Chapter 13 Properties of Solutions

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Chapter 13Properties of Solutions
Chemistry, The Central Science, 10th
edition Theodore L. Brown H. Eugene LeMay, Jr.
and Bruce E. Bursten

• John D. Bookstaver
• St. Charles Community College
• St. Peters, MO
• ? 2006, Prentice Hall, Inc.

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Solutions

• Solutions are homogeneous mixtures of two or more
  pure substances.
• In a solution, the solute is dispersed uniformly
  throughout the solvent.

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Solutions

• The intermolecular forces between solute and
  solvent particles must be strong enough to
  compete with those between solute particles and
  those between solvent particles.

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How Does a Solution Form?

• As a solution forms, the solvent pulls solute
  particles apart and surrounds, or solvates, them.
  Play animation

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How Does a Solution Form

• If an ionic salt is soluble in water, it is
  because the ion-dipole interactions are strong
  enough to overcome the lattice energy of the salt
  crystal.

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Energy Changes in Solution

• Simply put, three processes affect the energetics
  of the process
• Separation of solute particles
• Separation of solvent particles
• New interactions between solute and solvent

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Energy Changes in Solution

• The enthalpy change of the overall process
  depends on ?H for each of these steps.

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Why Do Endothermic Processes Occur?

• Things do not tend to occur spontaneously (i.e.,
  without outside intervention) unless the energy
  of the system is lowered.

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Why Do Endothermic Processes Occur?

• Yet we know that in some processes, like the
  dissolution of NH4NO3 in water, heat is absorbed,
  not released.

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Enthalpy Is Only Part of the Picture

• The reason is that increasing the disorder or
  randomness (known as entropy) of a system tends
  to lower the energy of the system.

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Enthalpy Is Only Part of the Picture

• So even though enthalpy may increase, the
  overall energy of the system can still decrease
  if the system becomes more disordered.

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Answer The entropy increases because each gas
eventually becomes dispersed in twice the volume
it originally occupied.
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Student, Beware!

• Just because a substance disappears when it
  comes in contact with a solvent, it doesnt mean
  the substance dissolved.

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Student, Beware!

• Dissolution is a physical changeyou can get back
  the original solute by evaporating the solvent.
• If you cant, the substance didnt dissolve, it
  reacted.

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Types of Solutions

• Saturated
• Solvent holds as much solute as is possible at
  that temperature.
• Dissolved solute is in dynamic equilibrium with
  solid solute particles.

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Types of Solutions

• Unsaturated
• Less than the maximum amount of solute for that
  temperature is dissolved in the solvent.

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Types of Solutions

• Supersaturated
• Solvent holds more solute than is normally
  possible at that temperature.
• These solutions are unstable crystallization can
  usually be stimulated by adding a seed crystal
  or scratching the side of the flask.

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Factors Affecting Solubility

• Chemists use the axiom like dissolves like
• Polar substances tend to dissolve in polar
  solvents.
• Nonpolar substances tend to dissolve in nonpolar
  solvents.

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Factors Affecting Solubility

• The more similar the intermolecular attractions,
  the more likely one substance is to be soluble in
  another.

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Factors Affecting Solubility

• Glucose (which has hydrogen bonding) is very
  soluble in water, while cyclohexane (which only
  has dispersion forces) is not.

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Factors Affecting Solubility

• Vitamin A is soluble in nonpolar compounds (like
  fats).
• Vitamin C is soluble in water.

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Solve C7H16 is a hydrocarbon, so it is molecular
and nonpolar. Na2SO4, a compound containing a
metal and nonmetals, is ionic HCl, a diatomic
molecule containing two nonmetals that differ in
electronegativity, is polar and I2, a diatomic
molecule with atoms of equal electronegativity,
is nonpolar. We would therefore predict that
C7H16 and I2 would be more soluble in the
nonpolar CCl4 than in polar H2O, whereas water
would be the better solvent for Na2SO4 and HCl.
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Answer  C5H12 lt C5H11 Cl lt C5H11 OH lt
C5H10(OH)2 (in order of increasing polarity and
hydrogen-bonding ability)
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Gases in Solution

• In general, the solubility of gases in water
  increases with increasing mass.
• Larger molecules have stronger dispersion forces.

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Gases in Solution

• The solubility of liquids and solids does not
  change appreciably with pressure.
• The solubility of a gas in a liquid is directly
  proportional to its pressure.

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Henrys Law

• Sg kPg
• where
• Sg is the solubility of the gas
• k is the Henrys law constant for that gas in
  that solvent
• Pg is the partial pressure of the gas above the
  liquid.play animation

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PRACTICE EXERCISE Calculate the concentration of
CO2 in a soft drink after the bottle is opened
and equilibrates at 25C under a CO2 partial
pressure of 3.0 ? 104 atm.
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Temperature

• Generally, the solubility of solid solutes in
  liquid solvents increases with increasing
  temperature.

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Temperature

• The opposite is true of gases
• Carbonated soft drinks are more bubbly if
  stored in the refrigerator.
• Warm lakes have less O2 dissolved in them than
  cool lakes.

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Ways of Expressing Concentrations of Solutions
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Mass Percentage
? 100

• Mass of A

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Parts per Million andParts per Billion
Parts per Million (ppm)
? 106

• ppm

Parts per Billion (ppb)
? 109
ppb
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PRACTICE EXERCISE (a) Calculate the mass
percentage of NaCl in a solution containing 1.50
g of NaCl in 50.0 g of water. (b) A commercial
bleaching solution contains 3.62 mass sodium
hypochlorite, NaOCl. What is the mass of NaOCl in
a bottle containing 2500 g of bleaching solution?
Answers (a) 2.91, (b) 90.5 g of NaOCl
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Mole Fraction (X)

• In some applications, one needs the mole fraction
  of solvent, not solutemake sure you find the
  quantity you need!

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Molarity (M)

• You will recall this concentration measure from
  Chapter 4.
• Because volume is temperature dependent, molarity
  can change with temperature.

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Molality (m)

• Because both moles and mass do not change with
  temperature, molality (unlike molarity) is not
  temperature dependent.

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PRACTICE EXERCISE What is the molality of a
solution made by dissolving 36.5 g of naphthalene
(C10H8) in 425 g of toluene (C7H8)?
Answer 0.670 m
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Solution Analyze We are asked to calculate the
concentration of the solute, HCl, in two related
concentration units, given only the percentage by
mass of the solute in the solution. Plan In
converting concentration units based on the mass
or moles of solute and solvent (mass percentage,
mole fraction, and molality), it is useful to
assume a certain total mass of solution. Lets
assume that there is exactly 100 g of solution.
Because the solution is 36 HCl, it contains 36 g
of HCl and (100 36) g 64 g of H2O. We must
convert grams of solute (HCl) to moles in order
to calculate either mole fraction or molality. We
must convert grams of solvent H2O to moles to
calculate mole fractions, and to kilograms to
calculate molality.
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PRACTICE EXERCISE A commercial bleach solution
contains 3.62 mass NaOCl in water. Calculate
(a) the molality and (b) the mole fraction of
NaOCl in the solution.
Answers (a) 0.505 m, (b) 9.00 ? 103
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Solution Analyze Our goal is to calculate the
molarity of a solution, given the masses of
solute and solvent and the density of the
solution. Plan The molarity of a solution is the
number of moles of solute divided by the number
of liters of solution (Equation 13.8). The number
of moles of solute (C7H8) is calculated from the
number of grams of solute and its molar mass. The
volume of the solution is obtained from the mass
of the solution (mass of solute mass of solvent
5.0 g 225 g 230 g) and its density.
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PRACTICE EXERCISE A solution containing equal
masses of glycerol (C3H8O3) and water has a
density of 1.10 g/mL. Calculate (a) the molality
of glycerol, (b) the mole fraction of glycerol,
(c) the molarity of glycerol in the solution.
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Changing Molarity to Molality

• If we know the density of the solution, we can
  calculate the molality from the molarity, and
  vice versa.

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Colligative Properties

• Changes in colligative properties depend only on
  the number of solute particles present, not on
  the identity of the solute particles.
• Among colligative properties are
• Vapor pressure lowering
• Boiling point elevation
• Melting point depression
• Osmotic pressure

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Vapor Pressure

• Because of solute-solvent intermolecular
  attraction, higher concentrations of nonvolatile
  solutes make it harder for solvent to escape to
  the vapor phase.

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Vapor Pressure

• Therefore, the vapor pressure of a solution is
  lower than that of the pure solvent.

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Raoults Law

• PA XAP?A
• where
• XA is the mole fraction of compound A
• P?A is the normal vapor pressure of A at that
  temperature
• NOTE This is one of those times when you want
  to make sure you have the vapor pressure of the
  solvent.

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Solution Analyze Our goal is to calculate the
vapor pressure of a solution, given the volumes
of solute and solvent and the density of the
solute. Plan We can use Raoults law (Equation
13.10) to calculate the vapor pressure of a
solution. The mole fraction of the solvent in the
solution, XA, is the ratio of the number of
moles of solvent (H2O) to total solution (moles
C3H8O3 moles H2O).
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PRACTICE EXERCISE The vapor pressure of pure
water at 110C is 1070 torr. A solution of
ethylene glycol and water has a vapor pressure of
1.00 atm at 110C. Assuming that Raoults law is
obeyed, what is the mole fraction of ethylene
glycol in the solution?
Answer 0.290
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Boiling Point Elevation and Freezing Point
Depression

• Nonvolatile solute-solvent interactions also
  cause solutions to have higher boiling points and
  lower freezing points than the pure solvent.

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Boiling Point Elevation

• The change in boiling point is proportional to
  the molality of the solution
• ?Tb Kb ? m
• where Kb is the molal boiling point elevation
  constant, a property of the solvent.

?Tb is added to the normal boiling point of the
solvent.
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Freezing Point Depression

• The change in freezing point can be found
  similarly
• ?Tf Kf ? m
• Here Kf is the molal freezing point depression
  constant of the solvent.

?Tf is subtracted from the normal freezing point
of the solvent.
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Boiling Point Elevation and Freezing Point
Depression

• Note that in both equations, ?T does not depend
  on what the solute is, but only on how many
  particles are dissolved.

• ?Tb Kb ? m
• ?Tf Kf ? m

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Solution Analyze We are given that a solution
contains 25.0 mass of a nonvolatile,
nonelectrolyte solute and asked to calculate the
boiling and freezing points of the solution. To
do this, we need to calculate the boiling-point
elevation and freezing-point depression. Plan In
order to calculate the boiling-point elevation
and the freezing-point depression using Equations
13.11 and 13.12, we must express the
concentration of the solution as molality. Lets
assume for convenience that we have 1000 g of
solution. Because the solution is 25.0 mass
ethylene glycol, the masses of ethylene glycol
and water in the solution are 250 and 750 g,
respectively. Using these quantities, we can
calculate the molality of the solution, which we
use with the molal boiling-point-elevation and
freezing-point-depression constants (Table 13.4)
to calculate ?Tb and ?Tf . We add ?Tb to the
boiling point and subtract ?Tf from the freezing
point of the solvent to obtain the boiling point
and freezing point of the solution.
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Comment Notice that the solution is a liquid
over a larger temperature range than the pure
solvent.
PRACTICE EXERCISE Calculate the freezing point of
a solution containing 0.600 kg of CHCl3 and 42.0
g of eucalyptol (C10H18O), a fragrant substance
found in the leaves of eucalyptus trees. (See
Table 13.4.)
Answer  65.6ºC
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Colligative Properties of Electrolytes

• Since these properties depend on the number of
  particles dissolved, solutions of electrolytes
  (which dissociate in solution) should show
  greater changes than those of nonelectrolytes.

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Colligative Properties of Electrolytes

• However, a 1 M solution of NaCl does not show
  twice the change in freezing point that a 1 M
  solution of methanol does.

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vant Hoff Factor

• One mole of NaCl in water does not really give
  rise to two moles of ions.

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vant Hoff Factor

• Some Na and Cl- reassociate for a short time,
  so the true concentration of particles is
  somewhat less than two times the concentration of
  NaCl.

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The vant Hoff Factor

• Reassociation is more likely at higher
  concentration.
• Therefore, the number of particles present is
  concentration dependent.

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The vant Hoff Factor

• We modify the previous equations by multiplying
  by the vant Hoff factor, i
• ?Tf Kf ? m ? i

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Osmosis

• Some substances form semipermeable membranes,
  allowing some smaller particles to pass through,
  but blocking other larger particles.
• In biological systems, most semipermeable
  membranes allow water to pass through, but
  solutes are not free to do so.

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Osmosis

• In osmosis, there is net movement of solvent
  from the area of higher solvent concentration
  (lower solute concentration) to the are of lower
  solvent concentration (higher solute
  concentration).

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Osmotic Pressure

• The pressure required to stop osmosis, known as
  osmotic pressure, ?, is

where M is the molarity of the solution
If the osmotic pressure is the same on both sides
of a membrane (i.e., the concentrations are the
same), the solutions are isotonic.play animation
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Osmosis in Blood Cells

• If the solute concentration outside the cell is
  greater than that inside the cell, the solution
  is hypertonic.
• Water will flow out of the cell, and crenation
  results.

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Osmosis in Cells

• If the solute concentration outside the cell is
  less than that inside the cell, the solution is
  hypotonic.
• Water will flow into the cell, and hemolysis
  results.

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Comment In clinical situations the
concentrations of solutions are generally
expressed as mass percentages. The mass
percentage of a 0.31 M solution of glucose is
5.3. The concentration of NaCl that is isotonic
with blood is 0.16 M because NaCl ionizes to
form two particles, Na and Cl (a 0.155 M
solution of NaCl is 0.310 M in particles). A 0.16
M solution of NaCl is 0.9 mass in NaCl. This
kind of solution is known as a physiological
saline solution.
PRACTICE EXERCISE What is the osmotic pressure at
20C of a 0.0020 M sucrose (C12H22O11) solution?
Answer 0.048 atm, or 37 torr
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Molar Mass from Colligative Properties

• We can use the effects of a colligative property
  such as osmotic pressure to determine the molar
  mass of a compound.

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Solution Analyze We must order five aqueous
solutions according to expected freezing points,
based on molalities and the solute
formulas. Plan The lowest freezing point will
correspond to the solution with the greatest
concentration of solute particles. To determine
the total concentration of solute particles in
each case, we must determine whether the
substance is a nonelectrolyte or an electrolyte
and consider the number of ions formed when it
ionizes. Solve CaCl2, NaCl, and HCl are strong
electrolytes, HC2H3O2, is a weak electrolyte, and
C12H22O11 is a nonelectrolyte. The molality of
each solution in total particles is as follows
Because the freezing points depend on the total
molality of particles in solution, the expected
ordering is 0.15 m NaCl (lowest freezing point),
0.10 m HCl, 0.050 m CaCl2, 0.10 m C12H22O11, and
0.050 m HC2H3O2, (highest freezing point).
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SAMPLE EXERCISE 13.10 continued
PRACTICE EXERCISE Which of the following solutes
will produce the largest increase in boiling
point upon addition to 1 kg of water 1 mol of
Co(NO3)2, 2 mol of KCl, 3 mol of ethylene glycol
(C2H6O2)?
Answer 2 mol of KCl because it contains the
highest concentration of particles, 2 m K and 2
m Cl, giving 4 m in all
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Solution Analyze Our goal is to calculate the
molar mass of a solute based on knowledge of the
boiling-point elevation of its solution in CCl4,
?Tb 0.357ºC, and the masses of solute and
solvent. Table 13.4 gives Kb for the solvent
(CCl4), Kb 5.02ºC/m. Plan We can use Equation
13.11, ?Tb Kbm, to calculate the molality of
the solution. Then we can use molality and the
quantity of solvent (40.0 g CCl4) to calculate
the number of moles of solute. Finally, the molar
mass of the solute equals the number of grams per
mole, so we divide the number of grams of solute
(0.250 g) by the number of moles we have just
calculated.
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PRACTICE EXERCISE Camphor (C10H16O) melts at
179.8C, and it has a particularly large
freezing-point-depression constant, Kf
40.0ºC/m. When 0.186 g of an organic substance
of unknown molar mass is dissolved in 22.01 g of
liquid camphor, the freezing point of the mixture
is found to be 176.7C. What is the molar mass of
the solute?
Answer 110 g/mol
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Comment Because small pressures can be measured
easily and accurately, osmotic pressure
measurements provide a useful way to determine
the molar masses of large molecules.
PRACTICE EXERCISE A sample of 2.05 g of
polystyrene of uniform polymer chain length was
dissolved in enough toluene to form 0.100 L of
solution. The osmotic pressure of this solution
was found to be 1.21 kPa at 25C. Calculate the
molar mass of the polystyrene.
Answer 4.20 ? 104 g/mol
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Colloids

• Suspensions of particles larger than individual
  ions or molecules, but too small to be settled
  out by gravity.

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Tyndall Effect

• Colloidal suspensions can scatter rays of light.
• This phenomenon is known as the Tyndall effect.

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Colloids in Biological Systems

• Some molecules have a polar, hydrophilic
  (water-loving) end and a nonpolar, hydrophobic
  (water-hating) end.

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Colloids in Biological Systems

• Sodium stearate is one example of such a
  molecule.

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Colloids in Biological Systems

• These molecules can aid in the emulsification of
  fats and oils in aqueous solutions.

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A kelvin has the same size as a degree Celsius.
(Section 1.4) Because the solution temperature
increases by 0.773C, the initial temperature was
27.0ºC 0.773ºC 26.2C.
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Table 13.4
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