Solutions

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The Properties of Mixtures:

Solutions and Colloids

Chapter 13

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The Properties of Mixtures: Solutions and Colloids

13.1 Types of Solutions: Intermolecular Forces and Predicting Solubility

13.2 Intermolecular Forces and Biological Macromolecules

13.3 Why Substances Dissolve: Understanding the Solution Process

13.5 Quantitative Ways of Expressing Concentration

13.6 Colligative Properties of Solutions

13.7 The Structure and Properties of Colloids

13.4 Solubility as an Equilibrium Process

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Solution: homogenous mixtures of two or more components that can be varied in compositionSolvent: component present in greatest amount; substance in which a solute dissolvesSolute: other components; substance that is dissolvedAqueous solutions: solution in which water is the solventSolvation: interaction between solute and solvent molecules; due to IMF; e.g. Na+ and Cl- ion surrounded byH2O moleculesHydration: solvation when solvent is water

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Types of Solutiondilute – solution with low solute concentrationconcentrated – one with high solute concentration

solubility: maximum amount of solute that can be dissolved in a given amount of solventsaturated solution – a solution that contains the maximum amount of solute the solvent can dissolves; no more solute can dissolve in itunsaturated solution – solution containing amount of solute less than its solubility; more solute can dissolve in itsupersaturated solution – solution containing an amount of solute greater than the solubility; unstable solution

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Factors Affecting Solubility1.TemperatureGases: ↑ T, ↓ solubilitySolids: ↑ T, ↑ solubility 1.Nature of Solute and Solvent (Solute-Solvent Interactions)miscible liquids – pair of liquids that mix in all proportion, e.g. ethanol-H 2O immiscible liquids – liquids that do not mix, e.g. oil-H2O

“Like dissolves like” Substances with similar intermolecular attractive forces tend to be soluble in one another: polar ↔ polar nonpolar ↔ nonpolar1.Pressure (for Gases)↑ pressure, ↑ solubility Henry’s Law: Cg = k Pg Cg ≡ solubility of gas in solution

K ≡ Henry’s law constantPg ≡ partial pressure of the gas

over solution

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Correlation Between Boiling Point and Solubility in Water

Table 13.3

Gas Solubility (M)* bp (K)

He

Ne

N2

CO

O2

NO

4.2 x 10-4 4.2

6.6 x 10-4 27.1

10.4 x 10-4 77.4

15.6 x 10-4 81.6

21.8 x 10-4 90.2

32.7 x 10-4 121.4

* At 273K and 1 atm

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Henry’s Law

Sgas = kH X Pgas

The solubility of a gas (Sgas) is directly proportional to the partial pressure of the gas (Pgas) above the solution.

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LIKE DISSOLVES LIKE

Substances with similar types of intermolecular forces dissolve in each other.

When a solute dissolves in a solvent, solute-solute interactions and solvent-solvent interactions are being replaced with solute-solvent interactions. The forces must be comparable in strength in order to have a solution occur.

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Figure 13.3 Like dissolves like: solubility of methanol in water.

water

methanol

A solution of methanol in water

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Figure 13.8 The structure and function of a soap.

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Figure 13.9 The structure of lecithin.

lecithinphospholipid found in all cell membranes

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(c) Ethanol - Diethyl ether can interact through a dipole and dispersion forces. Ethanol can provide both while water would like to H bond.

(b) Water - Hexane has no dipoles to interact with the -OH groups in ethylene glycol. Water can H bond to the ethylene glycol.

SAMPLE PROBLEM 13.1 Predicting Relative Solubilities of Substances

SOLUTION:

PROBLEM: Predict which solvent will dissolve more of the given solute:

(a) Sodium chloride in methanol (CH3OH) or in propanol (CH3CH2CH2OH)

(b) Ethylene glycol (HOCH2CH2OH) in hexane (CH3CH2CH2CH2CH2CH3)

or in water.

(c) Diethyl ether (CH3CH2OCH2CH3) in water or in ethanol (CH3CH2OH)

PLAN: Consider the intermolecular forces which can exist between solute molecules and consider whether the solvent can provide such interactions and thereby substitute.

(a) Methanol - NaCl is ionic and will form ion-dipoles with the -OH groups of both methanol and propanol. However, propanol is subject to the dispersion forces to a greater extent.

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Classification of SolutionsA. Based on elemental compositionOrganic – compounds containing carbon (except CO2, CO, carbonates and cyanides)Inorganic – compounds of the other elements including acids, bases, and saltsB. Based on Ionization/Electrolytic Property of SoluteElectrolytic property – the ability of the solution to conduct electricityElectrolytes – substances whose aqueous solutions contain ions and thus conduct electricitystrong electrolytes – substances which completely dissociates into ions

e.g. salts, strong acids, strong basesNaCl → Na+ (aq) + Cl- (aq)

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* weak electrolytes – substances which produce small amounts of ions; partially dissociated into ionse.g. weak acids, weak bases

CH3COOH ↔ CH3COO- (aq) + H+ (aq)•CH3COOH ionizes to form CH3COO- abd H+. While this happens, some ions also combine to form back CH3COOH. This results to partial dissociation

Nonelectrolytes – substances that does not dissociate into ions; form nonconducting solutionse.g. most molecular compounds are nonelectrolytes

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Figure 13.25 The three types of electrolytes.

STRONG

weak

nonelectrolyte

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Table 13.5 Concentration Definitions

Concentration Term Ratio

Molarity (M)amount (mol) of solutevolume (L) of solution

Molality (m)amount (mol) of solute

mass (kg) of solvent

percent by massmass of solute

mass of solution

Percent by volumevolume of solute

volume of solution

Mole fraction amount (mol) of solute

amount (mol) of solute + amount (mol) of solvent

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SAMPLE PROBLEM 13.3 Calculating Molality

PLAN:

SOLUTION:

PROBLEM: What is the molality of a solution prepared by dissolving 32.0 g of CaCl2 in 271 g of water?

We have to convert the grams of CaCl2 to moles and the grams of water to kg. Then substitute into the equation for molality.

molality =

= 0.288 mole CaCl2

271 g H2O

0.288 mole CaCl2

kg

103 gx

= 1.06 m CaCl2

32.0 g CaCl2mole CaCl2

110.98 g CaCl2x

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SAMPLE PROBLEM 13.4 Expressing Concentration in Parts by Mass, Parts by Volume, and Mole Fraction

PLAN:

PROBLEM: (a) Find the concentration of calcium (in ppm) in a 3.50-g pill that contains 40.5 mg of Ca.

(b) The label on a 0.750-L bottle of Italian chianti indicates “11.5% alcohol by volume”. How many liters of alcohol does the wine contain?

(c) A sample of rubbing alcohol contains 142 g of isopropyl alcohol (C3H7OH) and 58.0 g of water. What are the mole fractions of alcohol and water?

(a) Convert mg to g of Ca, find the ratio of g Ca to g pill and multiply by 106.

(b) Knowing the % alcohol and total volume, we can find volume of alcohol.

(c) Convert g of solute and solvent to moles; find the ratios of parts to the total.

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SAMPLE PROBLEM 13.4 Expressing Concentrations in Parts by Mass, Parts by Volume, and Mole Fraction

SOLUTION:

continued

(a)

3.5 g

103 mg

g40.5 mg Ca x

106x = 1.16x104 ppm Ca

(b) 11.5 L alcohol

100 L chianti0.750 L chianti x = 0.0862 L alcohol

(c) moles ethylene glycol = 142 gmole

60.09 g= 2.36 mol C3H8O

moles water = 38.0gmole

18.02 g= 3.22 mol H2O

2.39 mol C3H8O

2.39 mol C3H8O2+ 3.22 mol H2O

3.22 mol H2O

2.39 mol C3H8O2+ 3.22 mol H2O

= 0.423 C2H6O2= 0.577 H2O

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Interconverting Concentration Terms

To convert a term based on amount (mol) to one based on mass, you need the molar mass. These conversions are similar to mass-mole conversions.

To convert a term based on mass to one based on volume, you need the solution density. Working with the mass of a solution and the density (mass/volume), you can obtain volume from mass and mass from volume.

Molality involves quantity of solvent, whereas the other concentration terms involve quantity of solution.

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SAMPLE PROBLEM 13.5 Converting Concentration Units

PLAN:

SOLUTION:

PROBLEM: Hydrogen peroxide is a powerful oxidizing agent used in concentrated solution in rocket fuels and in dilute solution a a hair bleach. An aqueous solution H2O2 is 30.0% by mass and has a density of 1.11 g/mL. Calculate its

(a) Molality (b) Mole fraction of H2O2 (c) Molarity

(a) To find the mass of solvent we assume the % is per 100 g of solution. Take the difference in the mass of the solute and solution for the mass of peroxide.

(b) Convert g of solute and solvent to moles before finding .

(c) Use the density to find the volume of the solution.

(a) g of H2O = 100. g solution - 30.0 g H2O2 = 70.0 g H2O

molality =

30.0 g H2O2

34.02 g H2O2

mol H2O2

70.0 g H2Okg H2O103 g

= 12.6 m H2O2

0.882 mol H2O2

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SAMPLE PROBLEM 13.5 Converting Concentration Units

continued

(b)

0.882 mol H2O2

70.0 g H2Omol H2O

18.02 g H2O= 3.88 mol H2O

0.882 mol H2O2 + 3.88 mol H2O= 0.185 of H2O2

(c)100.0 g solution

mL

1.11 g= 90.1 mL solution

0.882 mol H2O2

90.1 mL solution L

103 mL

= 9.79 M H2O2

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Dilution of SolutionWhen a solution is diluted:the volume is increased by adding more solventthe concentration is decreased, andthe total amount of solute remains constant

Dilution Formula: M1V1 = M2V2where M1 ≡ initial concentration

M2 ≡ final concentrationV1 ≡ initial volumeV2 ≡ final volume

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Polyprotic Acids – conatins more than one 2 replaceable H+i. H3PO4 + 3 NaOH → Na3PO 4 + 3 H2O a = 3 eq/molii. H3PO4 + 2 NaOH → Na2HPO 4 + 2 H2O a = 2 eq/moliii. H3PO4 + NaOH → NaH2PO 4 + H2O a = 1 eq/mol

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COLLIGATIVE PROPERTIESproperties that are dependent only on the concentrations of solute, not on the natureapplicable only to non-volatile, non-electrolytic solutions

Vapor Pressure Lowering/DepressionBoiling Point Elevation Freezing Point Depression Osmotic Pressure, π

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Colligative Properties

Raoult’s Law (vapor pressure of a solvent above a solution, Psolvent)

Psolution = solvent X P0solvent

where P0solvent is the vapor pressure of the pure solvent

P0solvent - Psolution = P = solute x P0

solvent

Boiling Point Elevation and Freezing Point Depression

Tb = Kbm Tf = Kfm

Osmotic Pressure

M R T where M is the molarity, R is the ideal gas law

constant and T is the Kelvin temperature

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1.Vapor Pressure Lowering/DepressionWhen a liquid is placed in a sealed container, a certain amount will evaporate as vapor to completely occupy the container.

4

The vapors exerts a pressure (vapor pressure) over the liquid

The VP of a liquid depends on temperature: hi T → hi VPAddition of nonvolatile solute to a liquid reduces the capacity of the solvent molecules to evaporate (lower VP)Vapor pressure lowering α concentration of solute

RAOULT’S LAW – predicts the VP of solutions containing nonvolatile solutes

PA = χA PAo PA ≡ VP of solutionχA ≡ mole fraction of solventPA

o ≡ VP of pure solvent

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Figure 13.23 The effect of pressure on gas solubility.

Pure solventSolution

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Ideal Solution – a solution that obeys Raoult’s Law; ideal behavior can at low solute concentration, solute and solventIdea Solutions with two or more volatile components (A and B) → partial pressures of A and B

PA = χAPAo

PB = χBPBo

PTOTAL = PA + PB

PTOTAL = χAPAo + χBPB

o

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SAMPLE PROBLEM 13.6 Using Raoult’s Law to Find the Vapor Pressure Lowering

SOLUTION:

PROBLEM: Calculate the vapor pressure of the solution when 10.0 mL of glycerol (C3H8O3) is added to 500. mL of water at 50.0C. At this temperature, the vapor pressure of pure water is 92.5 torr and its density is 0.988 g/mL. The density of glycerol is 1.26 g/mL.

PLAN: Find the mol fraction, , of glycerol in solution and multiply by the vapor pressure of water.

10.0 mL C3H8O3

1.26 g C3H8O3

mL C3H8O3

mol C3H8O3

92.09 g C3H8O3

= 0.137 mol C3H8O3

500.0 mL H2O0.988 g H2O

mL H2O

mol H2O

18.02 g H2O= 27.4 mol H2O

P = 27.4 mol H2O

0.137 mol C3H8O3 + 27.4 mol H2O92.5 torrx

x

x

= 92.0 torr

= 0.00498

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Table of Constants:

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SAMPLE PROBLEM 13.7 Determining the Boiling Point Elevation and Freezing Point Depression of a Solution

SOLUTION:

PROBLEM: You add 1.00 kg of ethylene glycol (C2H6O2) antifreeze to your car radiator, which contains 4450 g of water. What are the boiling and freezing points of the solution?

PLAN: Find the # mols of ethylene glycol; m of the solution; multiply by the boiling or freezing point constant; add or subtract, respectively, the changes from the boiling point and freezing point of water.

1.00x103 g C2H6O2

mol C2H6O2

62.07 g C2H6O2

= 16.1 mol C2H6O2

Tbp = 0.512 0C/m

16.1 mol C2H6O2

4.450 kg H2O= 3.62 m C2H6O2

3.62mx = 1.850C

BP = 101.85 0C

Tfp = 1.86 0C/m 3.62mx

FP = -6.73 0C

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Freezing Point, oC

Kf, oC/m Boiling

Point, oC

Kb, oC/m

Acetic Acid

16.6 3.90 118.1 3.07

Benzene

5.51 4.90 80.1 2.53

Water 0.00 1.86 100.0 0.512

CCl4 -22.8 31.8 76.8 5.03

Ethanol -117.3 1.99 78.5 1.22

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Table 13.6 Molal Boiling Point Elevation and Freezing Point Depresssion Constants of Several Solvents

SolventBoilingPoint (0C)* Kb (0C/m) Kb (0C/m)

MeltingPoint (0C)

Acetic acid

Benzene

Carbon disulfide

Carbon tetrachloride

Chloroform

Diethyl ether

Ethanol

Water

117.9

80.1

46.2

76.5

61.7

34.5

78.5

100.0

3.07 16.6 3.90

2.53 5.5 4.90

2.34 -111.5 3.83

5.03 -23 30.

3.63 -63.5 4.70

2.02 -116.2 1.79

1.22 -117.3 1.99

0.512 0.0 1.86

*at 1 atm.

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nRT

V

4. Osmotic Pressure, π- pressure required to prevent osmosis from occurring

Osmosis – net movement of solvent molecules from a region of low solute concentration to a region of high solute concentration through a semi-permeable membrane(allows selective passage of water molecules)

π = MRT =

Where M = molarityR = gas constant = 0.0821 L·atm/K·molT = absolute temperature

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When Papplied < π: osmosis takes place in the normal way and water moves through the membrane in the solutionWhen Papplied > π: reverse osmosis occurs wherein water molecules move through the membrane from the solution to pure water.

*Note: Reverse osmosis is employed in making freshwater from seawater, where the applied pressure is large enough to reverse the normal precess.

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Figure 13.28 The development of osmotic pressure.

pure solvent

solution

net movement of solvent

semipermeable membrane

solvent molecules

solute molecules

osmotic pressure

Applied pressure needed to prevent volume increase

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Figure B13.3 Reverse osmosis for the removal of ions.

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SAMPLE PROBLEM 13.8 Determining Molar Mass from Osmotic Pressure

SOLUTION:

PROBLEM: Biochemists have discovered more than 400 mutant varieties of hemoglobin, the blood protein that carries oxygen throughout the body. A physician studing a variety associated with a fatal disease first finds its molar mass (M). She dissolves 21.5 mg of the protein in water at 5.00C to make 1.50 mL of solution and measures an osmotic pressure of 3.61 torr. What is the molar mass of this variety of hemoglobin?

PLAN: We know as well as R and T. Convert to atm and T to degrees K. Use the equation to find M and then the amount and volume of the sample to get to M.

M =

RT=

3.61 torr atm

760 torr

(0.0821 L*atm/mol*K)(278.1 K)

= 2.08 x10-4 M

2.08 x10-4 molL

(1.50 mL)103 mL

L= 3.12x10-8 mol

21.5 mg g103 mg

13.12 x10-8 mol

= 6.89 x104 g/mol

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DEFINITIONS:Isotonic Solutions – two solutions having the same osmotic pressure; equal concentrationsHypotonic Solutions – less concentrated solutionsHypertonic Solutions – more concentrated solutionsCrenation – caused by movement of water from a (hypotonic) cellHemolysis – caused by movement of water into a (hypertonic) cellActive transport – opposite of osmosis; movement of substances from a region of low concentration to a region of high concentration

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Colligative Properties of Electrolyte Solutions

For electrolyte solutions, the compound formulatells us how many particles are in the solution.

van’t Hoff factor (i)

i =measured value for electrolyte solution

expected value for nonelectrolyte solution

For vapor pressure lowering: P = i(solutex P0solvent)

For boiling point elevation: Tb = i(bm)

For freezing point depression: Tf = i(fm)

For osmotic pressure : = i(MRT)

The van’t Hoft factor, i, tells us what the “effective” number of ions are in the solution. The more ions, the larger the value of vant hoff facor.

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Colligative Properties of Electrolytic Solutionsdifferent from nonelectrolytes since ions (of opposing charges) have the tendency to stick together and from ION PAIRSformation of ion pairs causes slight changes in π, ΔTf, ΔTb, and VPvan’t Hoff factor, i – measure of the extent of dissociation of electrolytes

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Types of Homogeneous Mixture

Kind of Mixture Particle Size, nm

Examples Characteristics

Solution 0.2 – 2.0 Air, seawater, gasoline, wine

Transparent to light; does not separate on standing; nonfilterable

Colloid 2.0 – 1000 Butter, milk, fog, pearl

Often murky or opaque to light (Tyndall Effect); does not separate on standing; nonfilterable

Suspension > 1000 Blood, paint Murky or opaque to light; separate on standing; filterable

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Colloid – a dispersion of particles of one substance (the dispersed phase) throughout another substance (the continuous phase)Tyndal Effect – the scattering of light of by colloidal-sized particlesCoagulation – process by which the dispersed phase of a colloid is made to aggregate and thereby separate from the continuous phase, e.g. curdling of milk when in sours (lactose, milk sugar, ferments to lactic acid)

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Figure 13.32 Light scattering and the Tyndall effect.

Photo by C.A.Bailey, CalPoly SLO (Inlay Lake, Myanmar)

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SAMPLE PROBLEM 13.9 Depicting a Solution to Find ItsColligative Properties

PROBLEM: A 0.952-g sample of magnesium chloride is dissolved in 100. g of water in a flask.

(a) Which scene depicts the solution best?

(b) What is the amount (mol) represented by each green sphere?

(c) Assuming the solution is ideal, what is its freezing point (at 1 atm)?

PLAN: (a) Consider the formula for magnesium chloride, an ionic compound.

(b) Use the answer to part (a), the mass given, and the mol mass.

(c) The total number of mols of cations and anions, mass of solvent, and equation for freezing point depression can be used to find the new freezing point of the solution.

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continued

(a) The formula for magnesium chloride is MgCl2; therefore the correct depiction must be A with a ratio of 2 Cl-/ 1 Mg2+.

(b)mols MgCl2 = = 0.0100 mol MgCl2

0.952 g MgCl2

95.21 g MgCl2mol MgCl2

mols Cl- = 0.0100 mol MgCl2 x2 mols Cl-

1 mol MgCl2= 0.0200 mols Cl-

mols/sphere = 0.0200 mols Cl-

8 spheres= 2.50 x 10-3 mols/sphere

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continued

(c) molality (m) = 0.0100 mol MgCl2

100. g x103 g

1 kg

= 0.100 m MgCl2

Assuming this is an IDEAL solution, the van’t Hoff factor, i, should be 3.

Tf = i (Kfm) = 3(1.86 0C/m x 0.100 m) = 0.558 0C

Tf = 0.000 0C - 0.558 0C = - 0.558 0C

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