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Chapter 11Solutions

• The Composition of Solutions

• The Solution Process

• Factors that Affect Solubility

• Measuring Concentrations of Solutions

• Quantities for Reactions That Occur in Aqueous Solution

• Colligative Solutions

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Composition of Solutions• A solution is:

– A homogeneous mixture with uniform composition throughout

– Composed of:

• Solute

– Substance being dissolved

– Usually present in the lesser amount

• Solvent

– Substance doing the dissolving

– Usually present in the greater amount

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Composition of Solutions

• What is the solute? What is the solvent?

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Composition of Solutions

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Figure 11.5

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Solubility

• How readily or completely a solute dissolves in a solvent– A soluble ionic compound dissociates into

ions when dissolved in water.

– Insoluble compounds remain mostly undissociated in their ionic, crystalline structure.

• The solubilities for ionic compounds do not follow a predictable pattern– Must be recalled from experimental data

– Solubility rules are listed in Table 11.1

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Most carbonates, phosphates, sulfites, and silicates are insoluble.

Exceptions: Na+, K+, and NH4+

CO32-, PO4

3-, SO32-,

SiO32-

Most chromates are insoluble. Exceptions: Salts of Na+, K+, NH4+,

Mg2+, Ca2+, Al3+, and Ni2+CrO42-

Sulfides are insoluble. Exceptions: Salts of Na+, K+, NH4+, and the

alkaline earth metal ionsS2-

Oxides and hydroxides are insoluble. Exceptions: NaOH, KOH,

Ba(OH)2, & Ca(OH)2 (somewhat soluble)O2-, OH-

Silver salts, except AgNO3, are insoluble.Ag+

Most chlorides, bromides, and iodides are soluble. Exceptions: AgX, Hg2X, PbX2, & HgI2 (X = Cl, Br, or I)

Cl-, Br-, I-

Most sulfates are soluble. Exceptions: BaSO4, SrSO4, PbSO4,

CaSO4, Hg2SO4, & Ag2SO4

SO42-

All nitrates are soluble.NO3-

Most salts of sodium, potassium, and ammonium ions are soluble.Na+, K+, NH4+

RulesIons

Rules Used to Predict the Solubility of Ionic Salts

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Electrolytes• A solution that conducts electricity

• Two kinds of electrolytes:– Strong electrolyte

• A solute that dissociates completely into ions in aqueous solution

– Weak electrolyte• A solute that dissociates partially into ions

in aqueous solution

• Nonelectrolyte– A solute that does not dissociate into

ions in aqueous solution11-

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Electrolytes

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Practice – Composition of Solutions

• What ions, atoms, or molecules are

present when the following

substances are mixed with water?

Write a balanced equation to

represent the process of dissolving

each substance in water.

– ZnCl2(s)

– KNO3(s)

– C5H11OH(l)11-

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Practice Solutions – Composition of Solutions

– ZnCl2(s)

ZnCl2(s) + H2O(l) ���� Zn2+(aq) + 2 Cl-(aq)

– KNO3(s)

KNO3(s) + H2O(l) ���� K+(aq) + NO3-(aq)

– C5H11OH(l)

C5H11OH(l) + H2O(l) ���� C5H11O-(aq) + H+(aq)

Does not dissolve extensively…

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Like Dissolves Like• When the bonding in a solvent is similar to the bonding in a

solute, then the solute will dissolve in the solvent.

– Polar covalent solvents dissolve polar covalent solutes.

– Nonpolar covalent solvents dissolve nonpolar covalent solutes.

– See Table 11.2 below:

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InsolubleSolubleIonic

SolubleInsolubleNonpolar

InsolubleSolublePolar

NonpolarPolar

SolventSolute

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Solution Process

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Figure 11.12

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Solution Process• Ion-dipole force

– An attraction between an ion and a polar molecule

• When an ionic compound dissolves in water:– Ionic bonds in the solute

break.

– Hydrogen bonds between water molecules break.

– Ion-dipole forces form between ions and water molecules.

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Solution Process

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Heat of Solution

• The difference between the energy required for separation of solute and

solvent and the energy released upon formation of ion-dipole interactions

• When the solvation process provides

more energy than is needed to separate the pure solvent and solvent particles, the

heat of solution is negative, and the overall dissolving process is exothermic.

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Endothermic Solution Process

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Figure 11.14B

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Solution Process for Nonpolar-Nonpolar Interactions

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Entropy

• A measure of the tendency for matter to become disordered or random in its distribution

• Matter spontaneously changes from a state of order to a state of randomness, unless energy changes oppose it.

• Solutions form because the makeup of the solution naturally tends toward disorganization.

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Entropy

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Figure 11.16

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Practice – The Solution Process

• Iodine (I2) dissolves in hexane

(C6H14) to form a solution. Describe

the forces that must be broken for

the solution to form.

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Practice Solutions – The Solution Process

• Iodine (I2) dissolves in hexane

(C6H14) to form a solution. Describe

the forces that must be broken for

the solution to form.

London forces must be broken to

allow I2, the solute, to dissolve.

London forces must also be broken

to allow C6H14, the solvent, to

dissolve.11-

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Factors that Affect Solubility

• Three factors

affect

solubility:

– Structure

– Temperature

– Pressure

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Structure• For a solid to be soluble in a

given solvent, the energy released by the solvation process must compensate for all of the energy required to break up the forces at work both within the solute and the solvent.

– Entropy is also a factor

• The solubility (or miscibility) of liquids in other liquids depends largely on the polarity of the solute and solvent molecules.

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Temperature• Affects the solubility of most substances in

water– Most ionic solids are more soluble in water at higher

temperatures

– Gases are less soluble as temperature increases

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Figure 11.19

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Pressure

• Affects gases but not liquids or

solids

• The solubility of a gas in a liquid is

directly proportional to the pressure

of the gas above the liquid

• An increase in pressure results in an

increase in solubility of a gas

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Pressure

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Figure 11.20

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Measuring Concentrations of Solutions• Concentration

– The relative amounts of the solutes and solvent that make up the solution

• Solubility– The ratio that identifies the maximum amount of a

solute that will dissolve in a particular solvent to form a stable solution under specified conditions

– A measure of maximum concentration

– Three types of concentrated solution:

• Saturated solution

• Unsaturated solution

• Supersaturated solution11-

solvent)(or solution ofamount

solute ofamount ion Concentrat =

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Saturated Solution• A solution that contains the

maximum possible amount of solute

– When this type of solution forms, a dynamic equilibrium is established.

• Undissolved solute continues to dissolve, but at the same time, previously dissolved

solute is deposited from solution.

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Supersaturated Solution• A solution that contains more than the

maximum possible amount of solute• Unsaturated solution

– A solution that contains less than the maximum possible amount of solute

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Figure 11.23

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Table 11.3: Concentration Units

moles of solute

liters of solutionMolarity (M)

moles of solute

kilograms of solventMolality (m)

grams of solute x 109

grams of solutionParts per billion

grams of solute x 106

grams of solutionParts per million

grams of solute x 100%

volume of solutionMass/volume percent

volume of solute x 100%

volume of solutionPercent by volume

grams of solute x 100%

grams of solutionPercent by mass

DefinitionUnit

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Percent by Mass

• Is calculated by dividing the solute

mass by the mass of the solution

and multiplying it by 100%

Percent by mass = mass (solute) x 100%

mass (solution)

• To find the mass of the solution, we

can add the mass of the solute to the

mass of the solvent.

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Percent by Mass

• Example:

– What is the percent-by-mass concentration of acetic acid (CH3CO2H) in a vinegar solution

that contains 2.70 g of acetic acid and 122.8 g of water?

Percent by mass = mass (solute) x 100%

mass (solution)

Percent by mass = 2.70 g x 100%

(2.70 g + 122.8 g)

Percent by mass = 2.15%11-

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Percent by Volume

• Is calculated by dividing the solute

volume by the solution volume and

multiplying by 100%

Percent by volume = volume (solute) x 100%

volume (solution)

• The volume of the solution must be

measured. Volumes are not additive

like masses.

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Percent by Volume• Example:

– To prepare a solution, we dissolve 205 mL of ethanol in sufficient water to give a total volume of 235 mL. What is the percent by volume concentration of ethanol?

Percent by volume = volume (solute) x 100%

volume (solution)

Percent by volume = 205 mL x 100%

235 mL

Percent by volume = 87.2%

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Mass/Volume Percent

• Is calculated by dividing the solute mass by the volume of solution and multiplying

by 100%

Mass/Volume Percent = grams of solute x 100%

volume of solution

• Example:

– The concentration of NaCl in a saline bag is 0.9%. Therefore, for every 100 mL of

solution, the mass of NaCl present is 0.9 g.

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Parts per Million and per Billion

• Parts per million– Is calculated by dividing the mass of solute

by the mass of solution and multiplying by 1 million (106)

Parts per million = mass (solute) x 106

mass (solution)

• Parts per billion– Is calculated by dividing the mass of solute

by the mass of solution and multiplying by 1 billion (109)

Parts per billion = mass (solute) x 109

mass (solution)11-

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Molarity

• The ratio of moles of solute to the volume of solution in liters

Molarity (M) = moles (solute)

liters (solution)

• Example– A solution contains 117 g of potassium

hydroxide (KOH) dissolved in sufficient water to give a total volume of 2.00 L. The molar mass of KOH is 56.11 g/mol. What is the molarity of the aqueous potassium hydroxide solution?

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Molarity• Example

– A solution contains 117 g of potassium hydroxide (KOH)

dissolved in sufficient water to give a total volume of 2.00 L. The molar mass of KOH is 56.11 g/mol. What is the

molarity of the aqueous potassium hydroxide solution?

Molarity (M) = moles (solute)

liters (solution)

Moles KOH = 117 g KOH x 1 mol KOH = 2.09 mol KOH

56.11 g KOH

Molarity (M) = 2.09 mol KOH = 1.04 M KOH

2.00 L

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Molality

• If the temperature of a solution increases, the molarity of the solution changes

because the volume changes.

• Molality does not change with temperature because it divides moles of

solute with the mass of solution.

molality (m) = moles (solute)

kilograms (solution)

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Molality• Example

– A solution contains 56.7 g of potassium hydroxide (KOH)

dissolved in sufficient water to give a total mass of 245.8 g. The molar mass of KOH is 56.11 g/mol. What is the

molality of the aqueous potassium hydroxide solution?

molality (m) = moles (solute)

kilograms (solution)

moles (solute) = 56.7 g KOH x 1 mol KOH = 1.01 mol KOH

56.11 g KOH

kilograms (solvent) = 245.8 g x 1 kg = 0.2458 kg

1000 g

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Molality• Example

– A solution contains 56.7 g of potassium hydroxide (KOH) dissolved in sufficient water to give a total mass of 245.8 g. The molar mass of KOH is 56.11 g/mol. What is the molality of the aqueous potassium hydroxide solution?

molality (m) = moles (solute)

kilograms (solution)

molality (m) = 1.01 mol KOH = 4.11 m KOH

0.2458 kg KOH

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Precipitation Reactions• The cation from one reactant combines

with the anion from another reactant to form an insoluble compound, called a precipitate.

• Example– When a solution of lead(II) nitrate is mixed

with a solution of potassium chromate, a yellow precipitate forms according to the equation:

Pb(NO3)2(aq) + K2CrO4(aq) ���� 2 KNO3(aq) + PbCrO4(s)

What volume of 0.105 M lead(II) nitrate is required to react with 100.0 mL of 0.120 Mpotassium chromate? What mass of PbCrO4solid forms?

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Precipitation ReactionsPb(NO3)2(aq) + K2CrO4(aq) ���� 2 KNO3(aq) + PbCrO4(s)

What volume of 0.105 M lead(II) nitrate is required to react with 100.0 mL of 0.120 M potassium chromate? What mass of PbCrO4 solid forms?

To solve for volume of Pb(NO3)2:

M x V = moles100.0mL K2CrO4 x 0.120 moles K2CrO4 = 0.0120 moles K2CrO4

1000 mL K2CrO4

0.0120 moles K2CrO4 x 1 mol Pb(NO3)2 = 0.0120 moles Pb(NO3)2

1 mol K2CrO4

0.0120 moles Pb(NO3)2 x 1 L Pb(NO3)2 = 0.114 L Pb(NO3)2

0.105 mol Pb(NO3)2

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Precipitation ReactionsPb(NO3)2(aq) + K2CrO4(aq) ���� 2 KNO3(aq) + PbCrO4(s)

What volume of 0.105 M lead(II) nitrate is required to

react with 100.0 mL of 0.120 M potassium chromate? What mass of PbCrO4 solid forms?

To solve for volume of Pb(NO3)2:

M x V = moles0.0120 moles K2CrO4 * 1 mol PbCrO4 = 0.0120 moles PbCrO4

1 mol K2CrO4

0.0120 moles PbCrO4 * 323.2 g PbCrO4 = 3.88 g PbCrO4

1 mol PbCrO4

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Acid-Base Titrations• Titration

– The process of determining the concentration of one substance in solution by reacting it with a solution of another substance that has a known concentration

• Equivalence point

– The point in a titration at which the moles of the acid equal the moles of the base

• End point

– The point in a titration where the indicator changes color

– Slightly different than the equivalence point, but a good estimation

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Acid-Base Titrations

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Practice - Acid-Base Titrations

• We need 28.18 mL of 0.2437 M HCl solution to completely titrate 25.00 mL of

Ba(OH)2 solution of unknown concentration. What is the molarity of the

barium hydroxide solution?

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Practice Solutions - Acid-Base Titrations

• We need 28.18 mL of 0.2437 M HCl solution to completely titrate 25.00 mL of Ba(OH)2 solution of unknown concentration. What is the molarity of the barium hydroxide solution?

First, write the balanced chemical

equation:

2 HCl(aq) + Ba(OH)2(aq) ���� BaCl2(aq) + 2 H2O(l)

Next, find the moles of HCl:

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HCl moles 10 x 6.865 HCl mL 1000

HCl moles 0.2437 x HCl mL 17.28 3-

=

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Practice Solutions - Acid-Base

TitrationsHCl(aq) + Ba(OH)2(aq) ���� BaCl2(aq) + H2O(l)

Next, we need to find the volume (in L) of

Ba(OH)2:

Finally, divide the moles of Ba(OH)2 by the volume of Ba(OH)2:

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2

2

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2 Ba(OH) M 0.2746 Ba(OH) L 02500.0

Ba(OH) moles 10 x 6.865 ][Ba(OH)Molarity ==

2

2

22 Ba(OH) L 0.02500

Ba(OH) mL 1000

Ba(OH)liter 1 x Ba(OH) mL 00.25 =

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Acid-Base Titrations

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Volumeof

Reactant

Molesof

Reactant

Molesof

Product

Volumeof

Product

Mreactant

Mproduct

moleratio

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Colligative Properties

• A property that does not depend on the identity of a solute in solution

• Vary only with the number of solute particles present in a specific quantity of solvent

• 4 colligative properties:– Osmotic pressure

– Vapor pressure lowering

– Boiling point elevation

– Freezing point depression

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Osmotic Pressure• Osmosis

– A process in which solvent molecules diffuse through a barrier that does not allow the passage of solute particles

• The barrier is called a semipermeable membrane.– A membrane that allows the

passage of some substances but not others

• Osmotic Pressure– Pressure that can be exerted

on the solution to prevent osmosis

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Osmotic Pressure

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Figure 11.28

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Vapor Pressure Lowering• Solutes come in 2 forms:

– Volatile

• Solutes that readily form a gas

– Nonvolatile

• Solutes that DO NOT readily form a gas

• Generally, the addition of a solute lowers the vapor pressure of a solution when compared to the pure solvent.

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Vapor Pressure Lowering

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Phase Diagram

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Boiling Point Elevation• The addition of solute affects the boiling point

because it affects the vapor pressure.

• The boiling point is raised with the addition of solute in comparison to the pure solvent.

• An equation that gives the increase in boiling point:

∆Tb = Kbm

Where ∆Tb is the increase in temperature from the pure solvent’s boiling point, Kb is the boiling point constant, which is characteristic of a

particular solvent, and m is the molality (moles of solute per kg of solution)

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Freezing Point Depression

• The freezing point is lowered with the addition of solute in comparison to the pure solvent.

• An equation that gives the decrease in freezing point:

∆Tf = KfmWhere ∆Tf is the decrease in temperature

from the pure solvent’s freezing point, Kfis the freezing point constant, which is

characteristic of a particular solvent, and m is the molality (moles of solute per kg

of solution) 11-

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Practice – Freezing Point Depression

• What is the change in freezing point

for a 1.5 m solution of sucrose in

water?

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Practice Solutions – Freezing Point Depression

• What is the change in freezing point

for a 1.5 m solution of sucrose in

water?

Kf (water) = -1.86°C/m

∆Tf = Kfm

∆Tf = -1.86°C * 1.5 m = -2.79°C

m

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Colligative Properties and Strong Electrolytes

• Strong electrolytes dissociate most of the time into their constituent ions.

• Therefore, the number of particles (in this case ions) increases with the number of ions.

• Colligative properties are proportional to number of particles in solution.

• Example:

MgCl2(s) ���� Mg2+(aq) + 2 Cl-(aq)

In this case the number of particles increases to 3 particles. Therefore, we would multiply the colligative property amount by 3.

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Practice – Strong vs. Weak Electrolytes

• Which of the following aqueous

solutions is expected to have the

lowest freezing point?

– 0.5 m CH3CH2OH

– 0.5 m Ca(NO3)2

– 0.5 m KBr

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Practice Solutions – Strong vs. Weak Electrolytes

• Which of the following aqueous solutions is expected to have the lowest freezing point?

– 0.5 m CH3CH2OH– 0.5 m Ca(NO3)2

– 0.5 m KBr

Because the molalities are the same, we only need to concentrate on the number of particles generated by each solute. CH3CH2OH will not dissociate into ions, Ca(NO3)2 dissociates into 3 ions, and KBr dissociates into 2 ions. Therefore, the largest number of particles is generated by Ca(NO3)2 and it will have the lowest freezing point.

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