.86
C58.5
K
T
the thusKT
2
0
0
f
f
ff
m
mm
m
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Osmotic Pressure Osmosis is the net flow of a solvent
between two solutions separated by a semipermeable membrane.
The solvent passes from the lower concentration solution into the higher concentration solution.
Examples of semipermeable membranes include:
1. cellophane and saran wrap2. skin3. cell membranes
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Osmotic Pressure
H2O 2O
semipermeable membrane
H2O H2O
sugar dissolvedin water
H2O
H2O
H2O
H2O
net solvent flow
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Osmotic Pressure Osmosis is a rate controlled phenomenon.
The solvent is passing from the dilute solution into the concentrated solution at a faster rate than in opposite direction, i.e. establishing an equilibrium.
The osmotic pressure is the pressure exerted by a column of the solvent in an osmosis experiment.
M
M
RT
where: = osmotic pressure in atm
= molar concentration of solution
R = 0.0821L atmmol K
T = absolute temperature
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Osmotic Pressure
For very dilute aqueous solutions, molarity and molality are nearly equal. M m
m
for dilute aqueous solutions only
RT
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Osmotic Pressure Osmotic pressures can be very large.
For example, a 1 M sugar solution has an osmotic pressure of 22.4 atm or 330 p.s.i.
Since this is a large effect, the osmotic pressure measurements can be used to determine the molar masses of very large molecules such as:
1. Polymers
2. Biomolecules like proteins ribonucleotides
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Osmotic Pressure
Example : A 1.00 g sample of a biological material was dissolved in enough water to give 1.00 x 102 mL of solution. The osmotic pressure of the solution was 2.80 torr at 25oC. Calculate the molarity and approximate molecular weight of the material.
You do it!You do it!
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Osmotic Pressure
M M
M M
RT RT
atm = 2.80 torr1 atm
760 torr atm =
= atm
0.0821 KL atmmol K
? .
..
0 00368
0 00368298
150 10 4
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Osmotic Pressure
M M
M M
M
RT RT
atm = 2.80 torr1 atm
760 torr atm =
= atm
0.0821 K
g
mol
1.00 g
0.100 L
L
typical of small proteins
L atmmol K
gmol
? .
..
?
..
0 00368
0 00368
298150 10
1
150 106 67 10
4
44
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Application of Osmotic Pressure1. Water Purification by Reverse Osmosis If we apply enough external pressure to an
osmotic system to overcome the osmotic pressure, the semipermeable membrane becomes an efficient filter for salt and other dissolved solutes. Ft. Myers, FL gets it drinking water from the Gulf
of Mexico using reverse osmosis. Dialysis is another example of this phenomenon.
2) Isotonic solution: In the living cells, the osmotic pressure of solution is equal to the osmotic pressure of the cell.
e.g: NaCI (0.9%) has the same osmotic pressure as blood.
3) Hypertonic solution: A solution of higher osmotic pressure. In this solution red blood cells shrink. The cells are called plasmolysed.
4) Hypotonic solution: A solution of lower osmotic pressure. In this solution red blood cells swells up and burst. The cell is said to be haemolysed
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End of Chapter 2
Human Beings are solution chemistry in action!