### Chap 11 Example and Chem Tour PPT

```ChemTour: Lattice Energy
this ChemTour
Students learn to apply Coulomb’s law to calculate the exact
lattice energies of ionic solids. It includes Practice Exercises.
© 2012 by W. W. Norton & Company
ChemTour: Molecular Motion
this ChemTour
Students use an interactive graph to explore the relationship between
kinetic energy and temperature. It includes Practice Exercises.
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ChemTour: Raoult’s Law
this ChemTour
Students explore the connection between the vapor pressure of a
solution and its concentration as a gas above the solution. It includes
Practice Exercises.
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ChemTour: Fractional Distillation
this ChemTour
This ChemTour shows how differences in boiling points are used to
separate a mixture of volatile molecules, such as crude oil. It includes
follow-up questions.
© 2012 by W. W. Norton & Company
ChemTour: Boiling
and Freezing Points
this ChemTour
Students learn about colligative properties by exploring the relationship between solute
concentration and the temperature at which a solution will undergo phase changes. Interactive
exercises invite students to practice calculating the boiling and freezing points of different
solutions.
© 2012 by W. W. Norton & Company
ChemTour: Osmotic Pressure
this ChemTour
Students discover how a solute can build up pressure behind a
semipermeable membrane. This tutorial also discusses the osmotic
pressure equation and the van’t Hoff factor.
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Sample Exercise 11.1
Rank these three ionic compounds in order of
increasing lattice energy and increasing melting
point: NaF, KF, and RbF. Assume that these
compounds have the same solid structure, which
means they have the same value of k in Equation
11.4.
 Collect and Organize: We can determine relative
lattice energies from Equation 11.4. To do so we
need to establish the charges and radii of the ions in
each compound using Figure 10.2.
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Sample Exercise 11.1 (cont.)
 Analyze: All the cations are alkali metal cations
and have a charge of 1+; all the anions are
fluoride ions, with a charge of 1-. We are told
that k is the same for all three solids. Therefore,
any differences in lattice energy must be related
to differences in the nucleus-to-nucleus distance
d between ions. Because the fluoride ion is
common to all the salts, variations in d depend
only on the size of the cation.
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Sample Exercise 11.1 (cont.)
 Solve: Periodic trends in size predict, and Figure 11.1
confirms, that the cation sizes are Na+< K+< Rb+.
Therefore, the compounds in order of increasing value of
d are NaF < KF < RbF (Figure 11.1). As d increases,
lattice energy decreases, so RbF has the lowest lattice
energy, followed by KF, followed by NaF with the
highest. The same trend occurs in melting points: RbF <
KF < NaF.
 Think about It: We predicted the order of lattice
energies and melting points for three alkali metal
fluorides. This predicted order is confirmed by the
experimentally measured melting points: 775°C for RbF,
846°C for KF, and 988°C for NaF.
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Sample Exercise 11.2
In Sample Exercise 10.1, we predicted that the ion–
ion attraction in CaF2 is greater than in NaF. Confirm
this prediction by calculating the lattice energies of
(a) NaF and (b) CaF2 given the following
information:
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Sample Exercise 11.2 (cont.)
 Collect and Organize: We are to calculate the
lattice energy of two ionic compounds using
enthalpy values for processes that can be summed
to describe an overall process that produces a salt
from its constituent elements. We can use a Born–
Haber cycle to calculate the unknown lattice energy.
Table A4.3 in Appendix 4 contains standard
enthalpy of formation values for NaF and CaF2:
-575.4 kJ/mol and -1228.0 kJ/mol, respectively.
These enthalpy changes apply to the formation of
one mole of the two compounds by combining their
component elements in their standard states.
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Sample Exercise 11.2 (cont.)
 Analyze: The standard enthalpy of formation of NaF is
This value is the enthalpy change for the entire process,
ΔHrxn. The fluorine in the equation has a coefficient of ½; this
means that the energy needed to break a mole of F―F bonds
to make atomic fluorine must be multiplied by ½.
Following the same steps for part b:
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Sample Exercise 11.2 (cont.)
This value is the enthalpy change for the entire process. For
calcium, we must include both the first and second ionization
energies because Ca loses two electrons when it forms Ca2+
ions. Since two fluorine atoms are needed to react with a
single calcium atom, we do not need the factor of ½ in front of
the term for energy to break a mole of F―F bonds. However,
we need to multiply the electron affinity of F by 2 because two
moles of fluorine atoms gain two moles of electrons to form
two moles of fluoride ions.
Figure 11.3 summarizes the Born–Haber cycles for
calculating the lattice energies of NaF and CaF2. Based on
the charges on the ions, their ionic radii, and Equation 11.4,
the lattice energy of CaF2 should be greater than the lattice
energy of NaF.
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Sample Exercise 11.2 (cont.)
 Solve:
a. The overall enthalpy change in the reaction producing NaF
is
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Sample Exercise 11.2 (cont.)
b. The overall enthalpy change in the reaction producing CaF2
is
Substituting the values given:
Solving for U:
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Sample Exercise 11.2 (cont.)
 Think about It: Based on Equation 11.4, we
predicted that the lattice energy of CaF2 would
be larger than that of NaF. The calculated values
of U confirm that
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Sample Exercise 11.3
Use the appropriate enthalpy of hydration values
in Table 11.3 and ΔHsolution = 0.91 kJ/mol for NaF
to calculate the lattice energy of NaF.
 Collect and Organize: The enthalpies of
hydration for Na+ and F- are -418 kJ/mol and -502
kJ/mol, respectively. We can use the enthalpy of
solution and Equation 11.5 to calculate the lattice
energy.
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Sample Exercise 11.3 (cont.)
 Analyze: Before we can use Equation 11.5 to
calculate UNaF, we must add the values of ΔHhydration for
the two ions to obtain a value for ΔHhydration for NaF.
From our calculations in Sample Exercise 11.2, we
expect that the lattice energy of NaF should be close
to -927 kJ/mol.
 Solve: Solving for the enthalpy of hydration of NaF:
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Sample Exercise 11.3 (cont.)
Substituting ΔHhydration,NaF(aq) and ΔHsolution,NaF into
Equation 11.5:
Solving for UNaF:
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Sample Exercise 11.3 (cont.)
 Think about It: The value of UNaF calculated
from ΔHsolution and ΔHhydration is within 1% of the
value calculated in Sample Exercise 11.2 using
different kinds of measurements, suggesting that
the value calculated this way accurately reflects
the lattice energy of NaF.
© 2012 by W. W. Norton & Company
Sample Exercise 11.4
The liquid used in automobile cooling systems is
prepared by dissolving ethylene glycol
(HOCH2CH2OH, molar mass 62.07 g/mol) in
water. What is the vapor pressure of a solution
prepared by mixing 1.000 L of ethylene glycol
(density 1.114 g/mL) with 1.000 L of water
(density 1.000 g/mL) at 100.0°C? Assume that
the mixture obeys Raoult’s law.
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Sample Exercise 11.4 (cont.)
 Collect and Organize: We are asked to calculate
the vapor pressure of a solution of ethylene glycol in
water. The vapor pressure of a solution is a
colligative property that depends on the number of
solute particles, and hence the concentration.
Volume and density must be used to determine the
concentration of ethylene glycol. Ethylene glycol is a
nonvolatile solute and its solution obeys Raoult’s
law, which means it behaves ideally. The vapor
pressure curves in Figure 11.7 show that the vapor
pressure of pure water at 100°C (its normal boiling
point) is 1.00 atm, whereas that of ethylene glycol is
less than 0.05 atm.
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Sample Exercise 11.4 (cont.)
 Analyze: We have a mixture of equal volumes
of two liquids. The solvent is the one present in
the greater number of moles. We can determine
the numbers of moles of each by converting their
volumes into masses by multiplying by their
densities and then calculate the number of
moles of each by dividing by their molar masses.
From these values we can decide which liquid is
the solvent and calculate its mole fraction in the
mixture.
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Sample Exercise 11.4 (cont.)
 Solve: For ethylene glycol:
For water:
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Sample Exercise 11.4 (cont.)
The mole fraction for water is
Ethylene glycol is essentially nonvolatile, so the vapor
pressure of the solution is due only to the solvent, and Psolvent
= PH2O = 1.00 atm. Using this value and the calculated mole
fraction of water in the mixture yields
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Sample Exercise 11.4 (cont.)
 Think about It: The presence of a nonvolatile
solute causes the vapor pressure of the solution
to be less than 1 atm, giving us confidence in
our result. In addition, our selection of water as
the solvent is justified because it is the major
component in terms of numbers of moles of
substances present: 55.49 moles of water with
17.95 moles of ethylene glycol dissolved in it.
© 2012 by W. W. Norton & Company
Sample Exercise 11.5
At its normal boiling point of 126°C, n-octane,
C8H18, has a vapor pressure of 760 torr. What is
its vapor pressure at 25°C? The enthalpy of
vaporization of n-octane is 39.07 kJ/mol.
 Collect and Organize: The vapor pressure of
octane at any temperature can be calculated
using Equation 11.7, the Clausius–Clapeyron
equation, given the enthalpy of vaporization and
the vapor pressure of octane at its normal
boiling point.
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Sample Exercise 11.5 (cont.)
 Analyze: We can use the Clausius–Clapeyron equation, but
to do so we must convert the given temperature to kelvins.
Because we will use 8.314 J/(mol · K) for R, we must also
convert ΔHvap to joules per mole. Because vapor pressure
decreases as temperature decreases, we expect the vapor
pressure of octane to be significantly lower at 25°C than at its
boiling point, 126°C.
 Solve: The two temperatures are
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Sample Exercise 11.5 (cont.)
Inserting these values in Equation 11.7:
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Sample Exercise 11.5 (cont.)
 Think about It: We expect octane to have
a low vapor pressure at a temperature
much lower than its boiling point, so this
number seems reasonable.
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Sample Exercise 11.6
How would the graph in Figure 11.11 be different if (a) we
started with a mixture of 75 mL of n-octane (C8H18) and 25 mL
of n-heptane (C7H16) and (b) we started with a mixture of 75
mL of n-octane (C8H18) and 25 mL of n-nonane (C9H20)? The
normal boiling points of n-heptane, n-octane, and n-nonane
are 98°C, 126°C, and 151°C, respectively.
 Collect and Organize: The graph in Figure 11.11 is an
idealized plot of the temperature of the solution as a function
of the volume of distillate for a 50:50 mixture of C7H16:C8H18.
We are asked to describe the changes in the graph if we
change the ratio of the components and if we change the
identity of one of the components. We are given the normal
boiling points of all compounds.
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Sample Exercise 11.6 (cont.)
 Analyze: Fractional distillation separates
solutions of volatile liquids into pure substances
on the basis of their different boiling points. The
components distill in the order of their boiling
points, with the lowest-boiling component
distilling first. Ideally the total volume of distillate
equals the total volume of the solution, and the
volume of each fraction reflects the volume of
the component in the original solution.
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Sample Exercise 11.6 (cont.)
 Solve:
a. The first component that distills is the lower-boiling
n-heptane. If our system were perfect, 25 mL of nheptane would distill at 98°C. When the n-heptane
was completely removed from the solution, the only
remaining component (75 mL of n-octane) would distill
at 126°C. The idealized graph (Figure 11.13) shows
the boiling points of the two substances along the yaxis and the volume of each distilled along the x-axis.
The lengths of the horizontal lines at 98°C and 126°C
are in a 1:3 ratio reflecting the composition of the
mixture, 25 mL of C7H16 and 75 mL of C8H18.
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Sample Exercise 11.6 (cont.)
b. The first component that distills in the second
mixture is the lower boiling n-octane. If our system
were perfect, 75 mL of n-octane would distill at 126°C.
Once the n-octane is completely removed from the
solution, the only remaining component (25 mL of nnonane) would distill at 151°C. The idealized graph
(Figure 11.14) shows the boiling points of the two
substances along the y-axis and the volume of
distillate along the x-axis.
© 2012 by W. W. Norton & Company
Sample Exercise 11.6 (cont.)
 Think about It: Our graphs show the best
results we could achieve. In a real world system,
a small fraction that is a mixture of two
hydrocarbons distills at temperatures between
the boiling points of the two pure components.
© 2012 by W. W. Norton & Company
Sample Exercise 11.7
Calculate the vapor pressure of a solution prepared by
dissolving 13 g of n-heptane (C7H16) in 87 g of n-octane
(C8H18) at 25°C. By what factor does the concentration of the
more volatile component in the vapor exceed the
concentration of this component in the liquid? The vapor
pressures of n-octane and n-heptane at 25°C are 11 torr and
31 torr, respectively.
 Collect and Organize: We can use Equation 11.8 to
determine the total vapor pressure of the solution from the
vapor pressures of the components after we determine the
composition of the solution in terms of mole fractions.
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Sample Exercise 11.7 (cont.)
 Analyze: To calculate mole fractions, we need
the molar masses of n-heptane and n-octane.
The mole fraction is then equal to the number of
moles of each component divided by the total
number of moles of material in the solution. The
vapor pressure of the solution must lie between
the vapor pressures of the pure compounds.
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Sample Exercise 11.7 (cont.)
 Solve: The number of moles of each component is
The mole fraction of each component in the mixture is
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Sample Exercise 11.7 (cont.)
Using these mole fraction values and the vapor pressures of the two
hydrocarbons in Equation 11.8, we have
To calculate how enriched the vapor phase is in the more volatile
component (the component with the greater vapor pressure, nheptane), we need to recall Dalton’s law of partial pressures
(Section 6.7) and the concept that the partial pressure of a gas in a
mixture of gases is proportional to its mole fraction in the mixture.
Therefore, the ratio of the mole fraction of n-heptane to that of noctane is the ratio of their two vapor pressures:
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Sample Exercise 11.7 (cont.)
The mole ratio of n-heptane to n-octane in the liquid mixture is
Therefore, the vapor phase is enriched in n-heptane by a
factor of
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Sample Exercise 11.7 (cont.)
 Think about It: As expected, the vapor pressure
of the mixture (13.9 torr) is between the vapor
pressures of the separate components. This
result illustrates how fractional distillation works.
The vapor is enriched in the lower-boiling
component.
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Sample Exercise 11.8
How many grams of Na2SO4 should be added to
275 mL of water to prepare a 0.750 m solution of
Na2SO4? Assume the density of water is 1.000
g/mL.
 Collect and Organize: We are asked to
determine the mass of a solute, sodium sulfate,
needed to prepare a 0.750 m solution. Molality is
moles of solute per kilogram of solvent. The
volume of water, the solvent, is 275 mL.
© 2012 by W. W. Norton & Company
Sample Exercise 11.8 (cont.)
 Analyze: Following the procedure shown in
Figure 11.18, we use the density of water to
convert 275 mL of water to kilograms. We can
convert molality into a number of moles of
Na2SO4 needed, and then convert that number
into grams using the molar mass of Na2SO4. Our
goal is to prepare a relatively small volume (~ ⅓
L) of a dilute solution (< 1 m), so we predict that
the mass of solute needed will probably be less
than mol of solute, which has a molar mass of
142 g/mol, or less than 50 g.
© 2012 by W. W. Norton & Company
Sample Exercise 11.8 (cont.)
 Solve: Multiplying 275 mL of water by the density of water,
we find that the mass of water is
We can rearrange Equation 11.9 to determine the number of
moles of solute needed:
© 2012 by W. W. Norton & Company
Sample Exercise 11.8 (cont.)
The molar mass of Na2SO4 is 142.04 g/mol, so the
number of grams of Na2SO4 needed is
Dissolving 29.3 g of Na2SO4 in 275 mL of water
produces a 0.750 m solution.
© 2012 by W. W. Norton & Company
Sample Exercise 11.8 (cont.)
 Think about It: The calculated value of 29 g
Na2SO4 is consistent with our prediction that
less than 50 g of solute would be required.
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Sample Exercise 11.9
What is the boiling point of seawater if the
concentration of ions in seawater is 1.15 m?
 Collect and Organize: We are asked to
calculate the elevation in the boiling point of an
aqueous solution in which the ion concentration
is 1.15 m. The boiling-point-elevation constant of
water is Kb = 0.52°C/m, and the normal boiling
point of water is 100.0°C.
© 2012 by W. W. Norton & Company
Sample Exercise 11.9 (cont.)
 Analyze: The boiling point elevation is the product of the boilingpoint-elevation constant,0.52°C/m,and the solute concentration.
Since the concentration is close to 1 m, we predict that the boiling
point elevation will be close to 0.5°C.
 Solve:
The temperature at which this seawater boils is 0.60°C higher than
the normal boiling point of pure water: 100.0°C + 0.60°C = 100.6°C.
© 2012 by W. W. Norton & Company
Sample Exercise 11.9 (cont.)
 Think about It: As predicted, the boiling point of
the solution is only slightly higher than the
boiling point of the pure solvent.
© 2012 by W. W. Norton & Company
Sample Exercise 11.10
What is the freezing point of radiator fluid prepared
by mixing 1.00 L of ethylene glycol (HOCH2CH2OH,
density 1.114 g/mL) with 1.00 L of water (density
1.000 g/mL)? The freezing-point-depression
constant of water, Kf, is 1.86°C/m.
 Collect and Organize: We are asked to determine
the freezing point of a solution of ethylene glycol in
water. We are given the volumes of the two liquids,
their densities, and Kf for water, and we have
Equation 11.11.
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Sample Exercise 11.10 (cont.)
 Analyze: To determine the freezing point, we need to calculate ΔTf
for the solution and then subtract that value from water’s normal
freezing point. Because using Equation 11.11 requires us to know
m, the molality of the solution, we need to convert volumes of solute
and solvent into moles of solute and kilograms of solvent; knowing
their densities allows us to make both conversions, and we use
ethylene glycol’s formula to calculate its molar mass. Based on
Sample Exercise 11.4, we choose water as the solvent. It is a logical
choice, because ethylene glycol has a much higher molar mass and
a density very similar to water, so we can assume that 1 L of
ethylene glycol contains fewer moles of material than 1 L of water.
The relatively large volume (103 mL) and a density close to 1 g/mL
for the solute allow us to predict that the mass of solute will be on
the order of 103 g. With a solute molar mass of 62 g/mol, the
molality of the solution should be between 10 and 20 m. Considering
Kf = 1.86°C/m, we predict ΔTf to be in the range of 20° to 40°C.
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Sample Exercise 11.10 (cont.)
 Solve: The solvent mass is
After calculating the ethylene glycol molar mass to be 62.07
g/mol, we have for the solute
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Sample Exercise 11.10 (cont.)
Therefore the molal concentration is
Using Equation 11.11 gives
Subtracting this temperature change from the normal freezing
point of water, 0.0°C, we have
Freezing point of radiator fluid = 0.0°C - 33.3°C = -33.3°C
© 2012 by W. W. Norton & Company
Sample Exercise 11.10 (cont.)
 Think about It: The answer makes sense
because the reason we add antifreeze to
radiators is to depress the freezing point of
water, and the solution we evaluated here
certainly has a freezing point lower than that of
pure water. To be an effective radiator fluid, the
solution must have a freezing point sufficiently
below the coldest expected temperatures, so our
answer of -33°C is reasonable and matches our
prediction.
© 2012 by W. W. Norton & Company
Sample Exercise 11.11
The salt lithium perchlorate (LiClO4) is one of the most
water-soluble salts known. At what temperature does a
0.130 m solution of LiClO4 freeze? The Kf of water is
1.86°C/m; assume i = 2 for LiClO4 and the freezing point
of pure water is 0.00°C.
 Collect and Organize: We are asked to determine the
freezing point of a salt solution. We know the formula of
the solute, its molal concentration, and the Kf of the
solvent. We know that the freezing point of the pure
solvent is 0.00°C. Equation 11.13 relates freezing point
depression to solute concentration when we take the
van’t Hoff factor into consideration. We are given the
value of the van’t Hoff factor as i = 2.
© 2012 by W. W. Norton & Company
Sample Exercise 11.11 (cont.)
 Analyze: We need Equation 11.13 to solve for
the freezing point depression. The value of Kf is
close to 2, the value of i is 2, and the
concentration of solute is close to 0.1 m, so we
predict that the freezing point of the solution will
be about 0.4°C lower than that of pure water.
© 2012 by W. W. Norton & Company
Sample Exercise 11.11 (cont.)
 Solve:
The freezing point of the solution is (0.00 - 0.48)°C = -0.5°C.
 Think about It: Lithium perchlorate dissolves in aqueous
solution to form 2 moles of ions for every mole of solute that
dissolves: 1 mole of Li+ cations and 1 mole of ClO4- anions,
consistent with i = 2. The calculated value is consistent with
our prediction.
© 2012 by W. W. Norton & Company
Sample Exercise 11.12
The experimentally measured freezing point of a 1.90 m
aqueous solution of NaCl is -6.57°C. What is the value of
the van’t Hoff factor for this solution? Is the solution
behaving ideally, or is there evidence that solute
particles are interacting with one another? The freezingpoint-depression constant of water is Kf = 1.86°C/m, and
the freezing point of pure water is 0.00°C.
 Collect and Organize: We are asked to calculate the i
factor for a solution of known molality. We are given the
measured freezing point and the Kf value. Equation
11.13 relates ΔTf, Kf, the molality of the solution, and the
value of i. We are also asked whether the solution is
behaving ideally.
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Sample Exercise 11.12 (cont.)
 Analyze: Equation 11.13 tells us that
We can rearrange the equation to solve for i
before substituting the values for ΔTf, Kf, and m.
If the solution behaves ideally, we would expect i
= 2 for the strong electrolyte NaCl; however,
given the concentration of NaCl (1.90 m), we
predict a value less than 2.
© 2012 by W. W. Norton & Company
Sample Exercise 11.12 (cont.)
 Solve: Rearranging Equation 11.13 gives us
That i is not an integer tells us that the solution is not
behaving ideally. Ion pairs must be forming in solution.
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Sample Exercise 11.12 (cont.)
 Think about It: As predicted, the value of i for
this solution is less than the theoretical value of
2.
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Sample Exercise 11.13
At the beginning of this section, we mentioned that the
concentration of solutes in a red blood cell is about a
third of that of seawater—more precisely, about 0.30 M.
If red blood cells are bathed in pure water, they swell, as
shown in Figure 11.22(c). Calculate the osmotic
pressure at 25°C of red blood cells across the cell
membrane from pure water.
 Collect and Organize: We are to calculate an osmotic
pressure across a membrane that separates pure water
from red blood cells. The osmotic pressure of a solution
is related to the total particle concentration (0.30 M) and
temperature (25°C) of the solution by Equation 11.14.
We must convert the temperature to kelvins.
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Sample Exercise 11.13 (cont.)
 Analyze: Osmotic pressure is related to the total
concentration of all particles in solution and the absolute
temperature. The total particle concentration, 0.30 M,
represents the iM term in Equation 11.14.
 Solve: First we need to convert the temperature in degrees
Celsius to kelvins: T(°C) + 273 = T(K). Inserting the values of
iM, R, and T in Equation 11.14 we have
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Sample Exercise 11.13 (cont.)
 Think about It: The calculated pressure across
the membranes of red blood cells in pure water
is over 7 atmospheres: enough to rupture the
membranes.
© 2012 by W. W. Norton & Company
Sample Exercise 11.14
Red blood cells placed in seawater shrivel as shown in Figure
11.22(b). Calculate the osmotic pressure across the
semipermeable cell membrane, separating the solution inside
a red blood cell from seawater at 25°C if the total
concentration of all the particles inside the cell is 0.30 M and
the total concentration of ions in seawater is 1.15 M. Compare
the result with the answer from Sample Exercise 11.13.
 Collect and Organize: We are to calculate an osmotic
pressure across a membrane that separates two solutions
with different total concentrations of particles at a constant
temperature of 25°C. We can use Equation 11.15 to calculate
the difference in osmotic pressure between the two solutions.
As in Sample Exercise 11.13, we must convert the
temperature to units of kelvins.
© 2012 by W. W. Norton & Company
Sample Exercise 11.14 (cont.)
 Analyze: We know that the concentration of
particles in seawater is greater than inside a red
blood cell, so solvent will flow from the cell to the
seawater. Thus, the osmotic pressure in the cell
(πcell) will be less than the osmotic pressure of the
seawater (πseawater) and the osmotic pressure across
the membrane will be the difference between
πseawater and πcell, or Δπ. The value of Δπ should
be greater than the π calculated in Sample Exercise
11.13, where red blood cells were bathed in pure
water because the difference in solute
concentrations is greater.
© 2012 by W. W. Norton & Company
Sample Exercise 11.14 (cont.)
 Solve: First we need to convert the temperature in
degrees Celsius to kelvins: T(°C) + 273 = T(K).
Inserting the values of iM, R, and T in Equation 11.15
for seawater and cells we have
© 2012 by W. W. Norton & Company
Sample Exercise 11.14 (cont.)
 Think about It: As predicted, the calculated
pressure across a semipermeable membrane
separating seawater from red blood cells, 21
atm, is greater than theosmotic pressure across
a semipermeable membrane separating red
blood cells from pure water (7.3 atm).
© 2012 by W. W. Norton & Company
Sample Exercise 11.15
What is the reverse osmotic pressure required at 20°C to
purify brackish well water containing 0.355 M dissolved
particles if the product water is to contain no more than
87 mg of dissolved solids (as NaCl) per liter?
 Collect and Organize: We are asked to calculate the
reverse osmotic pressure needed to purify water to a
stated solute concentration. We are given the
temperature, the solute concentration in the water to be
purified, and the amount of solute (expressed as NaCl)
tolerable in the product water. We can adapt Equation
11.15 to calculate the difference in osmotic pressure
between the two solutions.
© 2012 by W. W. Norton & Company
Sample Exercise 11.15 (cont.)
 Analyze: We need the molarity of the less
concentrated solution, the drinkable water, to
calculate the osmotic pressure exerted by the
drinkable water and we can calculate that
molarity from the information given. We can also
calculate the osmotic pressure once we know
the difference in the molarities of the solutions
on the two sides of the semipermeable
membrane.
© 2012 by W. W. Norton & Company
Sample Exercise 11.15 (cont.)
 Solve: First we convert 87 mg NaCl/L to molarity:
The total ion concentration for a 1.5 ×10-3 M NaCl solution is
2(1.5 ×10-3 M) = 0.0030 M. Using Equation 11.15:
© 2012 by W. W. Norton & Company
Sample Exercise 11.15 (cont.)
If we maintain an osmotic pressure of exactly 8.47 atm
on the side of the membrane with the brackish well
water, no net flow of solvent will occur between the
two solutions. Pressures greater than 8.47 atm will
force water molecules from the well water through the
membrane to give water containing less than 87 mg
NaCl per liter. (The reason we can never obtain
absolutely pure water in this process is that, in
practice, some Na+ and Cl- ions pass through the
membrane from the well water to the drinkable water.)
© 2012 by W. W. Norton & Company
Sample Exercise 11.15 (cont.)
 Think about It: In the absence of any external
pressure, solvent will flow from the product water
to the well water, driven by the concentration
difference between the two solutions as shown
in Figure 11.23. However, if we apply sufficient
external pressure we can reverse the flow of
water. The value of the external pressure, about
8.5 atm, represents the minimum external
pressure that must be applied to make this
device function.
© 2012 by W. W. Norton & Company
Sample Exercise 11.16
Eicosene is a molecular compound and nonelectrolyte
with the empirical formula CH2. The freezing point of a
solution prepared by dissolving 100 mg of eicosene in
1.00 g of benzene was 1.75°C lower than the freezing
point of pure benzene. What is the molar mass of
eicosene? (Kf for benzene is 4.90°C/m.)
 Collect and Organize: We are asked to determine the
molar mass of a compound. We are given the mass of
the compound that lowers the freezing point of a solvent
by a known amount, and we have the Kf of the solvent.
Because the solute is a nonelectrolyte, i = 1. Equation
11.13 relates concentration to the change in freezing
point.
© 2012 by W. W. Norton & Company
Sample Exercise 11.16 (cont.)
 Analyze: The molar mass of a compound is expressed in
units of grams per mole. Our sample consists of 100 mg, or
0.100 g. To find the molar mass we need to determine how
many moles of eicosene are contained in the sample. The
concentration term in Equation 11.13 is molality, or moles of
solute per kilogram of solvent, so we can use the
experimental data to first calculate molality. When we know
the molality of the eicosene, we know the number of moles of
eicosene in 1 kg of benzene; we are given the mass of
benzene used, so we can calculate the number of moles of
eicosene in the sample, from which we may calculate the
molecular mass of eicosene. The empirical formula of
eicosene enables us to check the validity of our answer as the
molar mass of eicosene must be a whole-number multiple of
the mass of CH2 (14 g/mol).
© 2012 by W. W. Norton & Company
Sample Exercise 11.16 (cont.)
 Solve: The molality of the eicosene solution is
which means 0.357 mol of eicosene is dissolved per kilogram
of solvent. Only 1.00 g (1.00 × 10-3 kg) was used in this
sample. Calculating the moles of eicosene in the sample:
© 2012 by W. W. Norton & Company
Sample Exercise 11.16 (cont.)
Because the molar mass is the mass of one mole of
eicosene, and 100 mg of eicosene was used to prepare
the solution,
 Think about It: The molar mass of eicosene, 280 g/mol,
is reasonable since it corresponds to (CH2)n where n =
20. The molecular formula of eicosene is therefore
C20H40.
© 2012 by W. W. Norton & Company
Sample Exercise 11.17
A molecular compound that is a nonelectrolyte was
isolated from a South African tree. A 47 mg sample was
dissolved in water to make 2.50 mL of solution at 25°C,
and the osmotic pressure of the solution was 0.489 atm.
Calculate the molar mass of the compound.
 Collect and Organize: We are given the mass of a
substance, the volume of its aqueous solution, the
temperature, and its osmotic pressure. We can relate
these parameters using Equation 11.14, using the value
i = 1 because this is a nonelectrolyte.
© 2012 by W. W. Norton & Company
Sample Exercise 11.17 (cont.)
 Analyze: We can calculate the molar
concentration of the solution from the
information given and Equation 11.14. Since we
know the solution volume, we can calculate the
number of moles of the solute from the molarity.
Because we know the mass of this number of
moles, we can calculate the molar mass of the
solute.
© 2012 by W. W. Norton & Company
Sample Exercise 11.17 (cont.)
 Solve: Rearranging Equation 11.14 to isolate M and
substituting the given values:
Next we solve the defining equation for molarity, M = n/V, for
n, the number of moles of solute:
© 2012 by W. W. Norton & Company
Sample Exercise 11.17 (cont.)
We know that this number of moles of solute has a
mass of 47 mg. The molar mass of the solute is
therefore
© 2012 by W. W. Norton & Company
Sample Exercise 11.17 (cont.)
 Think about It: One of the advantages of
determining molar mass by osmotic pressure is
that only a small amount of material is required.
With only a 47 mg sample of a rather large
molecule (molar mass 940 g/mol), the osmotic
pressure is sufficiently large (0.489 atm) to
enable us to calculate the molar mass
accurately.
© 2012 by W. W. Norton & Company
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