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 REF: Page 442
8. 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 = cO2∙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 = csolvent∙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 = csolvent∙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 DTf = 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.
DTf = 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 i-factor.
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 non-electrolyte.
i = DTf(measured)/DTf(calculated)
We must first calculate the expected (calculated) temperature depression; for the situation where the substance is a non-electrolyte.
DTf = 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 = DTf(measured)/DTf(calculated) = 0.593/0.318 = 1.86
REF: Page 443
9. Osmotic pressure, P, 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 = P/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, P = i x cRT is the equation to use for osmotic pressure but the concentration is the total ion concentration.
P = 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 = c x Ptotal = 0.209 x (1.00 bar x (10 5pascals/bar)) = 20.9 kPa