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Chapter 17 - Properties of Solutions
17.1
Solution Composition
17.2
Thermodynamics of Solution Formation
17.3
Factors Affecting Solubility
17.4
Vapor Pressures of Solutions
17.5
Boiling-Point Elevation and Freezing-Point Depression
17.6
Osmotic Pressure
17.7
Colligative Properties of Electrolyte Solutions
17.8
Colloids
Definitions for Solutions
Solute - The smaller (in mass) of the components in a solution;
the material dispersed into a solvent.
Definition of Solution
(Ch. 16 dealt with pure substances rather than mixtures)
Solution – when the components of a mixture are uniformly
intermingled; the mixture is homogenous
Examples: air, seawater, steel
NOT: water (pure), wood (not homogeneous)
Colloid – a suspension of tiny particles in some medium is
called a colloidal dispersion, or a colloid; the mixture is
heterogeneous
Examples: gelatin, fog, butter
Solutions do not need to be liquid
Solvent - The major component of the solution; the material
that the solute is dissolved into.
Solubility - The maximum amount that can be dissolved into a
particular solvent to form a stable solution at a specified
temperature.
Miscibility - The ability of two substances to dissolve in one
another in any proportion.
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Types of Colloids
A quick note on “parts per…” notation
• One part per hundred: Typically represented by the percent (%)
symbol and denotes one part per 100 parts, one part in 102, and a
value of 1 × 10–2. This is equivalent to one drop of water diluted into
5 milliliters (one spoon-full) or one second of time in 1⅔ minutes.
• One part per million (ppm): Denotes one part per 1,000,000 parts,
one part in 106, and a value of 1 × 10–6. This is equivalent to one
drop of water diluted into 50 liters (roughly the fuel tank capacity of
a compact car), or one second of time in approximately 11½ days.
• One part per billion (ppb): Denotes one part per 1,000,000,000
parts, one part in 109, and a value of 1 × 10–9. This is equivalent to 1
drop of water diluted into 250 chemical drums (50 m3), or one
second of time in approximately 31.7 years.
• One part per trillion (ppt): Denotes one part per 1,000,000,000,000
parts, one part in 1012, and a value of 1 × 10–12. This is equivalent to
1 drop of water diluted into 20, two-meter-deep Olympic-size
swimming pools (50,000 m3), or one second of time in
approximately 31,700 years.
Methods for Quantifying How Much
Solute is in a Solvent
Remember !!
Concentrations are ratios. They are not additive!
Volumes are additive: V(total) = V1 + V2 + V3 +…
(in general)
Masses are additive:
m(total) = m1 + m2+ m3 + …
Moles are additive:
n(total) = n1 + n2 + n3 +…
Concentrations are not additive: c = c1 + c2 + c3 +…
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Calculating Molality - I
To calculate molality (mol solute/kg solvent), we need the
number of moles of solute and the mass of solvent used to
dissolve the solute! Normally, we are given the mass of solute and
mass of solvent, therefore we calculate the moles of the solute
from the mass, then use the mass of the solvent to calculate the
molality.
Mass (g) of solute
Remember:
Molality is different from
MM (g/mole)
molarity. Molality is based on
mass, and is independent of
Moles of Solute
temperature or pressure (unlike
molarity). Because 1 L of H2O
divide by kg water
weighs 1 kg, molality and
molarity of dilute aqueous
Molality (m) of solution
solutions are nearly identical.
Expressing Concentrations in Parts by Mass
Problem: Calculate the parts by mass of iron in a 1.85 g Fe
supplement pill that contains 0.0543 mg of Fe.
Plan: Convert mg Fe to grams and then use mass of Fe/mass of
pill.
Solution:
0.0543 mg Fe = 5.43 x 10 - 8 g Fe
5.43 x 10 - 8 g Fe
1.85 g pill
= 2.94 x 10 - 8
If 1 x 10-9 = 1 part per billion,
29.4 x 10 – 9 = 29.4 parts per billion (ppb)
Calculating Molality - II
Problem: Determine the molality and molarity of a solution
prepared by dissolving 75.0 g Ba(NO3)2 (s) in 374.0 g of water at 250C.
Plan: We convert the quantity of Ba(NO3)2 to moles using the molar
mass and then divide by the mass of H2O in kg or the volume of H2O
in liters (using water density = 0.99707 g/mL).
Solution: molar mass of Ba(NO ) = 261.32 g/mol
3 2
75.0 g
moles Ba(NO3)2 =
= 0.287 mole
261.32 g/mol
0.287 mole Ba(NO3)2
= 0.767 m
0.37400 kg H2O
molarity - we need the volume of solution, and can assume that
addition of the salt did not change the total volume.
374.0 g H2O
= 375.1 mL = 0.3751 L
0.99707 g/mL
0.287 mole
M=
= 0.765 M
0.3751 L
molality=
Expressing Concentrations in Parts by Volume
Problem: The label on a can of beer (340 mL) indicates “4.5%
alcohol by volume”. What is the volume in mL of alcohol in the
can?
Plan: We know the vol% and the total volume so we use the
definition of parts by volume to find the volume of alcohol.
Solution:
Vol. Alcohol =
x
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Expressing Concentrations in Mole Fraction
Problem: A sample of alcohol contains 118 g of ethanol
(C2H5OH), and 375.0g of water. What are the mole fractions of
each?
Plan: We know the mass and formula of each component so we
convert both to moles and apply the definition of mole fraction.
Solution:
Moles ethanol =
118 g ethanol
= 2.565 mol ethanol
46 g ethanol/mol
395 g H2O
18 g H2O/mol
2.565
Xethanol =
21.94 +2.565
21.94
Xwater =
21.94 +2.565
Moles water =
= 21.94 mol H2O
= 50.10
= 0.895
Converting Concentration Units - II
Problem: Commercial concentrated hydrochloric acid is 11.8 M
HCl and has a density of 1.190 g/ml. Calculate the (a) mass% HCl,
(b) molality and (c) mole fraction of HCl.
Plan: (b) From (a) we know moles of HCl, mass of solution, and
mass of HCl. Calculate mass of water and use definition of
molality.
Solution:
(b) mass of H2O = mass of solution - mass of HCl
= 1190 g solution - 430.2 g HCl
= 759.7 g H2O
11.8 moles HCl
0.7597 kg H2O
= 15.5 m HCl
Converting Concentration Units - I
Problem: Commercial concentrated hydrochloric acid is 11.8 M
HCl and has a density of 1.190 g/ml. Calculate the (a) mass% HCl,
(b) molality and (c) mole fraction of HCl.
Plan: We know Molarity and density. (a) For mass% HCl we need
the mass of HCl and water (the solvent). Assume 1L of solution:
from the density we know the mass of the solution and from the
molecular mass of HCl we can calculate its mass.
Solution:
(a) assume 1L of HCl solution
11.8 moles HCl
36.46 g HCl
= 430.2 g HCl
mole HCl
1 L solution x 1000 mL x 1.190 g soln = 1190. g solution
1 L solution
mL soln
11.8 moles HCl x
mass % HCl =
430.2 g HCl
x 100 = 36.2 % HCl
1190. g solution
Converting Concentration Units - III
Problem: Commercial concentrated Hydrochloric acid is 11.8 M
HCl and has a density of 1.190 g/ml. Calculate the (a) mass% HCl,
(b) molality and (c) mole fraction of HCl.
Plan: (c) From (a) we know the moles HCl and from (b) we know
the mass of water. Calculate moles of water then mole fractions.
Add the mole fractions to check!
Solution:
759.7 g H2O
(c)
= 42.17 mole H2O
18.016 g H2O/mol H2O
Total moles = 42.172 + 11.8 = 53.972 = 54.0
XHCl = 11.8 = 0.219
54.0
XH2O =42.17 = 0.781
54.0
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Chapter 17 - Properties of Solutions
17.1
Solution Composition
17.2
Thermodynamics of Solution Formation
17.3
Factors Affecting Solubility
17.4
Vapor Pressures of Solutions
17.5
Boiling-Point Elevation and Freezing-Point Depression
17.6
Osmotic Pressure
17.7
Colligative Properties of Electrolyte Solutions
17.8
Colloids
What happens when solutes dissolve?
Ch.17 Motto: Like Dissolves Like!
Polar/Ionic solutes - dissolve best in polar solvents.
Polar molecules can exhibit strong dipole-dipole
intermolecular interactions (ionic solutes: ion-dipole
interactions) with polar solvents, hence increasing their
solubility. Also called “hydrophilic” solutes.
Non-polar solutes - dissolve best in non-polar solvents.
To dissolve a non-polar molecule in a polar solvent requires
breaking strong dipole-dipole interactions of the solvent!
Also called “hydrophobic” solutes.
Key Point: Forces created between solute and
solvent must be comparable in strength to the forces
destroyed within both.
The formation of solutions from pure
substances is always favored by ENTROPY!
When two pure substances are mixed,
the disorder of the system is increased
As with any chemical reaction, solvation is a
balance between enthalpy and entropy
Gibbs Free Energy
∆G° = ∆H° - T∆S °
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A quantitative description of solvation:
The Solution Cycle
Step 1: Solute separates into components - overcoming attractions
in order to expand and make room for solvent -- Endothermic
A quantitative description of solvation:
The Solution Cycle
Figure 17.1
Step 2: Solvent expands - overcoming intermolecular attractions to
spread out and make room for the solute -- Endothermic
solvent (aggregated) + heat
solvent (separated)
DHsolvent > 0
Step 3: Solute and Solvent particles mix - particles attract each
other and give off energy as they interact -- Exothermic
solute (separated) + solvent (separated)
solution + heat
DHmix < 0
FIGURE 17.2
(a) Steps in an Exothermic Reaction
(b) Steps in an Endothermic Reaction
Enthalpy Balance
The Thermochemical Cycle: DHsolution= DHsolute + DHsolvent + DHmix
If DHEndothermic steps<DHExothermic steps
solution becomes warmer
If DHEndothermic steps >DHExothermic steps
solution becomes colder
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Solution Cycles and the Enthalpy
Components of the Heat of Solution
Endothermic Solution Process
Exothermic Solution Process
see Fig. 17.2
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Like dissolves like:
solubility of methanol in water
Hydration shells around an aqueous ion
 more charge density (charge/size ratio of ion)
results in greater hydration
Demo: non-additive volumes
250 mL ethanol + 250 mL water
Does 250 +250 always equal 500?
Effect of Structure on Solubility…
Predicting Relative Solubilities of Substances
Problem: Predict which solvent will dissolve more of the given
solute.
(a) Sodium Chloride
in methanol (CH3OH) or in propanol (CH3CH2CH2OH).
Nonpolar, fat-soluble
(builds up in fatty tissues –
pros and cons!)
Polar, water-soluble
(must be consumed
regularly – pros and
cons!)
Other examples:
Vitamins D, E, and K (fat soluble); Vitamin B (water-soluble)
DDT – pesticide (fat soluble)
Dioxins – group of molecules , herbicides and pollutants (fat
soluble)
BaSO4 – gastroenterology, enhances X-rays (insoluble)
(b) Ethylene glycol (HOCH2CH2OH)
in water or in hexane (CH3CH2CH2CH2CH2CH3).
(c) Diethyl ether (CH3CH2OCH2CH3)
in ethanol (CH3CH2OH) or in water.
Plan: Examine each solute and solvent to determine which
intermolecular forces will be active.
A solute tends to be more soluble in a solvent that has the same
type of intermolecular forces active.
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Predicting Relative Solubilities of Substances - II
Solution:
(a) _________________ - NaCl is an ionic compound that dissolves
through ion-dipole forces. Both methanol and propanol contain
a polar O-H group, but propanol’s longer hydrocarbon chain
would interact only weakly with the ions and be less effective in
stabilizing the ions.
(b) _______ - ethylene glycol has two O-H groups and is stabilized
by extensive H-bonding in water.
(c) _____________- diethyl ether shows both dipolar and dispersion
intermolecular forces and could form H bonds to both water and
ethanol. The ether would be more soluble in ethanol because
solvation in water must disrupt many more strong H-bonding
interactions.
Effect of Temperature on Solubility
Solubility increases with increasing temperature IF the solution
process is endothermic:
Solute + Solvent + Heat
Solution
Solubility decreases with increasing temperature IF the solution
process is exothermic:
Solute + Solvent
Solution + Heat
Recall Le Châtelier’s Principle
Figure 17.4:
Increase in gas solubility with increased pressure
Predicting the temperature dependence of solubility is very difficult.
The rate of dissolution might increase with temperature, but the
total amount of solute that will dissolve with increasing/decreasing
temperature is unpredictable. (You have to do an experiment!)
At equilibrium
Increase in P (and r)
Increase in [gas dissolved]
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Henry’s Law
The amount of gas dissolved in a solution is directly proportional
to the pressure of the gas above the solution.
P = kHX
P  partial pressure of the gaseous solute above the solution
X  mole fraction of the dissolved gas
kH a constant characteristic of a particular solution
Henry’s Law - Lake Nyos in Cameroon
CO2 buildup in colder, denser water layers near lake’s bottom (high
pressure, large mole fraction); “overturn” in 1986 sent water
supersaturated with CO2 to the surface and released an enormous
amount of CO2 into the area, suffocating thousands of humans and
animals.
Requirement:
This law holds only if there is no chemical reaction between the
solute and the solvent and if the solution is fairly dilute.
For all gases: ∆Hsolute ≈ 0, so ∆Hsoln < 0
Gas solubility always decreases with increasing T!
Colligative Properties
Properties of solutions that depend ONLY on the
number of molecules in a given volume of solvent
and NOT the properties (e.g. size or mass) of the
molecules.
Includes:
vapor-pressure lowering
boiling-point elevation
freezing-point depression
osmotic pressure
Colligative Properties of Solutions
17.4) Vapor Pressure Lowering - Raoult’s Law
17.5) Boiling-Point Elevation
and Freezing-Point Depression
17.6) Osmotic Pressure
17.7) Colligative Properties of Electrolyte Solutions
We’ll start by focusing on nonvolatile, nonelectrolytic solutes
(e.g. sucrose in H2O)
Later: volatile solutes (end of 17.4), electrolytic solutes (17.7)
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Pure solvent
increases Pvap to
get to equilibrium
Raoult’s Law:
Psolvent = xsolvent * P°solvent
Solution absorbs
vapor to lower Pvap to
get to equilibrium
Figure 17.8, Zumdahl
x is the solvent mole fraction
P°solvent is the vapor pressure
of pure solvent
See Figure 17.9
Example: Vapor Pressure Lowering
Problem: Calculate the vapor pressure lowering (DP) when 175 g of
sucrose is dissolved into 350.0 mL of water at 750C. The vapor
pressure of pure water at 75 oC is 289.1 mm Hg, and its density is
0.97489 g/mL.
Plan: Calculate the change in pressure from Raoult’s law using the
vapor pressure of pure water at 75oC. We calculate the mole fraction
of sugar in solution using the molecular formula of sucrose and
density of water at 75oC.
Solution: molar mass of sucrose (C12H22O11) = 342.30 g/mol
175 g sucrose
= 0.511 mol sucrose
342.30 g sucrose/mol
350.0 mL H2O  0.97489 g H2O = 341.2 g H2O
ml H2O
341.2 g H2O
= 18.94 mol H2O
18.02 g H2O/mol
Equilibrium can only be achieved after all the pure
solvent has been transferred to the solution.
Vapor Pressure Lowering (cont)
P = xsolvent * P°solvent
xwater =
=
moles water
moles of water + moles of sucrose
18.94 mol H2O
18.94 mol H2O + 0.511 mol sucrose
= 0.9737
P = xwater * P°solvent = 0.9737 x 289.1 mm Hg = 281.5 mm Hg
DP = P°solvent - P = 289.1 mm Hg - 281.5 mm Hg = 7.60 mm Hg
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Ideal/Non-ideal Solutions
Raoult’s Law is a
linear equation:
 Ideal behavior is approached when solute and solvent are
involved in similar intermolecular interactions (∆Hsoln = 0)
(i.e., follow Raoult’s law! – solute only dilutes the solvent, no
interaction)
Psolvent = xsolvent  P°solvent
y = x
*m
 When weaker solvent-solute interactions occur, heat is
removed upon dissolving (∆Hsoln > 0)
As nonvolatile solute is added
molessolute increases, so xsolvent
decreases.
 the observed vapor pressure is higher than ideal
(______________deviation from Raoult’s law)
xsolvent =
molessolvent
molessolute + molessolvent
 When stronger solute-solvent interactions occur (e.g. if Hbonding occurs), heat is released upon dissolving (∆Hsoln < 0)
Therefore, as xsolvent decreases,
Psolvent decreases.
 the observed vapor pressure is lower than ideal
(______________deviation from Raoult’s law)
Non-ideal solutions
Figure 17.11
Ideal
∆ Hsoln > 0
Summary of Ideal/Non-ideal Solutions
∆ Hsoln < 0
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What if the solute is volatile?
Ptotal = Psolute + Psolvent
Problem:
Dalton’s Law of Partial
Pressures – Chapter 5
Consider a solution having equal molar quantities of
acetone and chloroform, xacetone = xCHCl3 = 0.500.
At 35°C, the vapor pressures of the pure substances are:
Psolvent = xsolvent P osolvent
P°acetone = 345 torr
Psolute = xsolute P osolute
P°CHCl3 = 293 torr
What are the mole fractions, x, of each component in the
vapor phase?
Ptotal = (xsolute P osolute) + (xsolvent  P osolvent)
Determine the partial pressure of each component and the
vapor pressure of the solution.
Let’s condense the vapor and calculate the new mole fraction
of each component in the vapor phase.
Pacetone = xacetone  P0acetone = 0.500  345 torr = 172.5 torr
PCHCl3 = xCHCl3  P0CHCl3 = 0.500  293 torr = 146.5 torr
Pacetone = xacetone  P0acetone = 0.541  345 torr = 186.6 torr
PCHCl3 = xCHCl3  P0CHCl3 = 0.459  293 torr = 134.5 torr
Total Pressure = 172.5 + 146.5 = 319.0 torr
P
From Dalton’s law of partial pressures we know that xA = PA
Total
Pacetone 172.5 torr
xacetone = P
=
Total
319.0 torr
P
xCHCl3 = CHCl3 = 146.5 torr
PTotal
319.0 torr
= 0.541
= 0.459
Total Pressure = 186.6 + 134.5 = 321.1 torr
Pacetone 186.6 torr
xacetone = P
=
Total
321.1 torr
P
134.5 torr
CHCl
3 =
xCHCl3 =
PTotal
321.1 torr
= 0.581
= 0.419
Vapor is enriched in acetone!
Vapor is more enriched in acetone!
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Boiling Point Elevation
Recall: The boiling point (Tb) of a liquid is the
temperature at which its vapor pressure equals
atmospheric pressure.
Nonvolatile solutes lower the vapor pressure of a liquid,
making it more difficult for the solution to boil
 greater temperature required to reach boiling point
Boiling Point Elevation:
∆ Tb = Kbmsolute
Kb = molal boiling point constant for given solvent
msolute = solute concentration in molality
Kb = molal boiling point constant for given solvent
Kf = molal freezing point constant for given solvent
Freezing Point
Depression:
∆Tf = Kfmsolute
Presence of solute lowers the rate at
which molecules in the liquid return to
the solid; have to lower the temp to
decrease the rate of molecules leaving
the solid for the liquid until the rates
are equal (equilibrium)
See Figure 17.3
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See Figure 17.12
Example: Boiling Point Elevation and Freezing
Point Depression in an aqueous solution
Problem: We add 475 g of sucrose (sugar) to 600 g of water.
What will be the Freezing and Boiling points of the solution?
Plan: Find the molality of the sucrose solution and apply the
equations for FP depression and BP elevation, using the
constants from table 17.5.
Solution:
475 g sucrose
= 1.39 mole sucrose
342.30 g sucrose/mol
1.39 mole sucrose
molality =
= 2.31 m
0.600 kg H2O
0.51 oC
(2.31m)= 1.2 oC
m
BP = 100.00oC + 1.2oC
= 101.18oC
1.86 oC
∆Tf = Kf * m =
(2.31 m) = 4.3 oC
m
FP = 0.00oC - 4.3oC= - 4.3oC
∆Tb = Kb * m =
Effect of solute: extends the liquid range of the solvent
Example: Boiling Point Elevation and Freezing
Point Depression in a Non-aqueous Solution
Osmosis: The flow of solvent through a
semipermeable membrane into a solution
Problem: Calculate the Boiling Point and Freezing Point of a
solution having 257 g of naphthalene (C10H8) dissolved into
500.0 g of chloroform (CHCl3).
Plan: Just like the previous example.
Solution: MMnaphthalene= 128.16 g/mol; MMchloroform= 119.37 g/mol
The semipermeable membrane allows solvent
molecules to pass, but not solute molecules
molesnaph = 257 g nap
= 2.01 mol C10H8
128.16g/mol
molality = moles nap = 2.01 mol = 4.01 m
kg(CHCl3)
0.500 kg
o
∆Tb = Kb * m = 3.63 C (4.01m) = 14.6oC
m
4.70oC
*
∆Tf = Kf m =
(4.01m) = 18.9oC
m
See Figures
17.15 & 17.16
normal BP = 61.7oC
new BP = 76.3oC
normal FP = - 63.5oC
new FP = - 82.4oC
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Osmotic Pressure: π = MRT
Similar to ideal gas law!
Osmotic Pressure  Dialysis (does allow
the transfer of some solute particles)
R = gas constant
M = molar concentration of solute
Fig 17.14
Determining Molar Mass from
Osmotic Pressure
Problem: A physician studying hemoglobin dissolves 21.5 mg
of the protein in water at 5.0oC to make 1.5 mL of solution in
order to measure its osmotic pressure. At equilibrium, the
solution has an osmotic pressure of 3.61 torr. What is the molar
mass (MM) of the hemoglobin?
Plan: We know p, R, and T. We convert p from torr to atm, and
T from oC to K, and then use the osmotic pressure equation to
solve for molarity (M). Then we calculate the number of moles
of hemoglobin from the known volume and use the known
mass to find MM.
Solution:
1 atm
p = 3.61 torr *
= 0.00475 atm
760 torr
Molar Mass from Osmotic Pressure
Concentration from osmotic pressure:
M= π
= 0.00475 atm
= 2.08 x10 - 4 M
RT
0.0821 L atm (278.2 K)
mol K
Finding # moles of solute:
-4
n = M * V = 2.08 x10 mol
L soln
* 0.00150
L soln = 3.12 x10 - 7 mol
Calculating molar mass of hemoglobin (after changing mg to g):
MM =
0.0215 g
= 6.89 x104 g/mol
3.12 x10-7 mol
Temp = 5.00C + 273.15 = 278.2 K
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Colligative Properties of Ionic Solutions
For ionic solutions we must take into account the number of ions
present!
i = van’t Hoff factor = “ionic strength,” or the # of ions present
Non-ideal Behavior
of Electrolyte
Solutions
(0.05m aqueous)
For Electrolyte Solutions:
vapor pressure lowering:
∆P = i XsoluteP 0solvent
boiling point elevation:
∆Tb = i Kb m
freezing point depression: ∆Tf = i Kf m
osmotic pressure:
Most likely the result of
ion pairing in the solution
(ions periodically
interacting and behaving
as one particle).
π = i MRT
See Table 17.6
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