Chapter 12 Solutions

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Chapter 12 Solutions

Sherril Soman,

Grand Valley State University

Lecture Presentation

12.1 Thirsty Solutions: Why You Shouldn’t Drink Seawater 544

12.2 Types of Solutions and Solubility 546 12.3 Energetics of Solution Formation 551 12.4 Solution Equilibrium and Factors

Affecting Solubility 555

12.5 Expressing Solution Concentration 559 12.6 Colligative Properties: Vapor Pressure Lowering, Freezing Point

Depres sion, Boiling Point Elevation, and Osmotic Pressure 567

12.7 Colligative Properties of Strong Electrolyte Solutions 579

12.8 Colloids 582

Key Learning Outcomes 587

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12.1 Thirsty Solutions: Why You Shouldn’t Drink Seawater

Thirsty Seawater

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12.2 Types of Solutions and Solubility

When solutions with different solute concentrations

come in contact, they spontaneously mix to result in a

uniform distribution of solute throughout the solution.

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Common Types of Solutions

• A solution may be compost of a solid and a

liquid, a gas an a liquid, or other combinations.

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• The majority component of a solution is called the solvent.

• The minority component is called the solute.

Solutions

• In aqueous

solutions, water

is the solvent.

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• Many physical systems tend toward lower potential energy.

• But formation of a solution does not necessarily lower the potential energy of the system.

• When two ideal gases are put into the same container, they

spontaneously mix, even though the difference in attractive forces is negligible.

• The gases mix because the energy of the system is lowered through the release of entropy.

Nature’s Tendency Toward Mixing: Entropy

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• Energy changes in the formation of most solutions also involve differences in attractive forces

between the particles.

Solutions: Effect of Intermolecular Forces

• Chemist’s rule of thumb – like dissolves like

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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 12.1 Solubility

Vitamins are often categorized as either fat soluble or water soluble. Water-soluble vitamins dissolve in body fluids and are easily eliminated in the urine, so there is little danger of overconsumption. Fat-soluble vitamins, on the other hand, can accumulate in the body’s fatty deposits. Overconsumption of a fat-soluble vitamin can be dangerous to your health. Examine the structure of each vitamin shown here and classify it as either fat soluble or water soluble.

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Example 12.1 Solubility

a. The four —OH bonds in vitamin C make it highly polar and allow it to hydrogen bond with water. Vitamin C is water soluble.

b. The C — C bonds in vitamin K₃ are nonpolar and the C — H bonds are nearly so. The C O bonds are polar, but the bond dipoles oppose and largely cancel each other, so the molecule is dominated by the nonpolar bonds. Vitamin K₃ is fat soluble.

Solution

Continued

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Example 12.1 Solubility

Continued

c. The C — C bonds in vitamin A are nonpolar and the C — H bonds are nearly so. The one polar —OH bond may increase its water solubility slightly, but overall vitamin A is nonpolar and therefore fat soluble.

d. The three —OH bonds and one —NH bond in vitamin B₅ make it highly polar and allow it to hydrogen bond with water. Vitamin B₅ is water soluble.

For Practice 12.1

Determine whether each compound is soluble in hexane.

a. water (H₂O) b. propane (CH₃CH₂CH₃) c. ammonia (NH₃) d. hydrogen chloride (HCl)

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Solution Interactions

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• When the solute-to-solvent attractions are weaker than the sum of the solute-to-solute and solvent- to-solvent attractions, the solution will only form if the energy difference is small enough to be

overcome by the increase in entropy from mixing.

Relative Interactions and Solution

Formation

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• To make a solution you must

1. overcome all attractions between the solute particles;

therefore, ΔH

_solute

is endothermic.

2. overcome some attractions between solvent molecules;

therefore, ΔH

_solvent

is endothermic.

3. form new attractions between solute particles and solvent molecules; therefore, ΔH

_mix

is exothermic.

• The overall ΔH for making a solution depends on the relative sizes of the ΔH for these three processes.

ΔH

_sol’n

= ΔH

_solute

+ ΔH

_solvent

solvent, the overall process will be exothermic.

Energetics of Solution Formation

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If the total energy cost for

breaking attractions between particles in the pure solute and pure solvent is greater than the energy released in making the new attractions between the solute and

solvent, the overall process will be endothermic.

Energetics of Solution Formation

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Ion–Dipole Interactions

• When ions dissolve in water they become hydrated.

– Each ion is surrounded by water molecules.

• The formation of these ion–dipole attractions

causes the heat of hydration to be very exothermic.

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Heats of Solution for Ionic Compounds

• For an aqueous solution of an ionic compound, the ΔH _solution is the difference between the heat of

hydration and the lattice energy.

ΔH _solution = −ΔH lattice energy + ΔH _hydration ΔH _solution = ΔH _solute + ΔH _solvent + ΔH _mix

ΔH _solution = −ΔH lattice energy + ΔH _solvent + ΔH _mix

ΔH _solution = ΔH _hydration − ΔH lattice energy

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ΔH _solution = ΔH _hydration − ΔH lattice energy

Heat of Hydration

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12.4 Solution Equilibrium and Factors Affecting Solubility

Solution Equilibrium

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Adding a Crystal of NaC ₂ H ₃ O ₂ to a Supersaturated Solution

saturated. supersaturated.

unsaturated.

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Solubility Curves

• Solubility is

generally given in grams of solute that will

dissolve in

100 g of

water.

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Purification by Recrystallization

• One of the common operations performed by a chemist is

removing impurities from a solid compound.

• One method of purification involves dissolving a solid in a hot solvent until the solution is saturated.

• As the solution slowly cools,

the solid crystallizes out, leaving

impurities behind.

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Temperature Dependence of Solubility of

Gases in Water

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Pressure Dependence of Solubility of Gases in Water

• The larger the partial pressure of a gas in contact with a liquid, the more soluble the gas is in the

liquid.

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Henry’s Law

• The solubility of a gas (S _gas ) is directly

proportional to its partial pressure, (P _gas ).

S _gas = k _H P _gas

• k _H is called the Henry’s

law constant.

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Example 12.2 Henry’s Law

What pressure of carbon dioxide is required to keep the carbon dioxide concentration in a bottle of club soda at 0.12 M at 25 °C ?

Sort

You are given the desired solubility of carbon dioxide and asked to find the pressure required to achieve this solubility.

Given:

Find:

Strategize

Use Henry’s law to find the required pressure from the solubility. You will need the Henry’s law constant for carbon dioxide, which is listed in Table 12.4.

Conceptual Plan

Relationships Used

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Example 12.2 Henry’s Law

Continued

Solve

Solve the Henry’s law equation for and substitute the other quantities to calculate it.

Solution

Check

The answer is in the correct units and seems reasonable. A small answer (for example, less than 1 atm) would be suspect because you know that the soda is under a pressure greater than atmospheric pressure when you open it. A very large answer (for example, over 100 atm) would be suspect because an ordinary can or bottle probably could not sustain such high pressures without bursting.

For Practice 12.2

Determine the solubility of oxygen in water at 25 °C exposed to air at 1.0 atm. Assume a partial pressure for oxygen of 0.21 atm.

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• Solutions have variable composition.

• To describe a solution, you need to describe the components and their relative amounts.

• The terms dilute and concentrated can be

used as qualitative descriptions of the amount of solute in solution.

• Concentration = amount of solute in a given amount of solution.

– Occasionally amount of solvent

12.5 Expressing Solution Concentration

Concentrations

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Molarity and Molality

Slide 32 of 46 General Chemistry: Chapter 13

Molarity (M) = Amount of solute (in moles) Volume of solution (in liters)

Molality (m) = Amount of solute (in moles)

Mass of solvent (in kilograms)

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Parts per Million, Parts per Billion, and Parts per Trillion

Very low solute concentrations are expressed as:

ppm: parts per million (g/g, mg/L) ppb: parts per billion (ng/g, g/L) ppt: parts per trillion (pg/g, ng/L)

note that 1.0 L  1.0 g/mL = 1000 g

ppm, ppb, and ppt are properly m/m or v/v.

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Mole Fraction and Mole Percent



_i

= Amount of component i (in moles)

Total amount of all components (in moles)



₁

+ 

₂

+ 

₃

+ …

_n

= 1

Mole % i = 

_i

 100%

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Mass Percent (m/m) Volume Percent (v/v) Mass/Volume percent (m/v)

Isotonic saline is prepared by dissolving 0.9 g of NaCl in 100 mL of water and is said to be:

0.9% NaCl (mass/volume)

General Chemistry: Chapter 13 Slide 35 of 46

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Concentrations

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• Moles of solute per 1 liter of solution

• Describes how many molecules of solute in each liter of solution

• If a sugar solution concentration is 2.0 M,

– 1 liter of solution contains 2.0 moles of sugar – 2 liters = 4.0 moles sugar

– 0.5 liters = 1.0 mole sugar

Solution Concentration: Molarity

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• Moles of solute per 1 kilogram of solvent

– Defined in terms of amount of solvent, not solution

• Like the others

• Does not vary with temperature

– Because based on masses, not volumes

Solution Concentration: Molality, m

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• Need to know amount of solution and concentration of solution

• Calculate the mass of solute needed

– Start with amount of solution

– Use concentration as a conversion factor

• 5% by mass ⇒ 5 g solute ≡ 100 g solution

– “Dissolve the grams of solute in enough solvent to total the total amount of solution.”

Preparing a Solution

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Preparing a Solution

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• Parts can be measured by mass or volume.

• Parts are generally measured in the same units.

– By mass in grams, kilogram, lbs, etc.

– By volume in mL, L, gallons, etc.

– Mass and volume combined in grams and mL

Parts Solute in Parts Solution

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• Percentage = parts of solute in every 100 parts solution

– If a solution is 0.9% by mass, then there are 0.9 grams of solute in every 100 grams of solution (or 0.9 kg solute in every 100 kg solution).

• Parts per million = parts of solute in every 1 million parts solution

– If a solution is 36 ppm by volume, then there are 36 mL of solute in 1 million mL of solution.

Parts Solute in Parts Solution

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• Grams of solute per 1,000,000 g of solution

• mg of solute per 1 kg of solution

• 1 liter of water = 1 kg of water

– For aqueous solutions we often approximate the kg of the solution as the kg or L of water.

• For dilute solutions, the difference in density between the solution and pure water is usually negligible.

PPM

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Parts Per Billion Concentration

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Example 12.3 Using Parts by Mass in Calculations

What volume (in mL) of a soft drink that is 10.5% sucrose (C₁₂H₂₂O₁₁) by mass contains 78.5 g of sucrose? (The density of the solution is 1.04 g/mL.)

Sort

You are given a mass of sucrose and the concentration and density of a sucrose solution, and you are asked to find the volume of solution containing the given mass.

Given: 78.5 g C₁₂H₂₂O₁₁ 10.5% C₁₂H₂₂O₁₁ by mass density = 1.04 g/mL

Find: mL

Strategize

Begin with the mass of sucrose in grams. Use the mass percent concentration of the solution (written as a ratio, as shown under relationships used) to find the number of grams of solution containing this quantity of sucrose. Then use the density of the solution to convert grams to milliliters of solution.

Conceptual Plan

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Continued

Relationships Used

Solve

Begin with 78.5 g C₁₂H₂₂O₁₁ and multiply by the conversion factors to arrive at the volume of solution.

Solution

Check

The units of the answer are correct. The magnitude seems correct because the solution is approximately 10% sucrose by mass. Since the density of the solution is approximately 1 g/mL, the volume containing 78.5 g sucrose should be roughly 10 times larger, as calculated (719 ≈ 10 × 78.5).

Example 12.3 Using Parts by Mass in Calculations

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Continued

For Practice 12.3

What mass of sucrose (C₁₂H₂₂O₁₁), in g, is contained in 355 mL (12 ounces) of a soft drink that is 11.5% sucrose by mass? (Assume a density of 1.04 g/mL.)

For More Practice 12.3

A water sample is found to contain the pollutant chlorobenzene with a concentration of 15 ppb (by mass). What volume of this water contains 5.00 × 10² mg of chlorobenzene? (Assume a density of 1.00 g/mL.)

Example 12.3 Using Parts by Mass in Calculations

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• The mole fraction is the fraction of the moles of one component in the total moles of all the components of the solution.

• Total of all the mole fractions in a solution = 1.

• Unitless

• The mole percentage is the percentage of the moles of one component in the total moles of all the components of the solution.

– = mole fraction × 100%

Solution Concentrations: Mole Fraction, X _A

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1. Write the given concentration as a ratio.

2. Separate the numerator and denominator.

– Separate into the solute part and solution part

3. Convert the solute part into the required unit.

4. Convert the solution part into the required unit.

5. Use the definitions to calculate the new concentration units.

Converting Concentration Units

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Example 12.4 Calculating Concentrations

A solution is prepared by dissolving 17.2 g of ethylene glycol (C₂H₆O₂) in 0.500 kg of water. The final volume of the solution is 515 mL. For this solution, calculate the concentration in each unit.

a. molarity b. molality c. percent by mass

d. mole fraction e. mole percent

Solution

a. To calculate molarity, first find the amount of ethylene glycol in moles from the mass and molar mass. Then divide the amount in moles by the volume of the solution in liters

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Continued

Example 12.4 Calculating Concentrations

b. To calculate molality, use the amount of ethylene glycol in moles from part (a), and divide by the mass of the water in kilograms.

c. To calculate percent by mass, divide the mass of the solute by the sum of the masses of the solute and solvent and multiply the ratio by 100%.

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Continued

Example 12.4 Calculating Concentrations

d. To calculate mole fraction, first determine the amount of water in moles from the mass of water and its molar mass. Then divide the amount of ethylene glycol in moles (from part (a)) by the total number of moles.

e. To calculate mole percent, multiply the mole fraction by 100%.

For Practice 12.4

A solution is prepared by dissolving 50.4 g sucrose (C₁₂H₂₂O₁₁) in 0.332 kg of water. The final volume of the solution is 355 mL. Calculate the concentration of the solution in each unit.

a. molarity b. molality c. percent by mass

d. mole fraction e. mole percent

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Example 12.5 Converting between Concentration Units

What is the molarity of a 6.56% by mass glucose (C₆H₁₂O₆) solution? (The density of the solution is 1.03 g/mL.)

Sort

You are given the concentration of a glucose solution in percent by mass and the density of the solution. Find the concentration of the solution in molarity.

Given: 6.56% C₆H₁₂O₆ density = 1.03 g/mL Find: M

Strategize

Begin with the mass percent concentration of the solution written as a ratio, and separate the numerator from the denominator. Convert the numerator from g C₆H₁₂O₆ to mol C₆H₁₂O₆. Convert the denominator from g soln to mL of solution and then to L solution. Then divide the numerator (now in mol) by the denominator (now in L) to obtain molarity.

Conceptual Plan

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Example 12.5 Converting between Concentration Units

Continued

Relationships Used

Solve

Begin with the numerator (6.56 g C₆H₁₂O₆) and use the molar mass to convert to mol C₆H₁₂O₆.

Convert the denominator (100 g solution) into mL of solution (using the density) and then to L of solution.

Finally, divide mol C₆H₁₂O₆ by L solution to arrive at molarity.

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• Colligative properties are properties whose value depends only on the number of solute

particles, and not on what they are.

– Value of the property depends on the concentration of the solution.

12.6 Colligative Properties: Vapor Pressure

Lowering, Freezing Point Depression, Boiling Point

Elevation, and Osmotic Pressure

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Vapor Pressure of Solutions

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Thirsty Solutions • The vapor pressure of a volatile solvent above a solution is equal to its normal vapor pressure, P°, multiplied by its

mole fraction in the solution.

P solvent in solution = χ _solvent ∙P °

– Because the mole fraction is always less than 1, the vapor pressure of the

solvent in solution will always be less than the vapor pressure of the pure solvent.

Raoult’s Law

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Example 12.6 Calculating the Vapor Pressure of a Solution

Containing a Nonelectrolyte and Nonvolatile Solute

Calculate the vapor pressure at 25 °C of a solution containing 99.5 g sucrose (C₁₂H₂₂O₁₁) and 300.0 mL water. The vapor pressure of pure water at 25 °C is 23.8 torr. Assume the density of water to be 1.00 g/mL.

Sort

You are given the mass of sucrose and volume of water in a solution. You are also given the vapor pressure and density of pure water and asked to find the vapor pressure of the solution.

Given:

Find: P_solution

Strategize

Raoult’s law relates the vapor pressure of a solution to the mole fraction of the solvent and the vapor pressure of the pure solvent. Begin by calculating the amount in moles of sucrose and water.

Calculate the mole fraction of the solvent from the calculated amounts of solute and solvent.

Then use Raoult’s law to calculate the vapor pressure of the solution.

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Example 12.6 Calculating the Vapor Pressure of a Solution

Containing a Nonelectrolyte and Nonvolatile Solute

Continued

Conceptual Plan

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Example 12.6 Calculating the Vapor Pressure of a Solution

Containing a Nonelectrolyte and Nonvolatile Solute

Continued

Solve

Calculate the number of moles of each solution component.

Use the number of moles of each component to calculate the mole fraction of the solvent (H₂O).

Use the mole fraction of water and the vapor pressure of pure water to calculate the vapor pressure of the solution.

Solution

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Example 12.6 Calculating the Vapor Pressure of a Solution

Containing a Nonelectrolyte and Nonvolatile Solute

Continued

Check

The units of the answer are correct. The magnitude of the answer seems right because the calculated vapor pressure of the solution is just below that of the pure liquid, as you would expect for a solution with a large mole fraction of solvent.

For Practice 12.6

Calculate the vapor pressure at 25 °C of a solution containing 55.3 g ethylene glycol (HOCH₂CH₂CH₂OH) and 285.2 g water. The vapor pressure of pure water at 25 °C is 23.8 torr.

For Practice 12.6

A solution containing ethylene glycol and water has a vapor pressure of 7.88 torr at 10 °C. Pure water has a vapor pressure of 9.21 torr at 10 °C. What is the mole fraction of ethylene glycol in the solution?

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• The vapor pressure of a solvent in a solution is always lower than the vapor pressure of the pure solvent.

• The vapor pressure of the solution is directly proportional to the amount of the solvent in the solution.

• The difference between the vapor pressure of the pure solvent and the vapor pressure of the solvent in solution is called the vapor pressure lowering.

ΔP = P° _solvent − P _solution = χ _solute ∙ P° _solvent

Vapor Pressure Lowering

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Deviations from Raoult’s Law

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Example 12.7 Calculating the Vapor Pressure of a Two-Component Solution

Continued

Use the mole fraction of each component along with Raoult’s law to calculate the partial pressure of each component. The total pressure is the sum of the partial pressures.

Conceptual Plan

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Example 12.7 Calculating the Vapor Pressure of a Two-Component Solution

Continued

Relationships Used

Solve

Begin by converting the mass of each component to the amounts in moles.

Then calculate the mole fraction of carbon disulfide.

Calculate the mole fraction of acetone by subtracting the mole fraction of carbon disulfide from 1.

Calculate the partial pressures of carbon disulfide and acetone by using Raoult’s law and the given values of the vapor pressures of the pure substances.

Calculate the total pressure by summing the partial pressures.

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Example 12.7 Calculating the Vapor Pressure of a Two-Component Solution

Continued

Lastly, compare the calculated total pressure for the ideal case to the experimentally measured total pressure. Since the experimentally measured pressure is greater than the calculated pressure, we can conclude that the interactions between the two components are weaker than the interactions between the components themselves.

Solution

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Example 12.7 Calculating the Vapor Pressure of a Two-Component Solution

Continued

P_tot(ideal) = 285 torr + 148 torr

= 433 torr

P_tot(exp) = 645 torr P_tot(exp) > P_tot (ideal)

The solution is not ideal and shows positive deviations from Raoult’s law. Therefore, carbon disulfide–acetone interactions must be weaker than acetone–acetone and carbon disulfide–carbon disulfide interactions.

Check

The units of the answer (torr) are correct. The magnitude seems reasonable given the partial pressures of the pure substances.

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Example 12.7 Calculating the Vapor Pressure of a Two-Component Solution

Continued

For Practice 12.7

A solution of benzene (C₆H₆) and toluene (C₇H₈) is 25.0% benzene by mass. The vapor pressures of pure

benzene and pure toluene at 25 °C are 94.2 torr and 28.4 torr, respectively. Assuming ideal behavior, calculate the following:

a. The vapor pressure of each of the solution components in the mixture.

b. The total pressure above the solution.

c. The composition of the vapor in mass percent.

Why is the composition of the vapor different from the composition of the solution?

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Freezing Point Depression and Boiling Point Elevation

(FP _solvent – FP _solution ) = ΔT _f = m∙K _f

(BP _solution – BP _solvent ) = ΔT _b

= m∙K _b

with glucose,

which acts as an

antifreeze.

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Slide 70 of 33

Which of the following aqueous solutions will have the highest boiling point?

1. 3 molal glucose 2. 4 molal ethanol 3. 2.5 molal NaCl 4. 2.5 molal CaCl

₂

5. Cannot tell without K

_b

for water

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Slide 71 of 33

Which of the following aqueous solutions will have the highest boiling point?

1. 3 molal glucose 2. 4 molal ethanol 3. 2.5 molal NaCl 4. 2.5 molal CaCl

₂

5. Cannot tell without K

_b

for water

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Slide 72 of 33

A 2.0 molal aqueous solution of glucose (C

₆

O

^o

C

5. Cannot determine without K

_b

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Slide 73 of 33

A 2.0 molal aqueous solution of glucose (C

₆

O

^o

C

5. Cannot determine without K

_b

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Example 12.8 Freezing Point Depression

Calculate the freezing point of a 1.7 m aqueous ethylene glycol solution

Sort

You are given the molality of a solution and asked to find its freezing point.

Given: 1.7 m solution

Find: freezing point (from ΔT_f)

Strategize

To solve this problem, use the freezing point depression equation.

Conceptual Plan

Solve

Substitute into the equation to calculate ΔT_f.

The actual freezing point is the freezing point of pure water (0.00 °C) − ΔT_f.

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Example 12.8 Freezing Point Depression

Solution

Check

The units of the answer are correct. The magnitude seems about right. The expected range for freezing points of an aqueous solution is anywhere from −10 °C to just below 0 °C. Any answers out of this range would be suspect.

For Practice 12.8

Calculate the freezing point of a 2.6 m aqueous sucrose solution.

Continued

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Example 12.9 Boiling Point Elevation

What mass of ethylene glycol (C₂H₆O₂), in grams, must be added to 1.0 kg of water to produce a solution that boils at 105.0 °C?

Sort

You are given the desired boiling point of an ethylene glycol solution containing 1.0 kg of water and asked to find the mass of ethylene glycol you need to add to achieve the boiling point.

Given: ΔT_b = 5.0 °C, 1.0 kg H₂O Find: g C₂H₆O₂

Strategize

To solve this problem, use the boiling-point elevation equation to calculate the desired molality of the solution from ΔT_b.

Then use that molality to determine how many moles of ethylene glycol are needed per kilogram of water.

Finally, calculate the molar mass of ethylene glycol and use it to convert from moles of ethylene glycol to mass of ethylene glycol.

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Example 12.9 Boiling Point Elevation

Conceptual Plan

Relationships Used

C₂H₆O₂ molar mass = 62.07 g/mol ΔT_b = m × K_b (boiling point elevation)

Solve

Begin by solving the boiling point elevation equation for molality and substituting the required quantities to calculate m.

Continued

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Example 12.9 Boiling Point Elevation

Solution

Check

The units of the answer are correct. The magnitude might seem a little high initially, but the boiling point elevation constant is so small that a lot of solute is required to raise the boiling point by a small amount.

For Practice 12.9

Calculate the boiling point of a 3.60 m aqueous sucrose solution.

Continued

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K _f

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• The amount of pressure needed to keep osmotic flow from taking place is called the osmotic

pressure.

• The osmotic pressure, P, is directly proportional to the molarity of the solute particles.

– R = 0.08206 (atm∙L)/(mol∙K)

Π = MRT

Osmotic Pressure

• A semipermeable

membrane allows

solvent to flow

through it, but not

solute.

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• Osmosis is the flow of solvent from a solution of low concentration into a solution of high

concentration.

• The solutions may be separated by a semipermeable membrane.

• A semipermeable membrane allows solvent to flow through it, but not solute.

Osmosis

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Slide 82 of 33

A B A B

A B

To the right is a diagram of a closed system containing two salt water solutions. The solution labeled A is more concentrated than the one labeled B. Which of the diagrams below best represents the system at an infinite time after preparation?

1. 2. 3.

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Slide 83 of 33

A B A B

A B

To the right is a diagram of a closed system containing two salt water solutions. The solution labeled A is more concentrated than the one labeled B. Which of the diagrams below best represents the system at an infinite time after preparation?

1. 2. 3.

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Example 12.10 Osmotic Pressure

The osmotic pressure of a solution containing 5.87 mg of an unknown protein per 10.0 mL of solution is 2.45 torr at 25 °C. Find the molar mass of the unknown protein.

Sort

You are given that a solution of an unknown protein contains 5.87 mg of the protein per 10.0 mL of solution. You are also given the osmotic pressure of the solution at a particular temperature and asked to find the molar mass of the unknown protein.

Given: 5.87 mg protein 10.0 mL solution

Π = 2.45 torr T = 25 °C

Find: molar mass of protein (g/mol)

Strategize

Step 1: Use the given osmotic pressure and temperature to find the molarity of the protein solution.

Step 2: Use the molarity calculated in step 1 to find the number of moles of protein in 10 mL of solution.

Step 3: Finally, use the number of moles of the protein calculated in step 2 and the given mass of the protein in 10.0 mL of solution to find the molar mass

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Example 12.10 Osmotic Pressure

Conceptual Plan

Relationships Used

Π = MRT (osmotic pressure equation) Continued

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Example 12.10 Osmotic Pressure

Continued

Solve

Step 1: Begin by solving the osmotic pressure equation for molarity and substituting in the required quantities in the correct units to calculate M.

Step 2: Begin with the given volume, convert to liters, then use the molarity to find the number of moles of protein.

Step 3: Use the given mass and the number of moles from step 2 to calculate the molar mass of the protein.

Solution

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Example 12.10 Osmotic Pressure

Continued

Check

The units of the answer are correct. The magnitude might seem a little high initially, but proteins are large molecules and therefore have high molar masses.

For Practice 12.10

Calculate the osmotic pressure (in atm) of a solution containing 1.50 g ethylene glycol (C₂H₆O₂) in 50.0 mL of solution at 25 °C.

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12.7 Colligative Properties of Strong Electrolyte Solutions

van’t Hoff factor, i,

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Example 12.11 Van’t Hoff Factor and Freezing Point Depression

The freezing point of an aqueous 0.050 m CaCl₂ solution is −0.27 °C. What is the van’t Hoff factor (i) for CaCl₂ at this concentration? How does it compare to the expected value of i?

Sort

You are given the molality of a solution and its freezing point. You are asked to find the value of i, the van’t Hoff factor, and compare it to the expected value.

Given: 0.050 m CaCl₂ solution, ΔT_f = 0.27 °C

Find: i

Strategize

To solve this problem, use the freezing point depression equation including the van’t Hoff factor.

Conceptual Plan

ΔT_f = im × K_f

Solve

Solve the freezing point depression equation for i, substituting in the given quantities to calculate its value.

The expected value of i for CaCl₂ is 3 because calcium chloride forms 3 mol of ions for each mole of calcium chloride that dissolves. The experimental value is slightly less than 3, probably because of ion pairing.

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Example 12.11 Van’t Hoff Factor and Freezing Point Depression

Solution

Check

The answer has no units, as expected since i is a ratio. The magnitude is about right since it is close to the value you would expect upon complete dissociation of CaCl₂.

For Practice 12.11

Calculate the freezing point of an aqueous 0.10 m FeCl₃ solution using a van’t Hoff factor of 3.2.

Continued

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Example 12.12 Calculating the Vapor Pressure of a Solution Containing an Ionic Solute

A solution contains 0.102 mol Ca(NO₃)₂ and 0.927 mol H₂O. Calculate the vapor pressure of the solution at 55 °C.

The vapor pressure of pure water at 55 °C is 118.1 torr. (Assume that the solute completely dissociates.)

Sort

You are given the number of moles of each component of a solution and asked to find the vapor pressure of the solution. You are also given the vapor pressure of pure water at the appropriate temperature.

Given: 0.102 mol Ca(NO₃)₂ 0.927 mol H₂O

= 118.1 torr (at 55 C) Find: P_solution

Strategize

To solve this problem, use Raoult’s law as you did in Example 12.6. Calculate χ_solventfrom the given amounts of solute and solvent.

Conceptual Plan

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Example 12.12 Calculating the Vapor Pressure of a Solution Containing an Ionic Solute

Continued

Solve

The key to this problem is to understand the dissociation of calcium nitrate. Write an equation showing the dissociation.

Since 1 mol of calcium nitrate dissociates into 3 mol of dissolved particles, the number of moles of calcium nitrate must be multiplied by 3 when computing the mole fraction.

Use the mole fraction of water and the vapor pressure of pure water to calculate the vapor pressure of the solution.

Solution

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Example 12.12 Calculating the Vapor Pressure of a Solution Containing an Ionic Solute

Continued

Check

The units of the answer are correct. The magnitude also seems right because the calculated vapor pressure of the solution is significantly less than that of the pure solvent, as you would expect for a solution with a significant amount of solute.

For Practice 12.12

A solution contains 0.115 mol H₂O and an unknown number of moles of sodium chloride. The vapor pressure of the solution at 30 °C is 25.7 torr. The vapor pressure of pure water at 30 C is 31.8 torr. Calculate the number of moles of sodium chloride in the solution.