12.1 Thirsty Solutions: Why You Shouldn’t Drink Seawater ... · Vitamins are often categorized as...

© 2014 Pearson Education, Inc.

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


12.1 Thirsty Solutions: Why You Shouldn’t

Drink Seawater

Thirsty Seawater


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.


Common Types of Solutions

• A solution may be compost of a solid and a

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


• 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.


• 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


• 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

© 2014 Pearson Education, Inc. 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.


Nivaldo J. Tro

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


Nivaldo J. Tro

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)


Solution Interactions


• 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


• To make a solution you must

1. overcome all attractions between the solute particles;

therefore, ΔHsolute is endothermic.

2. overcome some attractions between solvent molecules;

therefore, ΔHsolvent is endothermic.

3. form new attractions between solute particles and solvent

molecules; therefore, ΔHmix is exothermic.

• The overall ΔH for making a solution depends on the relative

sizes of the ΔH for these three processes.

ΔHsol’n = ΔHsolute + ΔHsolvent + ΔHmix

12.3 Energetics of Solution Formation:

The Enthalpy of Solution


Step 1: Separating the solute into its constituent

particles

Solution Process


Step 2: Separating the solvent particles from each

other to make room for the solute particles

Solution Process


Step 3: Mixing the solute particles with the solvent

particles

Solution Process


If the total energy cost for

breaking attractions between

particles in the pure solute

and pure solvent is less than

the energy released in

making the new attractions

between the solute and

solvent, the overall process

will be exothermic.

Energetics of Solution Formation


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


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.


Heats of Solution for Ionic Compounds

• For an aqueous solution of an ionic compound, the

ΔHsolution is the difference between the heat of

hydration and the lattice energy.

ΔHsolution = −ΔHlattice energy + ΔHhydration

ΔHsolution = ΔHsolute+ ΔHsolvent + ΔHmix

ΔHsolution = −ΔHlattice energy+ ΔHsolvent + ΔHmix

ΔHsolution = ΔHhydration− ΔHlattice energy


ΔHsolution = ΔHhydration− ΔHlattice energy

Heat of Hydration


12.4 Solution Equilibrium and Factors

Affecting Solubility

Solution Equilibrium


Adding a Crystal of NaC2H3O2 to a

Supersaturated Solution

saturated. supersaturated. unsaturated.


Solubility Curves

• Solubility

is

generally

given in

grams of

solute that

will

dissolve in

100 g of

water.


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.


Temperature Dependence of Solubility of

Gases in Water


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.


Henry’s Law

• The solubility of a gas

(Sgas) is directly

proportional to its partial

pressure, (Pgas).

Sgas = kHPgas

• kH is called the Henry’s

law constant.


Nivaldo J. Tro

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


Nivaldo J. Tro

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.


• 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

Molarity and Molality

Copyright © 2011 Pearson Canada

Inc. 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)

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.

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%

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)


Inc. General Chemistry: Chapter 13 Slide 35 of 46


Concentrations


• 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


• 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


• 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


Preparing a Solution


• 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


• 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


• 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


Parts Per Billion Concentration


Nivaldo J. Tro

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


Nivaldo J. Tro

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).



Nivaldo J. Tro

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.)



• 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, XA


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


Nivaldo J. Tro

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


Nivaldo J. Tro

Continued


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%.


Nivaldo J. Tro

Continued


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


Nivaldo J. Tro

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


Nivaldo J. Tro

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.


• 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


Vapor Pressure of Solutions


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.

Psolvent in solution =

χsolvent∙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


Nivaldo J. Tro

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: Psolution

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.


Nivaldo J. Tro


Containing a Nonelectrolyte and Nonvolatile Solute

Continued

Conceptual Plan


Nivaldo J. Tro



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


Nivaldo J. Tro



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?


• 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 − Psolution = χsolute ∙ P°solvent

Vapor Pressure Lowering


Deviations from Raoult’s Law


Nivaldo J. Tro

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


Nivaldo J. Tro


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.


Nivaldo J. Tro


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


Nivaldo J. Tro


Solution

Continued

Ptot(ideal) = 285 torr + 148 torr

= 433 torr

Ptot(exp) = 645 torr

Ptot(exp) > Ptot (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.


Nivaldo J. Tro


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?


Freezing Point Depression and Boiling Point

Elevation (FPsolvent – FPsolution) = ΔTf =

m∙Kf

(BPsolution – BPsolvent) = ΔTb

= m∙Kb

with glucose, which acts as an antifreeze.

of 33 Copyright © 2011

Pearson Canada Inc.

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 CaCl2

5. Cannot tell without Kb for water


Pearson Canada Inc.

A 2.0 molal aqueous solution of

glucose (C6O6H12) is found to boil

at 101 oC. What would the boiling

point of a 2.0 molal solution of

sucrose(C12O11H22)

be?

glucose (C6O6H12)

sucrose (C12O11H22)

1. 102 oC

2. 100.5 oC

3. 101 oC

4. Slightly higher than 100.5 oC

5. Cannot determine without Kb


Nivaldo J. Tro

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 ΔTf)

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

Conceptual Plan

Solve Substitute into the equation to calculate ΔTf.

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


Nivaldo J. Tro

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


Nivaldo J. Tro

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: ΔTb = 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 ΔTb.

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.


Nivaldo J. Tro


Conceptual Plan

Relationships Used C2H6O2 molar mass = 62.07 g/mol

ΔTb = m × Kb (boiling point elevation)

Solve Begin by solving the boiling point elevation equation for molality and substituting the required quantities to

calculate m.

Continued


Nivaldo J. Tro


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


Kf


• 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.


• 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


Pearson Canada Inc.

A B 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.


Nivaldo J. Tro

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


Nivaldo J. Tro

Example 12.10 Osmotic Pressure

Conceptual Plan

Relationships Used Π = MRT (osmotic pressure equation)

Continued


Nivaldo J. Tro

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


Nivaldo J. Tro

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.


12.7 Colligative Properties of Strong

Electrolyte Solutions

van’t Hoff factor, i,


Nivaldo J. Tro

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,

ΔTf = 0.27 °C

Find: i

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

Conceptual Plan ΔTf = im × Kf

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.


Nivaldo J. Tro

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


Nivaldo J. Tro


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: Psolution

Strategize To solve this problem, use Raoult’s law as you did in Example 12.6. Calculate χsolvent from the given amounts of

solute and solvent.

Conceptual Plan


Nivaldo J. Tro



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


Nivaldo J. Tro



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.


isosmotic hyperosmotic solution

Conc.

hyposmotic solution

Dilu.

0.9 g NaCl per 100 mL of solution.

1 g NaCl per 100 mL of solution.

0.5 g NaCl per 100 mL of solution.


Pearson Canada Inc.

20 g C3H6O3 10 g C6H12O6

Two aqueous solutions of equal volume, one containing 20 g of

lactic acid (C3H6O3) and one containing 10 g of glucose

(C6H12O6) are separated by a semi permeable membrane

(allowing only H2O to pass). The net flow of water through

the membrane is?

1. Left

2. Right

3. There is no net flow.

4. Cannot tell without

knowing the

exact volume of the

solutions.


• Solutions = homogeneous

• Suspensions = heterogeneous, separate

on standing

• Colloids = heterogeneous, do not separate

on standing – Particles can coagulate

– Cannot pass through semipermeable membrane

– Hydrophilic

• Stabilized by attraction for solvent (water)

– Hydrophobic

• Stabilized by charged surface repulsions

• Show the Tyndall effect and Brownian motion.

12.8 Colloids

Mixtures


The Tyndall Effect


Brownian Motion


Soap

12.1 Thirsty Solutions: Why You Shouldn’t Drink Seawater ... · Vitamins are often categorized as...

Documents

Transcript of 12.1 Thirsty Solutions: Why You Shouldn’t Drink Seawater ... · Vitamins are often categorized as...