13.9

In predicting solubility,
remember the saying: “Like
dissolves like.” A nonpolar
solute will dissolve in a nonpolar solvent; ionic compounds will generally
dissolve in polar solvents due to favorable iondipole interactions;
solutes that can form hydrogen bonds with a solvent will also have high
solubility in the solvent. Naphthalene
and benzene are nonpolar, whereas CsF is ionic.


13.10

Strong hydrogen bonding (dipoledipole
attraction) is the principal intermolecular attraction in liquid ethanol,
but in liquid cyclohexane the intermolecular forces are dispersion forces
because cyclohexane is nonpolar.
Cyclohexane cannot form hydrogen bonds with ethanol, and therefore
cannot attract ethanol molecules strongly enough to form a solution.


13.11

The order
of increasing solubility is: O_{2}
< Br_{2} < LiCl < CH_{3}OH. Methanol is miscible with water because
of strong hydrogen bonding. LiCl is
an ionic solid and is very soluble because of the high polarity of the
water molecules. Both oxygen and
bromine are nonpolar and exert only weak dispersion forces. Bromine is a larger molecule and is
therefore more polarizable and susceptible to dipoleinduced dipole
attractions.


13.12

The longer the CC chain, the more the
molecule "looks like" a hydrocarbon and the less significant the OH group becomes. Hence, as the CC
chain length increases, the molecule becomes less polar. Since “like dissolves like”, as the
molecules become more nonpolar, the solubility in polar water
decreases. The OH group of the
alcohols can form strong hydrogen bonds with water molecules, but this
property diminishes in importance as the chain length increases.


13.16

a.

The percent by mass is defined by Equation 13.2 as
Substituting in the percent by mass of solute and the
mass of solute, we can solve for the mass of solvent (water).
(0.162)(mass
of water) = 5.00 g 
(0.162)(5.00g)
mass
of water = 25.9 g

b.

Similar to part (a),
mass
of water = 1.72 ´
10^{3} g



13.17

a.

The molality is the number of
moles of sucrose (molar mass 342.3 g/mol) divided by the mass of the
solvent (water) in kg.
molality
=0.0610 m

b.

molality
=2.04 m



13.18

a.

molality
=2.74 m

b.

100 g of the solution
contains 45.2 g KBr and 54.8 g H_{2}O.
molality
=6.93 m



13.19

In each case we consider one liter of solution.
mass of solution
= volume ´
density
a.


b.

mass
solvent (H_{2}O) = 1040 g 
35 g = 1005 g
= 1.005 kg

c.

mass
solvent (H_{2}O) = 1190 g 
440 g = 750 g
= 0.75 kg



13.20

Let’s assume
that we have 1.0 L of a 0.010 M solution.
Assuming
solution density of 1.0 g/mL, the mass of 1.0 L (1000 mL) of the solution
is 1000 g or
1.0 ´
10^{3} g.
The mass of 0.010 mole of urea is:
The mass of the solvent is:
(solution
mass) 
(solute mass) = (1.0 ´ 10^{3} g) 
(0.60 g) = 1.0 ´ 10^{3} g =
1.0 kg


13.21

We find the volume of ethanol in 1.00 L of 75 proof
gin. Note that 75 proof means


13.22

a.

Convert mass percent to
molality, where
.
We first convert 98.0 g H_{2}SO_{4}
to moles of H_{2}SO_{4} using its molar mass, then we
convert 2.0 g of H_{2}O to units of kilograms.
Lastly, we divide moles of solute by mass of solvent in
kg to calculate the molality of the solution.

b.

Convert molality to
molarity.
From part (a), we know the
moles of solute (0.999 mole H_{2}SO_{4}) and the mass of
the solution (100.0 g). To solve
for molarity, we need the volume of the solution, which we can calculate
from its mass and density.
First, we use the
solution density as a conversion factor to convert to volume of solution.
Since we already know moles
of solute from part (a), 0.999 mole H_{2}SO_{4}, we
divide moles of solute by liters of solution to calculate the molarity of
the solution.



13.23

Strategy:

In this problem, the masses of both solute and solvent
are given, and the density is given.

Solution:

In order to determine molarity, we need to know the
volume of the solution. We add the
masses of solute and solvent to get the total mass, and use the density
to determine the volume.
35.0 g NH_{3} + 75.0 g H_{2}O = 110.0
g soln
Volume of soln = 110.0 g soln
We then convert the mass of NH_{3} to moles,
and divide by the volume in liters to get molarity.
35.0 g NH_{3}
Molarity =18.3 M
To
calculate molality, we use Equation 13.1.
Molality = 27.4 m

Think About It:

Remember in
these calculations to express volume of solution in L and mass of solvent
in kg.



13.24

Assume 100.0 g of solution.
a.

The mass of ethanol in the
solution is 0.150 ´ 100.0 g = 15.0 g. The mass of the water is
100.0 g 
15.0 g = 85.0 g = 0.0850 kg. The
amount of ethanol in moles is:
Note that we retain an extra digit
until the end of the calculation, to avoid rounding error.
molality = 3.83 m

b.

The volume of the
solution is:
The amount of ethanol in
moles is 0.3256 mole [part (a)].
molarity = 3.19 M

c.

solution
volume = 0.0784 L or 78.4 mL



13.25

Assuming that N_{2} and O_{2} are the only
dissolved gases, the mole fractions in the total dissolved gas can be
determined as follows:
We multiply each partial pressure by the corresponding Henry’s law
constant (solubility).
For O_{2}, we have
0.20 atm ´ 1.3 ´
10^{3}
mol/L×atm
= 2.6 ´
10^{4}
mol/L
For N_{2}, we have
0.80 atm ´ 6.8 ´
10^{4}
mol/L×atm
= 5.44 ´
10^{4}
mol/L
In a liter of water, then, there will be 2.6 ´ 10^{4}
mol O_{2} and 5.44 ´ 10^{4}
mol N_{2}. The mole
fractions are
c(N_{2}) = 0.677
c(O_{2}) = 0.323
Due to the
greater solubility of oxygen, it has a larger mole fraction in solution than
it does in the air.


13.32

The amount of salt dissolved in 100 g of water is:
Therefore, the solubility of the salt is 35.2 g
salt/100 g H_{2}O.


13.33

At 75°C,
155 g of KNO_{3} dissolves in 100 g of water to form 255 g of solution. When cooled to 25°C, only 38.0 g of KNO_{3}
remain dissolved. This means that
(155 
38.0) g = 117 g of KNO_{3} will crystallize. The amount of KNO_{3} formed when
100 g of saturated solution at 75°C is cooled to 25°C
can be found by a simple unit conversion.


13.34

The mass of
KCl is 10% of the mass of the whole sample or 5.0 g. The KClO_{3} mass is 45 g. If 100 g of water will dissolve 25.5 g of
KCl, then the amount of water to dissolve 5.0 g KCl is:
The 20 g of
water will dissolve:
The KClO_{3}
remaining undissolved will be:
(45  1.4) g KClO_{3} = 44
g KClO_{3}


13.35

Strategy:

Consider the number of polar groups, and determine
whether each molecule is predominantly polar or predominantly
nonpolar. Predominantly polar
molecules tend to be water soluble.
Predominantly nonpolar molecules tend to be fat soluble.

Setup:

Polar groups include those containing O atoms or N
atoms (atoms with lone pairs).
Vitamin C contains six polar groups (six O atoms) making it
predominately polar. Vitamin E
contains only two polar groups and is therefore predominately nonpolar.

Solution:

a.

Water soluble.

b.

Fat soluble.




13.36

Strategy:

Consider the number of polar groups, and determine
whether each molecule is predominantly polar or predominantly
nonpolar. Predominantly polar
molecules tend to be water soluble.
Predominantly nonpolar molecules tend to be fat soluble.

Setup:

Polar groups include those containing O atoms or N
atoms (atoms with lone pairs).
Vitamin D contains only one polar group and is therefore
predominately nonpolar. Vitamin B_{2}
contains 10 polar groups (six O atoms and four N atoms) making it
predominately polar.

Solution:

a.

Fat soluble.

b.

Water soluble.




13.37

Strategy:

The given solubility allows us to calculate Henry's
law constant (k), which can then be used to determine the
concentration of CO_{2} at 0.0003 atm.

Solution:

First, calculate the Henry's law constant, k, using
the concentration of CO_{2} in water at 1 atm.
For atmospheric conditions we write:
c =
kP = (0.034 mol/L×atm)(0.00030
atm) = 1.0 ´
10^{5}
mol/L



13.38

We first find
the value of k for Henry's law
Next, we can calculate the
concentration of N_{2} in blood at 4.0 atm using k
calculated above.
c = kP
c =
(7.0 ´
10^{4}
mol/L×atm)(4.0
atm) = 2.8 ´ 10^{3}
mol/L
From each of the concentrations of N_{2} in blood, we can calculate
the number of moles of N_{2} dissolved by multiplying by the total
blood volume of 5.0 L. Then, we can
calculate the number of moles of N_{2} released when the diver
returns to the surface.
The number of moles of N_{2}
in 5.0 L of blood at 0.80 atm is:
(5.6
´
10^{4}
mol/L )(5.0 L) = 2.8 ´ 10^{3}
mol
The number of moles of N_{2}
in 5.0 L of blood at 4.0 atm is:
(2.8
´
10^{3}
mol/L)(5.0 L) = 1.4 ´ 10^{2}
mol
The amount of N_{2} released in moles when the diver
returns to the surface is:
(1.4 ´ 10^{2}
mol) 
(2.8 ´
10^{3}
mol) = 1.1 ´ 10^{2}
mol
Finally, we
can now calculate the volume of N_{2} released using the ideal gas
equation. The total pressure pushing
on the N_{2} that is released is atmospheric pressure (1 atm).
The volume of
N_{2} released is:


13.39

Strategy:

The
mass of CO_{2} liberated from the container is given by the
difference of the two mass measurements. Assume that, at the time of the
second mass measurement, all the CO_{2} has been released and
that the amount of CO_{2} in the vapor phase of the unopened soda
is negligible. Then, we can use the following conversions to calculate
the pressure of CO_{2} in the unopened soda.

Solution:

Mass of CO_{2} lost =
853.5 g – 851.3 g = 2.2 g CO_{2}
moles CO_{2} =
Using Henry’s law for the final conversion:
This pressure is
only an estimate since we ignored the amount of CO_{2} that was
present in the unopened container in the gas phase.



13.56

The
first step is to find the number of moles of sucrose and of water.
The mole
fraction of water is:
The
vapor pressure of the solution is found as follows:


13.57

Strategy:

From the vapor pressure of
water at 20°C
and the change in vapor pressure for the solution
(2.0 mmHg), we can solve for the mole fraction of sucrose using Equation
13.5 of the text. From the mole
fraction of sucrose, we can solve for moles of sucrose. Lastly, we convert form moles to grams
of sucrose.

Solution:

Using Equation 13.5 of the
text, we can calculate the mole fraction of sucrose that causes a
2.0 mmHg drop in vapor pressure.
4
From the definition of mole fraction, we can calculate moles of
sucrose.
moles of water
= 552 g30.63 mol H_{2}O
n_{sucrose}
=
3.94 mol sucrose
Using the molar
mass of sucrose as a conversion factor, we can calculate the mass of
sucrose.
mass
of sucrose =3.94 mol sucrose1.3 ´
10^{3} g sucrose



13.58

Let us call benzene
component 1 and camphor component 2.


13.59

For any solution the sum of the mole fractions of the
components is always 1.00, so the mole fraction of
1propanol
is 0.700. The partial pressures are:
Think About It:

Is the vapor phase
richer in one of the components than the solution? Which component? Should this always be true for ideal
solutions?



13.60

a.

First find the mole fractions of the solution components.
The vapor
pressures of the methanol and ethanol are:
P_{methanol}
= (0.489)(94 mmHg) =
46 mmHg
P_{ethanol} =
(0.511)(44 mmHg) = 22 mmHg

b.

Since n = PV/RT
and V and T are the same for both vapors, the number of
moles of each substance is proportional to the partial pressure. We can then write for the mole
fractions:
X_{ethanol} =
1 
C_{methanol} = 0.32

c.

The two components could be separated
by fractional distillation.



13.61

This
problem is very similar to Problem 13.57.
2.50 mmHg
= X_{urea}(31.8
mmHg)
X_{urea} =
0.0786
The number
of moles of water is:
n_{water} = 658 g H_{2}O36.51 mol H_{2}O
0.0786 =
n_{urea} =
3.11 mol
mass of urea = 3.11 mol urea187 g urea


13.62

DT_{b} = K_{b}m =
(2.53°C/m)(3.12
m) = 7.89°C
The new boiling
point is 80.1°C + 7.89°C
= 88.0°C
DT_{f} = K_{f}m =
(5.12°C/m)(3.12
m) = 16.0°C
The new freezing point is 5.5°C

16.0°C
= 10.5°C


13.63


13.64

We want a freezing point depression of 20°C.
The mass of ethylene glycol (EG) in 6.5 L or 6.5 kg of water is:
The volume of EG needed is:
Finally, we calculate the boiling point:
DT_{b} = mK_{b} =
(10.8 m)(0.52°C/m) =
5.6°C
The boiling point of the solution will be 100.0°C
+ 5.6°C
= 105.6°C.


13.65

We first find the number of moles of gas using the ideal gas
equation.
molality
=
DT_{f} =
K_{f}m
= (5.12°C/m)(2.13
m) = 10.9°C
freezing
point = 5.5°C 
10.9°C = 5.4°C


13.66

p =
MRT = (1.57 mol/L)(0.0821 L×atm/K×mol)(27.0
+ 273)
K =
38.7 atm


13.67

CaCl_{2} is an ionic compound
and is therefore an electrolyte in water.
Assuming that CaCl_{2} is a strong electrolyte and
completely dissociates (no ion pairs, van't Hoff factor i = 3), the
total ion concentration will be 3 ´
0.35 = 1.05 m, which is larger than the urea (nonelectrolyte)
concentration of 0.90 m.
a.

The CaCl_{2} solution will show a
larger boiling point elevation.

b.

The freezing point of the
urea solution will be higher because the CaCl_{2} solution
will have a larger freezing point depression.

c.

The CaCl_{2} solution will have a
larger vapor pressure lowering.



13.68

Boiling point, vapor
pressure, and osmotic pressure all depend on particle concentration. Therefore, these solutions also have the
same boiling point, osmotic pressure, and vapor pressure.


13.69

Assume that all the salts are completely dissociated. Calculate the molality of the ions in the
solutions.
0.10 m
Na_{3}PO_{4}: 0.10
m ´
4 ions/unit = 0.40 m
0.35 m
NaCl: 0.35
m ´
2 ions/unit = 0.70 m
0.20 m
MgCl_{2}: 0.20
m ´
3 ions/unit = 0.60 m
0.15 m
C_{6}H_{12}O_{6}: nonelectrolyte,
0.15 m
0.15
m CH_{3}COOH: weak
electrolyte, slightly greater than 0.15 m
The solution
with the lowest molality will have the highest freezing point (smallest
freezing point depression): 0.15 m
C_{6}H_{12}O_{6} > 0.15 m CH_{3}COOH
> 0.10 m Na_{3}PO_{4} > 0.20 m MgCl_{2}
> 0.35 m NaCl.


13.70

The freezing
point will be depressed most by the solution that contains the most solute
particles. Classify each solute as a
strong electrolyte, a weak electrolyte, or a nonelectrolyte. All three solutions have the same
concentration, so comparing the solutions is straightforward. HCl is a strong electrolyte, so under
ideal conditions it will completely dissociate into two particles per
molecule. The concentration of
particles will be 1.00 m.
Acetic acid is a weak electrolyte, so it will only dissociate to a
small extent. The concentration of
particles will be greater than 0.50 m, but less than 1.00 m. Glucose is a nonelectrolyte, so glucose
molecules remain as glucose molecules in solution. The concentration of particles will be
0.50 m. For these solutions,
the order in which the freezing points become lower is:
0.50
m glucose > 0.50 m acetic acid >
0.50 m HCl
The HCl
solution will have the lowest freezing point (greatest freezing point
depression).


13.71

a.

NaCl is a strong
electrolyte. The concentration of
particles (ions) is double the concentration of NaCl. Note that because the density of water
is 1 g/mL, 135 mL of water has a mass of 135 g.
The number of moles of NaCl is:
Next, we can find the changes in boiling and freezing
points (i = 2)
DT_{b} =
iK_{b}m
= 2(0.52°C/m)(2.70
m) = 2.8°C
DT_{f} =
iK_{f}m
= 2(1.86°C/m)(2.70
m) = 10.0°C
The boiling point is 102.8°C; the freezing point is 10.0°C.

b.

Urea is a nonelectrolyte.
The particle concentration is just equal to the urea
concentration.
The molality of the urea
solution is:
DT_{b} =
iK_{b}m
= 1(0.52°C/m)(3.84
m) = 2.0°C
DT_{f} =
iK_{f}m
= 1(1.86°C/m)(3.84
m) = 7.14°C
The boiling point is 102.0°C; the freezing point is 7.14°C.



13.72

Using Equation 13.5 of the text, we can find the mole fraction of
the NaCl. We use subscript 1 for H_{2}O
and subscript 2 for NaCl.
Let’s assume
that we have 1000 g (1 kg) of water as the solvent, because the definition
of molality is moles of solute per kg of solvent. We can find the number of moles of
particles dissolved in the water using the definition of mole fraction.
Since NaCl dissociates to form two particles (ions), the number of
moles of NaCl is half of the above result.
The molality of the solution is:


13.73

Both NaCl and CaCl_{2}
are strong electrolytes. Urea and
sucrose are nonelectrolytes. The
NaCl or CaCl_{2} will yield more particles per mole of the solid
dissolved, resulting in greater freezing point depression. Also, sucrose and urea would make a mess
when the ice melts.


13.74

We want to calculate the osmotic pressure of a NaCl solution. Since NaCl is a strong electrolyte, i
in the van't Hoff equation is 2.
p = iMRT
Since, R is a constant and T is given, we need to
first solve for the molarity of the solution in order to calculate the
osmotic pressure (p). If we assume a
given volume of solution, we can then use the density of the solution to
determine the mass of the solution.
The solution is 0.86% by mass NaCl, so we can find grams of NaCl in
the solution.
To calculate molarity, assume that we have 1.000 L of solution
(1.000 ´
10^{3} mL). We can use the
solution density as a conversion factor to calculate the mass of 1.000 ´
10^{3} mL of solution.
Since the solution is 0.86% by mass NaCl, the mass of NaCl in the
solution is:
The molarity of the solution is:
Since NaCl is a strong electrolyte, we assume that the van't Hoff
factor is 2. Substituting i, M,
R, and T into the equation for osmotic pressure gives:


13.75

The temperature
and molarity of the two solutions are the same. If we divide Equation 13.11 of the text
for one solution by the same equation for the other, we can find the ratio
of the van't Hoff factors in terms of the osmotic pressures (i = 1
for urea).


13.76

From Table 13.3 of the text, i = 1.3 for MgSO_{4}. Using Equation 13.11,
p = iMRT
p =
1.6 atm


13.77

For this problem, we first need to calculate the pressure necessary
to support a column of water 105 m high.
We use the equation P = hdg, where h, d,
and g are the column height in m, the density of water in kg/m^{3},
and the gravitational constant, respectively.
P = 105 m ´ 1.00 ´
10^{3} ´
9.81= 1.03 ´ 10^{6} kg/m×s^{2}
= 1.03 ´
10^{6} Pa ´10.16 atm
Because atmospheric pressure contributes 1 atm, the osmotic pressure
required is only 9.16 atm.


13.78

Convert both glucose levels to molar concentrations and use molarity
to calculate osmotic pressure.
p = MRT =
(0.00971 M)(0.0821 L×atm/mol×K)(310
K) = 0.247 atm
p = MRT =
(0.0047 M)(0.0821 L×atm/mol×K)(310
K) = 0.120 atm
Because of its higher glucose
concentration, the blood plasma of a diabetic patient has a much higher
osmotic pressure. The water in the
body moves via osmosis into the plasma to dilute the concentrated glucose. The need to replenish the water used to
dilute the plasma makes the patient thirsty.


13.82

METHOD 1:
Assume 100 g of compound.
Thus, we arrive at the formula C_{3.33}H_{6.6}O_{3.3},
which gives the identity and the ratios of atoms present. Dividing each subscript by the smallest
subscript gives whole numbers.
This gives us the empirical
formula, CH_{2}O.
Now, we can use the
freezing point data to determine the molar mass. First, calculate the molality of the
solution.
Multiplying the molality by
the mass of solvent (in kg) gives moles of unknown solute. Then, dividing the mass of solute (in g)
by the moles of solute, gives the molar mass of the unknown solute.
= 0.00542 mol solute
Finally, we compare the
empirical molar mass to the molar mass above.
empirical molar mass =
12.01 g + 2(1.008 g) + 16.00 g
= 30.03 g/mol
The number of (CH_{2}O) units
present in the molecular formula is:
Thus, there are four CH_{2}O
units in each molecule of the compound, so the molecular formula is (CH_{2}O)_{4},
or C_{4}H_{8}O_{4}.
METHOD 2:
We use the freezing point data to
determine the molar mass. First,
calculate the molality of the solution.
Multiplying the molality by
the mass of solvent (in kg) gives moles of unknown solute. Then, dividing the mass of solute (in g)
by the moles of solute, gives the molar mass of the unknown solute.
= 0.00542 mol solute
Next, we multiply the mass
% (converted to a decimal) of each element by the molar mass to convert to
grams of each element. Then, we use
the molar mass to convert to moles of each element.
Since we used the molar mass to calculate
the moles of each element present in the compound, this method directly
gives the molecular formula. The
formula is C_{4}H_{8}O_{4}.


13.83

First, from the freezing point depression we can calculate the
molality of the solution. See Table
13.2 of the text for the normal freezing point and K_{f}
value for benzene.
DT_{f} =
(5.5 
4.3)°C =
1.2°C
Multiplying
the molality by the mass of solvent (in kg) gives moles of unknown
solute. Then, dividing the mass of
solute (in g) by the moles of solute, gives the molar mass of the unknown
solute.
= 0.0058
mol solute
The empirical molar mass of C_{6}H_{5}P is 108.1
g/mol. Therefore, the molecular
formula is (C_{6}H_{5}P)_{4} or C_{24}H_{20}P_{4}.


13.84

The experimental data indicate
that the benzoic acid molecules are associated together in pairs in
solution due to hydrogen bonding.


13.85

Strategy:

We are asked to calculate the
molar mass of the polymer. Grams
of the polymer are given in the problem, so we need to solve for moles of
polymer.
From the osmotic pressure of
the solution, we can calculate the molarity of the solution. Then, from the molarity, we can
determine the number of moles in 0.8330 g of the polymer. What units should we use for π and
temperature?

Solution:

First, we calculate the molarity using Equation 13.8 of
the text.
p =
MRT
Multiplying the molarity by the volume of solution (in
L) gives moles of solute (polymer).
?
mol of polymer = (2.80 ´
10^{4}
mol/L)(0.170 L) = 4.76 ´
10^{5}
mol polymer
Lastly, dividing the mass of polymer (in g) by the
moles of polymer, gives the molar mass of the polymer.



13.86

Method
1: First, find the concentration
of the solution, then work out the molar mass. The concentration is:
The solution volume is 0.3000 L so the number of moles of solute is:
The molar mass is then:
The empirical formula can be found most easily by assuming a 100.0 g
sample of the substance.
The gives the formula: C_{3.48}H_{4.7}O_{2.33}N_{1.16}. Dividing through by the smallest
subscript (1.16) gives the empirical formula, C_{3}H_{4}O_{2}N,
which has a mass of 86.0 g per formula unit. The molar mass is five times this amount
(430 ¸
86.0 = 5.0), so the molecular formula is (C_{3}H_{4}O_{2}N)_{5}
or C_{15}H_{20}O_{10}N_{5}.
METHOD 2: Use
the molarity data as above to determine the molar mass.
molar mass =
430 g/mol
Multiply the mass % (converted to a decimal) of each element
by the molar mass to convert to grams of each element. Then, use the molar mass to convert to
moles of each element.
Since we used the molar mass to calculate the moles of each element
present in the compound, this method directly gives the molecular
formula. The formula is C_{15}H_{20}O_{10}N_{5}.


13.87

We use the osmotic pressure data to determine the molarity.
Next we use the density and the solution mass to find the volume of
the solution.
mass
of soln = 6.85 g + 100.0 g = 106.9 g soln
Multiplying the molarity by the volume (in L) gives moles of solute
(carbohydrate).
mol
of solute = M ´ L =
(0.192 mol/L)(0.1044 L)
= 0.02004 mol solute
Finally, dividing mass of carbohydrate by moles of carbohydrate
gives the molar mass of the carbohydrate.
molar
mass =342 g/mol


13.88

Use the osmotic pressure to calculate the molar concentration of
dissolved particles in the HNO_{2} solution.
M =
% ionization = 5.6%


13.89

Use the osmotic pressure to calculate the molar concentration of
dissolved particles in the HB solution.
M =
When an HB molecule ionizes it produces one H_{2}B^{+}
ion and one OH^{} ion.
HB(aq) + H_{2}O(l)
⇄
H_{2}B^{+}(aq) + OH^{}(aq)
The concentration of HB is originally 0.100 M. It decreases by x and the
concentrations of H_{2}B^{+} and OH^{}
each increase by x. Thus, the
concentration of dissolved particles is 0.100  x + 2x
= 0.100 + x. We solve for x
and determine what percentage x is of the original concentration,
0.100 M.
% ionization = 15.7%


13.90

Strategy:

Use the osmotic pressure and Equation 13.8 to
determine the molar concentration of the particles in solution. Compare the concentration of particles
to the nominal concentration
(0.15 M) to determine what
percentage of the original HF molecules are ionized.

Setup:

R = 0.08206
L · atm/K · mol, and T = 298 K.

Solution:

Rearranging Equation 13.8 to solve for molarity,
The concentration of dissolved particles is 0.159 M.
Consider the ionization of HF:
According to this equation, if x HF molecules ionize, we get x H^{+} ion and x
F^{–} ions. Thus, the
total concentration of particles in solution will be the original
concentration of HF minus x,
which gives the concentration of intact HF molecules, plus 2x, which is the concentration of
ions (H^{+} and F^{–}):
(0.15 – x) + 2x = 0.15 + x
Therefore, 0.159 = 0.15 + x and x =
0.009. Because we earlier defined x as the amount of HF ionized, the
percent ionization is given by
percent
ionization =
At this concentration HF is 6 percent ionized.



13.94

For this problem we must find the solution mole fractions, the
molality, and the molarity. For
molarity, we can assume the solution to be so dilute that its density is
1.00 g/mL. We first find the number
of moles of lysozyme and of water.
Vapor pressure lowering:
Freezing point depression:
Boiling point elevation:
Osmotic pressure: As stated above, we assume the density of
the solution is 1.00 g/mL. The
volume of the solution will be 150 mL.
With the exception of osmotic pressure, these changes in
colligative properties are essentially negligible.


13.95

Convert each concentration to moles glucose per mL and to
total moles and total grams in 5.0 L.
Levels before:
7.77 ´ 10^{6}
mol glucose/mL blood
3.89 × 10^{–3} mol glucose
7.00 g glucose
Levels after:
1.33 ´ 10^{5}
mol glucose/mL blood
6.66 × 10^{–2} mol glucose
12.0 g glucose


13.96

DT_{f}
= K_{f}m
m =
This is an extremely high molal concentration.


13.97

Water migrates
through the semipermiable cell walls of the cucumber into the concentrated
salt solution.
Think About It:

When we go swimming in
the ocean, why don't we shrivel up like a cucumber? When we swim in fresh water pool, why don't
we swell up and burst?



13.98

a.

We use Equation 13.4 of the text to calculate the vapor
pressure of each component.
First, you must calculate the mole fraction of
each component.
Similarly,
C_{B} =
0.500
Substitute
the mole fraction calculated above and the vapor pressure of the pure
solvent into
Equation 13.4 to calculate the vapor pressure of each component of the
solution.
The total vapor pressure is the sum of the vapor
pressures of the two components.
P_{Total} =
P_{A} + P_{B} =
38 mmHg + 66 66 mmHg = 104 mmHg

b.

This problem is solved similarly to part a.
Similarly,
C_{B} =
0.714
P_{Total} =
P_{A} + P_{B} =
22 mmHg + 94 mmHg = 116 mmHg



13.99

DT_{f} =
iK_{f}m


13.100

From the osmotic pressure, you can calculate the molarity
of the solution.
Multiplying molarity by the volume of solution in liters
gives the moles of solute.
(1.58
´
10^{3}
mol solute/L soln) ´ (0.262 L soln) =
4.14 ´
10^{4}
mol solute
Divide the grams of solute by the moles of solute to
calculate the molar mass.


13.101

One manometer has pure water over the mercury, one manometer
has a 1.0 m solution of NaCl and the other manometer has a 1.0 m
solution of urea. The pure water
will have the highest vapor pressure and will thus force the mercury column
down the most; column X. Both the
salt and the urea will lower the overall pressure of the water. However, the salt dissociates into sodium
and chloride ions (van't Hoff factor i = 2), whereas urea is a
molecular compound with a van't Hoff factor of 1. Therefore the urea solution will lower
the pressure only half as much as the salt solution. Y is the NaCl solution and Z is the urea
solution.
X = water, Y = NaCl,
Z = urea
Assuming
that you knew the temperature, could you actually calculate the distance
from the top of the solution to the top of the manometer?


13.102

Solve
Equation 13.7 of the text algebraically for molality (m), then
substitute DT_{f}
and K_{f} into the equation to calculate the molality. You can find the normal freezing point
for benzene and K_{f} for benzene in Table 13.2 of the text.
DT_{f} =
5.5°C

3.9°C =
1.6°C
Multiplying
the molality by the mass of solvent (in kg) gives moles of unknown
solute. Then, dividing the mass of
solute (in g) by the moles of solute, gives the molar mass of the unknown
solute.
= 2.5 ´ 10^{3}
mol solute
The molar mass of cocaine C_{17}H_{21}NO_{4}
= 303 g/mol, so the compound is not
cocaine. We assume in our analysis
that the compound is a pure, monomeric, nonelectrolyte.


13.103

The pill is in a
hypotonic solution. Consequently, by
osmosis, water moves across the semipermeable membrane into the pill. The increase in pressure pushes the
elastic membrane to the right, causing the drug to exit through the small
holes at a constant rate.


13.104

The molality
of the solution assuming AlCl_{3} to be a nonelectrolyte is:
The molality
calculated with Equation 13.7 of the text is:
The ratio is 4. Thus each AlCl_{3} dissociates as
follows:
AlCl_{3}(s) ® Al^{3+}(aq) + 3Cl^{}(aq)


13.105

Reverse osmosis involves no phase
changes and is usually cheaper than distillation or freezing.
To reverse
the osmotic migration of water across a semipermeable membrane, an external
pressure exceeding the osmotic pressure must be applied. To find the osmotic pressure of 0.70 M
NaCl solution, we must use the van’t Hoff factor because NaCl is a strong
electrolyte and the total ion concentration becomes 2(0.70 M) =
1.4 M.
The osmotic
pressure of sea water is:
p = iMRT =
2(0.70 mol/L)(0.0821 L×atm/mol×K)(298
K) = 34 atm
To cause
reverse osmosis a pressure in excess of 34 atm must be applied.


13.106

First, we
tabulate the concentration of all of the ions. Notice that the chloride concentration
comes from more than one source.
MgCl_{2}: If [MgCl_{2}] =
0.054 M, [Mg^{2+}]
= 0.054 M [Cl^{}]
= 2 ´
0.054 M
Na_{2}SO_{4}: if [Na_{2}SO_{4}]
= 0.051 M, [Na^{+}]
= 2 ´
0.051 M [SO] = 0.051 M
CaCl_{2}: if [CaCl_{2}] =
0.010 M, [Ca^{2+}]
= 0.010 M [Cl^{}]
= 2 ´
0.010 M
NaHCO_{3}: if [NaHCO_{3}] = 0.0020
M [Na^{+}]
= 0.0020 M [HCO] = 0.0020 M
KCl: if [KCl] = 0.0090 M [K^{+}] =
0.0090 M [Cl^{}]
= 0.0090 M
The subtotal
of chloride ion concentration is:
[Cl^{}] =
(2 ´
0.0540) + (2 ´
0.010) + (0.0090) = 0.137 M
Since the
required [Cl^{}] is 2.60 M, the
difference (2.6 
0.137 = 2.46 M) must come from NaCl.
The subtotal
of sodium ion concentration is:
[Na^{+}] =
(2 ´
0.051) + (0.0020) = 0.104 M
Since the
required [Na^{+}] is 2.56 M, the difference (2.56 
0.104 = 2.46 M) must come from NaCl.
Now,
calculating the mass of the compounds required:
NaCl:
MgCl_{2}:
Na_{2}SO_{4}:
CaCl_{2}:
KCl:
NaHCO_{3}:


13.107

Using Equation 13.8 of the text, we find the molarity of the
solution.
M = 0.295 M
This is the combined concentrations of all the ions. For dilute solutions, M »
m. Therefore, the freezing
point of blood can be calculated using Equation 13.7.
DT_{f} =
K_{f} m = (1.86°C/m)(0.295
m) = 0.55°C
Thus, the freezing point of blood should be about 0.55°C.


13.108

In order to increase the concentration in the intracellular volume
by 8.0 ´
10^{3}
M, the antibiotic must transport 1.6 ´ 10^{15}
mol Na^{+} ions.
time =19 s


13.109

a.

Using Equation 13.8 of the text, we find the molarity of the
solution.
This is
the combined concentrations of all the ions. The amount dissolved in 10.0 mL
(0.01000 L) is
Since
the mass of this amount of protein is 0.225 g, the apparent molar mass is

b.

We need to use a van’t Hoff
factor to take into account the fact that the protein is a strong
electrolyte. The van’t Hoff factor
will be i = 21 (why?).
This is
the actual concentration of the protein.
The amount in 10.0 mL (0.0100 L) is
Therefore
the actual molar mass is:



13.110

Solution A: Let molar
mass be M.
(760  754.5) = C_{A}(760)
C_{A} =
7.237 ´
10^{3}
M
= 124 g/mol
Solution
B: Let molar mass be M
C_{B} = 7.237 ´ 10^{3}
M
= 248 g/mol
The molar mass in benzene is about
twice that in water. This suggests
some sort of dimerization is occurring in a nonpolar solvent such as
benzene.


13.111

2H_{2}O_{2} ® 2H_{2}O + O_{2}
a.

Using the ideal gas law:

b.

The ratio of the volumes:

Think About It:

Could we have made the calculation in part (a) simpler
if we used the fact that 1 mole of all ideal gases at STP occupies a
volume of 22.4 L?






13.112

As the chain becomes longer, the
alcohols become more like hydrocarbons (nonpolar) in their properties. The alcohol with five carbons (npentanol)
would be the best solvent for iodine (a) and npentane (c)
(why?). Methanol (CH_{3}OH)
is the most waterlike and is the best solvent for an ionic solid like KBr.


13.113

a.

At reduced
pressure, the solution is supersaturated with CO_{2}.

b.

As the escaping CO_{2} expands
it cools, condensing water vapor in the air to form fog.



13.114

I_{2} 
H_{2}O: Dipole  induced
dipole.
I  H_{2}O: Ion  dipole. Stronger interaction causes more I_{2}
to be converted to I.


13.115

a.

Runoff of the
salt solution into the soil increases the salinity of the soil. If the soil becomes hypertonic relative
to the tree cells, osmosis would reverse, and the tree would lose water
to the soil and eventually die of dehydration.

b.

Assuming the
collecting duct acts as a semipermeable membrane, water would flow from
the urine into the hypertonic fluid, thus returning water to the body.



13.116

20.3 mL O_{2}
20.3 mL O_{2} ´ 0.95 = 19 mL O_{2}
in 100 mL blood in the lungs
20.3 mL O_{2} ´ 0.74 = 15 mL O_{2}
in 100 mL blood in the capillaries
The volume of
O_{2} released by hemoglobin when 100 mL of blood flows from the
lungs to the capillaries is 19  15 = 4 mL.


13.117

At
equilibrium, the concentrations in the 2 beakers are equal. Let x L be the change in volume.
The final
volumes are:
(50  16.7) mL = 33
mL
(50 + 16.7) mL = 67 mL


13.118

a.

If the membrane is permeable to all the ions and to the water,
the result will be the same as just removing the membrane. Eventually, the concentrations will
become uniform throughout the sample.

b.

This part is tricky. The
movement of one ion but not the other would result in one side of the
apparatus acquiring a positive electric charge and the other side
becoming equally negative. This
has never been known to happen, so we must conclude that migrating ions
always drag other ions of the opposite charge with them. In this hypothetical situation only
water would move through the membrane from the dilute to the more
concentrated side.

c.

This is the classic osmosis situation. Water would move through the membrane
from the dilute to the concentrated side.



13.119

First, we calculate the number of moles of HCl in 100 g
of solution.
Next, we
calculate the volume of 100 g of solution.
Finally, the
molarity of the solution is:


13.120

a.

Seawater has a larger number of ionic compounds dissolved in it;
thus the boiling point is elevated.

b.

Carbon dioxide escapes from an opened soft drink bottle because
gases are less soluble in liquids at lower pressure (Henry’s law).

c.

As you proved in Problem 13.20, at dilute concentrations molality
and molarity are almost the same because the density of the solution is
almost equal to that of the pure solvent.

d.

For colligative properties we are concerned with the number of
solute particles in solution relative to the number of solvent
particles. Since in colligative
particle measurements we frequently are dealing with changes in
temperature (and since density varies with temperature), we need a
concentration unit that is temperature invariant. We use units of moles per kilogram of
mass (molality) rather than moles per liter of solution (molarity).

e.

Methanol is very water soluble (why?) and effectively lowers the
freezing point of water. However
in the summer, the temperatures are sufficiently high so that most of the
methanol would be lost to vaporization.



13.121

Let the mass
of NaCl be x g. Then, the
mass of sucrose is (10.2  x)g.
We know that
the equation representing the osmotic pressure is:
p = MRT
p,
R, and T are given.
Using this equation and the definition of molarity, we can calculate
the percentage of NaCl in the mixture.
Remember that
NaCl dissociates into two ions in solution; therefore, we multiply the
moles of NaCl by two.
mol solute = 0.03422x + 0.02980 
0.002921x
mol solute = 0.03130x + 0.02980
Substitute
molarity into the equation for osmotic pressure to solve for x.
p = MRT
0.0753 = 0.03130x + 0.02980
x = 1.45 g
= mass of NaCl


13.122

DT_{f} =
5.5 
2.2 = 3.3°C C_{10}H_{8}:
128.2 g/mol
C_{6}H_{12}:
84.16 g/mol
Let x
= mass of C_{6}H_{12} (in grams).
Using,
x =
0.47 g


13.123

a.

Solubility
decreases with increasing lattice energy.

b.

Ionic compounds
are more soluble in a polar solvent.

c.

Solubility
increases with enthalpy of hydration of the cation and anion.



13.124

The completed
table is shown below:
Attractive
Forces Deviation
from Raoult’s DH_{solution}
A
«
A, B «
B >
A «
B Positive Positive (endothermic)
A «
A, B « B < A « B Negative Negative (exothermic)
A «
A, B « B = A « B Zero Zero
The first row
represents a Case 1 situation in which A’s attract A’s and B’s attract B’s
more strongly than A’s attract B’s.
This results in positive deviation from Raoult’s law (higher vapor
pressure than calculated) and positive heat of solution (endothermic).
In the second
row a negative deviation from Raoult’s law (lower than calculated vapor
pressure) means A’s attract B’s better than A’s attract A’s and B’s attract
B’s. This causes a negative
(exothermic) heat of solution.
In the third
row a zero heat of solution means that AA, BB,
and AB
interparticle attractions are all the same.
This corresponds to an ideal solution which obeys Raoult’s law
exactly.
Think About It:

What sorts
of substances form ideal solutions with each other?



13.125

Let's assume
we have 100 g of solution. The 100 g
of solution will contain 70.0 g of HNO_{3} and 30.0 g of H_{2}O.
To calculate
the density, let's again assume we have 100 g of solution. Since,
we know the
mass (100 g) and therefore need to calculate the volume of the
solution. We know from the molarity
that 15.9 mol of HNO_{3} are dissolved in a solution volume of 1000
mL. In 100 g of solution, there are
1.11 moles HNO_{3} (calculated above). What volume will 1.11 moles of HNO_{3}
occupy?
Dividing the
mass by the volume gives the density.


13.126

_{}
P_{ethanol} =
(0.62)(108 mmHg) = 67.0 mmHg
P_{1propanol} =
(0.38)(40.0 mmHg) = 15.2 mmHg
In the vapor
phase:


13.127

NH_{3} can form hydrogen bonds
with water; NCl_{3} cannot.
(Like dissolves like.)


13.128

Since the total volume is less than the
sum of the two volumes, the ethanol and water must have an intermolecular
attraction that results in an overall smaller volume.


13.129

We can
calculate the molality of the solution from the freezing point depression.
DT_{f} = K_{f}m
0.203
= 1.86 m
The molality
of the original solution was 0.106 m. Some of the solution has ionized to H^{+}
and CH_{3}COO^{}.
CH_{3}COOH ⇄ CH_{3}COO^{} +
H^{+}
Initial 0.106 m 0 0
Equil. 0.106 m 
x x x
At
equilibrium, the total concentration of species in solution is 0.109 m.
(0.106  x) + 2x =
0.109 m
x
= 0.003 m
The percentage of acid that has undergone ionization is:


13.130

Strategy:

Consider the number of polar groups, and determine
whether each molecule is predominantly polar or predominantly
nonpolar. Predominantly polar
molecules tend to be water soluble.
Predominantly nonpolar molecules tend to be fat soluble.

Setup:

Polar groups include those containing O atoms or N
atoms (atoms with lone pairs).
Pantothenic acid contains 7 polar groups (five O atoms and two N
atoms) making it predominantly polar.
Nicotinic acid is a small molecule containing 3 polar groups (two
O atoms and one N atom) making it predominately polar. Vitamin K is a large molecule that
contains only two polar groups and is therefore predominately nonpolar.

Solution:

Pantothenic
acid: water soluble
Nicotinic
acid: water soluble
Vitamin K: fat soluble



13.131

Egg yolk contains lecithins which
solubilize oil in water (See Figure 13.18 of the text). The nonpolar oil becomes soluble in water
because the nonpolar tails of lecithin dissolve in the oil, and the polar
heads of the lecithin molecules dissolve in polar water (like dissolves
like).


13.132

First, we can
calculate the molality of the solution from the freezing point depression.
DT_{f} =
(5.12)m
(5.5  3.5) =
(5.12)m
m =
0.39
Next, from the definition of molality, we can calculate
the moles of solute.
mol solute =
0.031 mol
The molar
mass (M) of the solute is:
The molar
mass of CH_{3}COOH is 60.05 g/mol.
Since the molar mass of the solute calculated from the freezing
point depression is twice this value, the structure of the solute most
likely is a dimer that is held together by hydrogen bonds.


13.133

192 mg =
192 ´
10^{6}
g or
1.92 ´
10^{4}
g
Safety limit: 0.050 ppm implies a mass of 0.050 g Pb
per 1 ´
10^{6} g of water. 1 liter
of water has a mass of 1000 g.
Yes, the concentration of lead calculated above
(7.4 ´
10^{5}
g/L) exceeds the safety limit of 5.0 ´ 10^{5}
g/L. (Don’t drink the water!)


13.134

a.

DT_{f} =
K_{f}m
2 =
(1.86)(m)
molality =
1.1 m
This concentration
is too high and is not a reasonable physiological concentration.

b.

As soon as a microscopic ice crystal begins to form, the protein
binds to it, blocking the addition of other water molecules to the
crystal and thus halting its growth.



13.135

As the water
freezes, dissolved minerals in the water precipitate from solution. The minerals refract light and create an
opaque appearance.


13.136

If the can is
tapped with a metal object, the vibration releases the bubbles and they
move to the top of the can where they join up to form bigger bubbles or mix
with the gas at the top of the can.
When the can is opened, the gas escapes without dragging the liquid
out of the can with it. If the can
is not tapped, the bubbles expand when the pressure is released and push
the liquid out ahead of them.


13.137

At equilibrium, the vapor pressure of benzene over each beaker
must be the same. Assuming ideal
solutions, this means that the mole fraction of benzene in each beaker must
be identical at equilibrium.
Consequently, the mole fraction of solute is also the same in each
beaker, even though the solutes are different in the two solutions. Assuming the solute to be nonvolatile,
equilibrium is reached by the transfer of benzene, via the vapor phase,
from beaker A to beaker B.
The mole fraction of naphthalene in beaker A at equilibrium
can be determined from the data given.
The number of moles of naphthalene is given, and the moles of
benzene can be calculated using its molar mass and knowing that 100 g 
7.0 g = 93.0 g of benzene remain in the beaker.
Now, let the number of moles of unknown compound be n. Assuming all the benzene lost from beaker
A is transferred to beaker B, there are 100 g + 7.0 g = 107 g of benzene in
the beaker. Also, recall that the
mole fraction of solute in beaker B is equal to that in beaker A at
equilibrium (0.112). The mole
fraction of the unknown compound is:
n =
0.1728 mol
There are 31 grams of the unknown compound dissolved in
benzene. The molar mass of the
unknown is:
Temperature is assumed constant and ideal behavior is also
assumed. Both solutes are assumed to
be nonvolatile.


13.138

a.

or
(1)
If we assume 1 L of
solution, then we can calculate the mass of solution from its density and
volume (1000 mL), and the mass of solute from the molarity and its molar
mass.
Substituting these expressions into Equation (1)
above gives:
or
(2)
From the definition of molality (m), we
know that:
(3)
Assuming 1 L of solution, we also know that mol
solute (n) = Molarity (M), so Equation (3) becomes:
Substituting back into
Equation (2) gives:
Taking the inverse of
both sides of the equation gives:
or

b.

The density
of a dilute aqueous solution is approximately 1 g/mL, because the density
of water is approximately 1 g/mL.
In dilute solutions, .
Consider a 0.010 M NaCl solution.
With , the derived equation reduces to:
Because d
»
1 g/mL. m » M.



13.139

To solve for the molality of the solution, we need the
moles of solute (urea) and the kilograms of solvent (water). If we assume that we have 1 mole of
water, we know the mass of water.
Using the change in vapor pressure, we can solve for the mole
fraction of urea and then the moles of urea.
Using Equation 13.5 of the text, we solve for the
mole fraction of urea.
DP =
23.76 mmHg  22.98 mmHg =
0.78 mmHg
Assuming that we have 1 mole of water, we can now
solve for moles of urea.
0.033n_{urea}
+ 0.033 = n_{urea}
0.033 =
0.967n_{urea}
n_{urea} =
0.034 mol
1 mole of water has a mass of 18.02 g or 0.01802 kg. We now know the moles of solute (urea)
and the kilograms of solvent (water), so we can solve for the molality of
the solution.


13.140

The desired
process is for (fresh) water to move from a more concentrated solution
(seawater) to pure solvent. This is
an example of reverse osmosis, and external pressure must be provided to
overcome the osmotic pressure of the seawater. The source of the pressure here is the
water pressure, which increases with increasing depth. The osmotic pressure of the seawater is:
p
= MRT
p
= (0.70 M)(0.0821 L×atm/mol×K)(293
K)
p
= 16.8 atm
The water pressure at the membrane depends on the height of
the sea above it, i.e. the depth.
P = hdg where h, d, and g are the
height of the sea above the membrane, the density of the solution, and the
gravitational constant, respectively;
and fresh water will begin to pass through the membrane when P
= p. Substituting p = P into the equation gives:
p
= hdg
and
Before substituting into the equation to solve for h,
we need to convert atm to pascals, and the density to units of kg/m^{3}. These conversions will
give a height in units of meters.
1
Pa = 1 N/m^{2} and 1 N = 1 kg×m/s^{2}. Therefore, we can write 1.70 × 10^{6}
Pa as 1.70 × 10^{6} kg/m×s^{2}
h =168 m


13.141

The total
vapor pressure depends on the vapor pressures of A and B in the mixture,
which in turn depends on the vapor pressures of pure A and B. With the total vapor pressure of the two
mixtures known, a pair of simultaneous equations can be written in terms of
the vapor pressures of pure A and B.
We carry 2 extra significant figures throughout this calculation to
avoid rounding errors.
For the
solution containing 1.2 moles of A and 2.3 moles of B,
C_{B} =
1  0.3429 = 0.6571
Substituting in P_{total} and
the mole fractions calculated gives:
Solving for ,
(1)
Now, consider the solution with the
additional mole of B.
C_{B} =
1  0.2667 = 0.7333
Substituting in P_{total} and
the mole fractions calculated gives:
(2)
Substituting Equation (1) into Equation (2)
gives:
Substitute the value of into Equation (1) to solve for .


13.142

Starting with n
= kP and substituting into the ideal gas equation (PV = nRT),
we find:
PV = (kP)RT
V = kRT
This equation
shows that the volume of a gas that dissolves in a given amount of solvent
is dependent on the temperature, not the pressure of the gas.


13.143

To calculate the freezing point of the solution, we need the
solution molality and the freezingpoint depression constant for water (see
Table 13.2 of the text). We can first
calculate the molarity of the solution using Equation 13.8 of the text: p =
MRT. The solution molality
can then be determined from the molarity.
Let’s assume that we have 1 L (1000 mL) of solution. The mass of 1000 mL of solution is:
The mass of the solvent (H_{2}O) is:
mass
H_{2}O = mass soln  mass solute
The molality of the
solution is:
The freezing point
depression is:
DT_{f} = K_{f}m =
(1.86°C/m)(0.396 m) =
0.737°C
The solution will freeze at 0°C

0.737°C
= 0.737°C


13.144

Strategy:

Because the solutes are all ionic, each solute
concentration will have to be multiplied by the appropriate van’t Hoff
factor. (Because we are assuming
no ion pairing, we can use calculated van’t Hoff factors.) Sum the concentrations for all solutes
to determine the total concentration of dissolved particles, and use
Equation 13.8 to calculate osmotic pressure.

Setup:

Because the volume of the solution described is 1 L,
the number of moles is also the molarity for each solute. R
= 0.08206 L · atm/K · mol, T =
310 K, and the van’t Hoff factors for the solutes in Ringer’s lactate are
as follows:
NaCl(s) → Na^{+}(aq) + Cl^{–}(aq) i = 2
KCl(s) → K^{+}(aq) + Cl^{–}(aq) i =
2
CaCl_{2}(s) → Ca^{2+}(aq) + 2Cl^{–}(aq) i
= 3
NaCH_{3}CH_{2}COO(s) → Na^{+}(aq) + CH_{3}CH_{2}COO^{–}(aq) i = 2

Solution:

The total concentration of ions in solution is the sum
of the individual concentrations.
total
concentration = 2[NaCl] + 2[KCl] + 3[CaCl_{2}] + 2[NaCH_{3}CH_{2}COO]
= 2(0.102 M) + 2(4 × 10^{–3} M) + 3(1.5 × 10^{–3} M) + 2(2.8 × 10^{–2} M)
= 2.73 × 10^{–1}
M
π = MRT = (0.273 M)(0.08206 L · atm/K · mol)(310 K)
= 6.94 atm



13.145

Strategy:

From the figure, note
that the normal boiling point of the solution is elevated by about 1°C.
Use the boiling point elevation equation (Equation 13.6) with K_{b}
= 2.53 °C/m (Table 13.2).

Solution:

The solution concentration is about 0.4 molal.



13.146

The interior of the
molecule is lined with polar C=O and NH functional groups, and a K^{+}
ion readily “dissolves” into this polar environment. One the other hand,
the exterior of the molecule has a large number of nonpolar CH_{3}
groups (one at the end of each “Y”), making the entire molecule (and its K^{+}
“cargo”) soluble in the nonpolar cell membrane. As the molecule
incorporates a K^{+} ion into its interior, it deforms slightly,
presenting even more nonpolar groups to the exterior side, thus enhancing
the “loaded” molecule’s solubility in nonpolar environments.


13.147

Strategy:

Use Equation 13.2 and the definition of ppm to
calculate the percent F^{–} by mass.
Use the molar mass of NaF to determine the mass of NaF
in a gram of solution. Subtract
the mass of NaF from the mass of the solution to determine the mass of
the solvent (water). Use Equation
13.1 to determine the molality.

Setup:

The molar mass of NaF is 41.99 g/mol.
Consider the ionization of NaF:
According to this equation, if 1.0 mol NaF molecules
ionize, we get 1.0 mol Na ^{+} ion and
1.0 mol F^{–} ions.

Solution:

To calculate percent mass by F^{–},
1.0 ppm F^{–}
=
To calculate molality, first determine the number of
moles and mass of NaF per gram of solution.
Finally, calculate the molality.



13.148

a.

Strategy:

Use Equation 13.9 to calculate the van’t Hoff factor
of fluorosilicic acid.

Setup:

K_{f}
for water (Table 13.2) is 1.86°C/m.
The normal freezing point of water is 0°C. Therefore, ∆T_{f} is 0°C – (–15.5°C)
= 15.5°C.

Solution:

Solve Equation 13.9 for the van’t Hoff factor:


b.

Strategy:

Use Equation 13.11, p =
iMRT, to
calculate the molarity of the solution.
Then, determine the grams of acid in the solution using the
density. Multiply the grams of
acid by one over the molarity to determine the molar mass.

Setup:

The van’t Hoff factor (calculated in part a) is
5.22.
The density of 23% fluorosilicic acid is 1.19 g/mL.
Converting to g/L gives:
Since the solution is 23% fluorosilicic acid,
there are 23/100 or 0.23 g acid for every gram of solution.

Solution:

The molarity is given by:
The molar mass is:




13.149

a.

Strategy:

The mole fraction of a component is the number of
moles of the component divided by the total number of moles in the
mixture:

Setup:

The solution is prepared by mixing equal masses of A
and B. Assume a mass of 1.00 g
for each component.
The molar mass of A is 100 g/mol and the molar mass
of B is 110 g/mol.

Solution:



b.

Strategy:

According to Raoult’s Law, the vapor pressure of the
solution is the sum of the individual partial pressures exerted by the
solution components. Use
Equation 13.4 to calculate the partial pressures of A and B.

Setup:

At 55°C, = 98 mm Hg and = 42 mm Hg.
The mole fractions of A and B were calculated in
part (a): χ_{A}= 0.52 and χ_{B}
=0.48.

Solution:



c.

Strategy:

The total pressure is given by Dalton’s law of
partial pressures, P_{T}
= P_{A} + P_{B}. The mole fraction is then equal to
the partial pressure of a component divided by the total pressure
according to Equation 11.16:

Setup:

Use the partial pressures of A and B (calculated in
part (b)) to determine the total pressure, P_{T}.
P_{T} = 51 mm Hg + 20 mm
Hg = 71 mm Hg

Solution:



d.

Strategy:

Use Equation 13.4 to calculate the partial pressures
of A and B.

Setup:

The mole fraction of each component above the
condensed liquid was calculated in part (c): χ_{A}
= 0.72 and χ_{B}
= 0.28

Solution:




