12.1 thirsty solutions: why you shouldn’t drink seawater ... · vitamins are often categorized as...
TRANSCRIPT
© 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
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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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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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• 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, 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
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Step 1: Separating the solute into its constituent
particles
Solution Process
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Step 2: Separating the solvent particles from each
other to make room for the solute particles
Solution Process
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Step 3: Mixing the solute particles with the solvent
particles
Solution Process
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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
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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
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
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12.4 Solution Equilibrium and Factors
Affecting Solubility
Solution Equilibrium
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Adding a Crystal of NaC2H3O2 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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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
(Sgas) is directly
proportional to its partial
pressure, (Pgas).
Sgas = kHPgas
• kH 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
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
Copyright © 2011 Pearson Canada
Inc. Slide 33 of 46 General Chemistry: Chapter 13
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
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Inc. Slide 34 of 46 General Chemistry: Chapter 13
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)
Copyright © 2011 Pearson Canada
Inc. General Chemistry: Chapter 13 Slide 35 of 46
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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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• 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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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, XA
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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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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
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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: 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.
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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 − Psolution = χsolute ∙ P°solvent
Vapor Pressure Lowering
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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
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.
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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 (FPsolvent – FPsolution) = ΔTf =
m∙Kf
(BPsolution – BPsolvent) = ΔTb
= m∙Kb
with glucose, which acts as an antifreeze.
Slide 70 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
Slide 71 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
Slide 72 of 33 Copyright © 2011
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
Slide 73 of 33 Copyright © 2011
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
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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 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.
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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: 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.
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Example 12.9 Boiling Point Elevation
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
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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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• 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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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.
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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.
© 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition
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
© 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
Example 12.10 Osmotic Pressure
Conceptual Plan
Relationships Used Π = MRT (osmotic pressure equation)
Continued
© 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition
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
© 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition
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.
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12.7 Colligative Properties of Strong
Electrolyte Solutions
van’t Hoff factor, i,
© 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition
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.
© 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition
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
© 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
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: 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
© 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
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
© 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
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.
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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.
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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.
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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.
© 2014 Pearson Education, Inc.
• 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