Chapter 12: Solutions and Their Behaviour
A: MULTIPLE CHOICE QUESTIONS
1. Which one of the following is not true regarding a saturated solution?
a) At the given temperature, no more of the solute will dissolve
b) Evidence of saturation is the presence of undissolved solute in the solution
c) Further addition of solid solute will result in supersaturation
d) A decrease in temperature can lead to crystallization
ANS: c
REF: Page 430
2. With regard to the dissolution process and a soluble ionic solid, which one of the following is not true?
a) The lattice enthalpy is positive
b) Dissolution may be exothermic or endothermic
c) Enthalpy of aquation is negative
d) Salts composed of smaller ions with high charge density typically have lower lattice energies
ANS: d
REF: Page 432
3. Which one of the following is a solubility property of gases?
a) Solubility increases with increasing temperature
b) Solubility increases with increasing pressure
c) Solubility varies with the square of the partial pressure of the gas
d) Henry’s Law constants are the same for all gases at the same temperature
ANS: b
REF: Pages 432, to 435
4. Which one of the following has units of moles of solute per kilogram of solvent?
a) Molarity
b) Molality
c) Mole fraction
d) Mass percent
ANS: b
REF: Section 12.5
5. Vapour pressure of an ideal electrolyte solution differs from the vapour pressure of a real solution in
which one of the following ways?
a) If solvent-solute interactions are weaker than solvent-solvent, vapour pressure is higher than
calculated by Raoult’s Law
b) If solvent-solute interactions are weaker than solvent-solvent, vapour pressure is lower than
calculated by Raoult’s Law
c) If solvent-solute interactions are stronger than solvent-solvent interactions, vapour pressure is
higher than calculated by Raoult’s Law
d) Vapour pressure of the real solution is always higher than predicted by Raoult’s Law
ANS: a
REF: Section 12.6
6. When a solution containing dissolved food coloring is gradually frozen, which one of the following
best describes what is likely to occur?
a) The integrity of the solution will be unaffected and the concentration of the solute will remain
consistent throughout the solution as it solidifies together
b) The molecules of food coloring will always remain in solution as water solidifies as ice
c) The molecules of food coloring may precipitate or crystallize while the water solidifies separately
as ice
d) The molecules of food coloring will form small crystals throughout a network of solid ice that is
indistinguishable from a uniform solid.
ANS: c
REF: Page 440
7. Which one of the following 0.05 molal solutions would most significantly depress the freezing point
of water?
a) NaCl
b) Al2(NO3)3
c) Ca(NO3)2
d) Ca(OH)2
ANS: b
8.
REF: Page 442
A solution contains a Lewis acid that is a neutral complex. Anions in solution are attracted to its
surface through their unbonded electrons. The anionic shell around the molecules further attracts
cations in solution until a large positive shell surrounds each Lewis acid molecule. Repulsions
between cations on nearby complexes of similar construction prevent their aggregation and they
remain in solution. This best describes which one of the following?
a) A hydrophilic emulsion
b) A colloidal dispersion
c) A hydrophobic colloid
d) A hydrophilic colloid
ANS: c
REF: Section 12.7
9. Reverse osmosis utilizes which one of the following.
a) Difference in concentration to drive net water movement
b) Difference in temperature to drive net water movement
c) Ionic strength to drive net water movement
d) Physical pressure to drive net water movement
ANS: d
REF: Page447
10. A student was asked to prepare a 1.0 molal solution of sodium hydroxide. They weighed 4.0 grams
of sodium hydroxide (0.10 moles) and completely dissolved it in water to a final volume of 1.00 L
(density of water is 997 g/L) at 25 0C. This is not a 0.1 molal solution for which of the following
reasons?
a) It is a 0.10 M solution
b) The sodium hydroxide is an electrolyte
c) The dissolution reaction is exothermic and the rise in temperature effects the volume
d) Hydroxide is a strong base and removes protons from water
ANS: a
REF: Section 12.5
B: SHORT ANSWER
1. Potassium bromide and potassium iodide are dissolved in water. Of the two, which one would you
expect to require the most energy to separate into ions and, which one would release the most energy
upon aquation?
ANS:
The smaller ion pair would be the most difficult to separate. It would have a higher lattice
enthalpy. Of the two, potassium bromide fits this description. The potassium and bromide ions
are closer to each other. Coulomb’s Law predicts the force of attraction will be greater than for
potassium iodide. This attractive force must be overcome as the ions become aquated.
The smaller ions are also more strongly attracted to the water molecules upon hydration. The
potassium is a constant for both salts. Only the anion varies between the compounds. The
bromide should have a higher enthalpy of aquation than the iodide since it is smaller.
REF: Page 432
2. Henry’s constant for hydrogen is 8.5 x 10 -4 molL-1atm-1 at 20 0C in water. The partial pressure of
hydrogen gas in a fuel cell is 15 atmospheres. What is the hydrogen gas concentration in g/L in the
aqueous solution of the fuel cell?
ANS:
The equation for solubility, using Henry’s Law is; s = kh x p where kh is Henry’s constant and p is
the partial pressure of the gas above the solvent.
s = 8.5 x 10 -4 molL-1 atm-1 x 15 atm = 0.01275 g/L
REF: Page 4343
3. A patient is suffering from bronchial pneumonia. The attending physician administers a mix of
oxygen an helium called heliox. 91.9% by mass is oxygen, the rest is helium. The gas enters the
lungs at atmospheric pressure of 101.3 kPa and room temperature. What is the partial pressure oxygen
in the mix?
ANS:
The pressure contributed by the oxygen is determined by the mole fraction and the overall
pressure according to the equation; PO2 = O2∙Ptotal
The mole fraction is; moles of O2/total moles
We will assume 100 g of gas. If the composition is 91.9% by mass O2 then 91.9 grams of O2 are
in the 100 g.
91.9 g/ (32.00 g/mole) = 2.87 moles of O2
The helium is the remaining mass = (100 – 91.9)g = 8.1 g
8.1 g/ (4.00 g/mole) = 2.025 moles
Mole fraction O2 = (2.87 moles) /(2.87 + 2.025) = 0.586
Therefore; partial pressure O2 = 0.586 x 101.3 kPa = 59.39 kPa = 59.4 kPa
REF: Section 12.5
4. Referring to question #3, if the Henry’s Law constant for helium is 3.7 x 10 -4 mol∙L-1∙atm-1 at 20 0C,
what will the solubility of helium be in the aqueous film found in the patients alveoli and, by
inference, in their bloodstream?
ANS:
s = kH x p
partial pressure helium is 1.00 – partial pressure O2 = 101.3 kPa – 59.4 kPa = 41.9 kPa
41.9 kPa x(1.00 atm/101.3 kPa) = 0.4136 atm
s = (3.7 x 10 -4 mol∙L-1∙atm-1 ) x 0.4136 atm = 1.53 x 10 -4 M
REF: Page 434
5. The helium solubility will actually be lower in the blood than expected by the prior calculation. With
respect to the pressure, temperature, solubility characteristics of gases, speculate as to the basis for
this.
ANS:
The temperature of the blood is higher than the air temperature. Solubility of gases at higher
temperatures decreases. This means less of the helium would diffuse into the water/blood than
predicted. The prediction is based upon a Henry’s constant for 20 0C.
REF: section 12.4
6. A mixture of water and methanol is prepared to use as antifreeze in the water system of a camper
trailer. 25.00 L of methanol are mixed with 35.00 L of water. The density of methanol is 791.3
kg/m3. The density of water at this temperature is 0.997 g/mL. What is the reduced vapour pressure
of the water if the normal vapour pressure at this temperature is17.54 torr? Assume that the methanol
is non-volatile. (It is not! The actual vapour pressure will be the sum of the vapour pressure of the
water and the vapour pressure of the methanol)
ANS:
Vapour pressure lowering can be calculated using Raoult’s Law, Preduced = solvent∙Ppure solvent
To arrive at the mole fraction solvent we need to calculate the mole fraction water in the
water/methanol mix.
Moles of methanol = (25.00 L) x (791.3 kg/m3) x (1000g/kg) x (1 m3/1000 L) x ( 1mole of
methanol/16.05 g methanol) = 1232 moles
Moles of water = (35.00 L) x (1000 mL/L) x (0.997 g/mL) x (1 mole/18.02 g) = 1936 moles
Mole fraction water = 1936/(1936 + 1232) = 0.61
Then;
Preduced = solvent∙Ppure solvent = 0.61 x 17.54 torr = 10.7 torr
REF: section 12.6
7. The freezing point depression constant, Kf, for ethylene dibromide (CH2BrCH2Br) is 11.8 K/molal. A
solution of water and ethylene dibromide is prepared through mixing of 15 grams of the hydrocarbon
in 10.00 L of water. By how many degrees will the freezing point of water be depressed? Assume
the water has a mass of 1.0 g/mL.
ANS:
Freezing point depression can be calculated for a dilute non-electrolyte solution using the
equation Tf = Kf x msolute where Kf is the freezing point depression constant of the solute and m
is the molality of that solute. Molality is expressed in moles of solute per kg of solvent.
Tf = Kf x msolute = 11.8 K/molal x( (15 g/187.76 g/mole) / 1.0 kg) = 0.94 K
REF: Pages; 440, 441
8.
A 1% by mass solution of sodium chloride in water has a freezing point that is depressed from the
normal freezing point of water by 0.593 0C. Kf for water is 1.86 K∙kg∙mol-1. Estimate the van’t Hoff ifactor.
ANS:
Sodium chloride is an electrolyte. As such, the freezing point depression uses the same equation
as for a non-electrolytes with one difference. The (Kf x m) expression is multiplied by a factor, i,
called the van’t Hoff factor. It has a predicted value equal to the number of moles of ions in
solution per mole of molecular units. In this case, NaCl produces 2 moles of ions per mole of
NaCl, and predicted-i has an expected value of 2.
The true value of the van’t Hoff factor is less than predicted. It’s actual value is the ratio
of the measured temperature depression to the expected temperature depression of a nonelectrolyte.
i = Tf(measured)/Tf(calculated)
We must first calculate the expected (calculated) temperature depression; for the situation
where the substance is a non-electrolyte.
Tf = Kf x m = (1.86 K∙kgwater∙molsolute-1) x [(0.01kg NaCl/ kg water) x (1mole
NaCl/0.05844 kg NaCl)] = 0.318 K
Then;
i = Tf(measured)/Tf(calculated) = 0.593/0.318 = 1.86
REF: Page 443
9. Osmotic pressure, , of a solution = cRT. What is the molar mass of an immunoglobulin if 0.65 g of
the Ig dissolved in 0.200 L of water results in an osmotic pressure of 310 pascals at 20 0C?
ANS:
We can calculate the molar mass from the concentration. Concentration has units of moles/L.
The mass of immunoglobulin in solution divided by moles in solution will give us the molar
mass.
c = /RT = 0.310 kPascals /(8.314 L∙kPa∙mol-1∙K-1)(293.15K) = 1.27 x 10 -4 moles/L
The volume used was 0.200L therefore; 0.200L x (1.27 x 10 -4 moles/L) =2.54 x 10 -5 moles of Ig.
Mm = m/n = 0.65g/2.54 x 10 -5 moles = 25,551 g/mole = 25,000 g/mole = 25 kdaltons
(one Dalton = 1 amu)
REF: Exercise 12.18
10. The osmotic pressure of a solution of electrolyte must consider the concentration of hydrated ions in
solution and not simply the concentration of the overall salt. For example; a 10.0 M solution of NaF
is 10.0 M in Na+ and 10.0M in F- for a total ion concentration of 20.0M. What is the estimated
osmotic pressure of a 0.035 M solution of K2SO4? Assume that the van’t Hoff factor for this salt is
1.89 and the temperature is 25 0C..
ANS:
In this case, since we have an electrolyte,  = i x cRT is the equation to use for osmotic pressure
but the concentration is the total ion concentration.
 = 1.89 x (3 x 0.035 / L) x (8.314 L∙kPa∙mol-1∙K-1) x 298.15K = 163.97 kPa = 160 kPa
REF: Page 447
C: LONG-ANSWER QUESTIONS
1. Henry’s Law constants for water at 30 0C and 5 0C are 4.0 x 10 -4 and 1.1 x 10 -3, mol∙L-1∙kPa-1,
respectively. Water, running through a hydroelectric dam is heated from as low as 40C , at the top, up
to temperatures approaching 30 0C, after running through the turbines. This deprives the water of
oxygen needed by fish downstream of the dam. Assume 1.00 bar air pressure and that oxygen in the
dry air is 20.9% by volume. How much more oxygen, by percent, is there in a litre of water at the
bottom of the lake behind the dam compared to the water leaving the turbines?
ANS:
Solubility can be calculated using the formula; s = kH x p where kH is Henry’s constant, p is the
pressure of the gas above the solvent. In this case p is the partial pressure of oxygen in the air.
As 20.9% by volume it is also 20.9% by moles because all ideal gases occupy the same volume
per mole.
Thus the mole fraction is 0.209 since in 1 mole of air 20.9% is oxygen by mole.
Then; using the equation PO2 = x Ptotal = 0.209 x (1.00 bar x (10 5pascals/bar)) = 20.9 kPa
Then;
For water at 40C; s = kH x p = (1.1 x 10 -3 mol∙L-1∙kPa-1) x (20.9 kPa) = 0.02299 moles/L
For water at 300C; s == kH x p = (4.0 x 10 -4 mol∙L-1∙kPa-1) x (20.9 kPa) = 0.00836 moles/L
By percent the reserve of oxygen in the colder water is;
(0.02299 – 0.00836)/0.02299 x 100 = 63.6%
Since oxygen absorption by the fish is through concentration dependent transfer across the gill
membranes, this can starve fish for oxygen and lead to fish kills.
REF: Section 12.4
2. A solution is prepared by dissolving 15.0 g of sucrose, C12H22O11, in 150 mL of water at 21 0C.
The molal freezing point depression constant of water is 1.86 K∙kg∙mol-1. The density of water is 997
g/L. Answer the following;
a) The freezing point depression of the above solution
b) The osmotic pressure of the solution
c) Would the freezing point increase or decrease if sucrose dissociated into a cation and an anion
(which it does not).
ANS:
a) For non-electrolytes Tf = Kf x m(solute)
Molality is moles of solute/kg of solvent = [15.0 g / (342.34 g/mole) ] / (0.150 L x 0.997
kg/L) = 0.293 moles/kg
Tf = Kf x m(solute) = (1.86 K∙kg∙mol-1) x (0.293 moles/kg) = 0.545 K = 0.55K
b) The osmotic pressure, equals cRT
= [(15.0 g)/(342.34 g/mole) / 0.150L ](8.314 L∙kPa∙mol-1∙K-1)(294.15K) = 714.36 kPa = 710
kPa
c) The freezing point would decrease because the effective solute concentration would be higher
than for a non electrolyte given that there would be two ions per molecule.
REF: Section 12.6
3. In chemistry it is common to hear the phrase “likes dissolve likes”. Compounds with similar chemical
and physical properties will often dissolve in each other. Dissimilar compounds are less soluble or
not at all soluble. For each of the following, predict whether it will be more soluble in ethanol
(CH3CH2OH), carbon tetrachloride, or benzene (C6H6) and explain your rationale.
a) Dibromoethane, CH2BrCH2Br
b) Methane gas, CH4
c) Ammonia, NH3
d) Water, H2O
ANS:
a) It is a polar molecule because of the carbon-bromine bonds. Although, free rotation around
the C-C bond can orient the bromines in diametrically opposed positions that nullify the
polarity. There will be a relatively low energy barrier to rotation around the C-C bond,
however that can permit the establishment of molecular dipoles in all other orientations
around that bond. This makes the solution, net polar. Of the three solvents, ethanol is the
most polar, carbon tetrachloride is non-polar because the dipoles cancel in this tetrahedral
arrangement, and benzene is non-polar. It is expected that the dibromoethane would be most
soluble in the ethanol.
b) Methane gas has a tetrahedral shape and is non-polar. The C-H bonds are considered nonpolar, and regardless, the tetrahedral arrangement would nullify any bond dipoles at the
molecular level. It is likely to be most soluble in the benzene. The carbon tetrachloride is
non-polar but it also has each carbon surrounded by four chlorine atoms, each of which has
three unbonded electron pairs. These electrons lend the CCl4 a tendency to behave as a Lewis
base. This promotes interaction with polar molecules more-so than benzene will do. Of the
two, benzene is most like the methane.
c) Ammonia has a trigonal pyramidal structure when describing bonded atoms. An unbonded
electron pair is in a fourth position. Altogether, the electron domains occupy a tetrahedral
arrangement. The unbonded pair is available for polar interactions. The N-H bonds are
polar. The solubility will be high in a polar solvent. Ethanol is the preferred choice.
d) Water, is obviously polar. It will dissolve best in the ethanol. In fact, it is soluble at any
concentration.
REF: Section 12.2
4. When lithium sulfate, Li2SO4, dissolves, the solution temperature rises . Answer the following;
a) Thermodynamically, how would you describe the enthalpy of solution of this salt?
b) Does the lattice enthalpy have a positive or a negative sign?
c) Does the enthalpy of aquation have a negative or a positive sign?
d) Which change is greater, the enthalpy of aquation or the lattice enthalpy?
ANS:
a) Since the temperature rises, energy is released during the process. The products have less
stored chemical energy than did the reactants. In this case the reactant is the solid and the
product is the solvated ions.
b) The lattice enthalpy will always have a positive sign. It requires energy to separate the ions.
Two lithium ions must be separated from each sulfate polyatomic ion.
c) The enthalpy of aquation is a type of bond formation so it is energy releasing. It is the sum of
the enthalpies of aquation for the lithium ion s and the sulfate ions.
d) Since the reaction is exothermic, energy is released. For energy to be released, the energy
change for aquation must be greater than the lattice energy change.
REF: Section 12.3
5.
If a glass of red wine is placed in the freezer of your fridge, it will be exposed to a temperature of
about -20 0C, assuming it is a standard refrigerator/freezer unit. The wine will not freeze solid.
Some of the water will solidify as ice until a point is reached where the % ethanol in the wine has
depressed the freezing point of the water to below -20 0C. The density of water is assumed to be
0.998 g/mL. Answer the following;
a) What is the molality of the ethanol (CH3CH2OH) in the remaining liquid solution? The freezing
point depression constant of water is 1.86 K∙kg∙mol-1.
b) What do you expect would happen if you continued to drop the temperature. At what point would
you expect the entire liquid to freeze solid.
c) Would any ice form in a fortified wine such as brandy at 19% ethanol vol/vol?
ANS:
a) Freezing point depression, Tf = Kf x msolute
Rearranged; msolute = Tf/Kf = 20 K/(1.86 K.kg.mol-1) = 10.75 moles ethanol/kg water
b) As the temperature dropped, more and more water would separate from the solution as solid
ice. As the percent alcohol continued to rise, the freezing point would continue to drop until
the freezing point of the ethanol was reached. At that point, virtually no water would remain
(in theory).
c) Ethanol has a density of 0.785 g/mL at 25 0C. A 19% vol/vol solution would have 19 mL of
ethanol for every 81 mL of water. 19 mL x 0.785 g/mL = 14.915 g of ethanol
At a molecular mass of 46.08 g/mole this is; 0.3237 moles
There are 0.3237 moles of ethanol in 100 mL of the wine. If we assume a density of 0.998
g/mL for the water then 1.0 kg of water is; 1.0 kg/(0.998 x 10 -3kg/mL) = 1002.00 ml
1002 mL x 0.3237 moles of ethanol/100 mL = 3.24 moles of ethanol/kg of water
This is less than the molality of the non-frozen liquid in part “a”. Therefore, this would
permit formation of some ice crystals.
REF: Section 12.6