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CHAPTER ELEVEN PROPERTIES OF SOLUTIONS For Review 1. Mass percent: the percent by mass of the solute in the solution. Mole fraction: the ratio of the number of moles of a given component to the total number of moles of solution. Molarity: the number of moles of solute per liter of solution. Molality: the number of moles of solute per kilogram of solvent. Volume is temperature dependent, whereas mass and the number of moles are not. Only molarity has a volume term so only molarity is temperature dependent. 2. KF(s) K + (aq) + F (aq) H = H soln ; K + (g) + Cl (g) → K + (aq) + F (aq) H = H hyd KF(s) K + (g) + F (g) H 1 = H LE K + (g) + F (g) K + (aq) + F (aq) H 2 = H hyd __________________________________________________________ KF(s) K + (aq) + F (aq) H = H soln = H LE + H hyd It is true that H 1 and H 2 have large magnitudes for their values; however, the signs are opposite (H 1 is large and positive because it is the reverse of the lattice energy and H 2 , the hydration energy, is large and negative). These two H values basically cancel out each other giving a H soln value close to zero. 372

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CHAPTER ELEVEN

PROPERTIES OF SOLUTIONS

For Review

1. Mass percent: the percent by mass of the solute in the solution.

Mole fraction: the ratio of the number of moles of a given component to the total number of moles of solution.

Molarity: the number of moles of solute per liter of solution.

Molality: the number of moles of solute per kilogram of solvent.

Volume is temperature dependent, whereas mass and the number of moles are not. Only molarity has a volume term so only molarity is temperature dependent.

2. KF(s) → K+(aq) + F(aq) H = Hsoln; K+(g) + Cl (g) → K+(aq) + F(aq) H = Hhyd

KF(s) → K+(g) + F(g) H1 = HLE

K+(g) + F(g) → K+(aq) + F(aq) H2 = Hhyd

__________________________________________________________ KF(s) → K+(aq) + F(aq) H = Hsoln = HLE + Hhyd

It is true that H1 and H2 have large magnitudes for their values; however, the signs are opposite (H1 is large and positive because it is the reverse of the lattice energy and H2, the hydration energy, is large and negative). These two H values basically cancel out each other giving a Hsoln value close to zero.

3. “Like dissolves like” refers to the nature of the intermolecular forces. Polar solutes and ionic solutes dissolve in polar solvents because the types of intermolecular forces present in solute and solvent are similar. When they dissolve, the strength of the intermolecular forces in solution are about the same as in pure solute and pure solvent. The same is true for nonpolar solutes in nonpolar solvents. The strength of the intermolecular forces (London dispersion forces) are about the same in solution as in pure solute and pure solvent. In all cases of like dissolves like, the magnitude of Hsoln is either a small positive number (endothermic) or a small negative number (exothermic). For polar solutes in nonpolar solvents and vice versa, Hsoln is a very large, unfavorable value (very endothermic). Because the energetics are so unfavorable, polar solutes do not dissolve in nonpolar solvents and vice versa.

4. Structure effects refer to solute and solvent having similar polarities in order for solution formation to occur. Hydrophobic solutes are mostly nonpolar substances that are “water-fearing.” Hydrophilic solutes are mostly polar or ionic substances that are “water-loving.”Pressure has little effect on the solubilities of solids or liquids; it does significantly affect the solubility of a gas. Henry’s law states that the amount of a gas dissolved in a solution is

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373 CHAPTER 11 PROPERTIES OF SOLUTIONS

directly proportional to the pressure of the gas above the solution (C = kP). The equation for Henry’s law works best for dilute solutions of gases that do not dissociate in or react with the solvent. HCl(g) does not follow Henry’s law because it dissociates into H+(aq) and Cl(aq) in solution (HCl is a strong acid). For O2 and N2, Henry’s law works well since these gases do not react with the water solvent.

An increase in temperature can either increase or decrease the solubility of a solid solute in water. It is true that a solute dissolves more rapidly with an increase in temperature, but the amount of solid solute that dissolves to form a saturated solution can either decrease or increase with temperature. The temperature effect is difficult to predict for solid solutes. However, the temperature effect for gas solutes is easier to predict as the solubility of a gas typically decreases with increasing temperature.

5. Raoult’s law: Psoln = ; When a solute is added to a solvent, the vapor pressure of a solution is lowered from that of the pure solvent. The quantity solvent, the mole fraction of solvent, is the fraction that the solution vapor pressure is lowered.

For the experiment illustrated in Fig. 11.9, the beaker of water will stop having a net transfer of water molecules out of the beaker when the equilibrium vapor pressure, , is reached. This can never happen. The beaker with the solution wants an equilibrium vapor pressure of

, which is less than . When the vapor pressure over the solution is above , a net transfer of water molecules into the solution will occur in order to try to reduce the vapor pressure to . The two beakers can never obtain the equilibrium vapor pressure they want. The net transfer of water molecules from the beaker of water to the beaker of solution stops after all of the water has evaporated.

If the solute is volatile, then we can get a transfer of both the solute and solvent back and forth between the beakers. A state can be reached in this experiment where both beakers have the same solute concentration and hence the same vapor pressure. When this state is reached, no net transfer of solute or water molecules occurs between the beakers so the levels of solution remain constant.

When both substances in a solution are volatile, then Raoult’s law applies to both. The total vapor pressure above the solution is the equilibrium vapor pressure of the solvent plus the equilibrium vapor pressure of the solute. Mathematically:

where A is either the solute or solvent and B is the other one.

6. An ideal liquid-liquid solution follows Raoult’s law:

A nonideal liquid-liquid solution does not follow Raoult’s law, either giving a total pressure greater than predicted by Raoult’s law (positive deviation) or less than predicted (negative deviation).In an ideal solution, the strength of the intermolecular forces in solution are equal to the strength of the intermolecular forces in pure solute and pure solvent. When this is true, Hsoln

= 0 and Tsoln = 0. For positive deviations from Raoult’s law, the solution has weaker

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CHAPTER 11 PROPERTIES OF SOLUTIONS 374

intermolecular forces in solution than in pure solute and pure solvent. Positive deviations have Hsoln > 0 (are endothermic) and Tsoln < 0. For negative deviations, the solution has stronger intermolecular forces in solution than in pure solute or pure solvent. Negative deviations have Hsoln < 0 (are exothermic) and Tsoln > 0. Examples of each type of solution are:

ideal: benzene-toluenepositive deviations: ethanol-hexanenegative deviations: acetone-water

7. Colligative properties are properties of a solution that depend only on the number, not the identity, of the solute particles. A solution of some concentration of glucose (C6H12O6) has the same colligative properties as a solution of sucrose (C12H22O11) having the same con-centration.

A substance freezes when the vapor pressure of the liquid and solid are identical to each other. Adding a solute to a substance lowers the vapor pressure of the liquid. A lower temperature is needed to reach the point where the vapor pressures of the solution and solid are identical. Hence, the freezing point is depressed when a solution forms.

A substance boils when the vapor pressure of the liquid equals the external pressure. Because a solute lowers the vapor pressure of the liquid, a higher temperature is needed to reach the point where the vapor pressure of the liquid equals the external pressure. Hence, the boiling point is elevated when a solution forms.

The equation to calculate the freezing point depression or boiling point elevation is:

T = Km

where K is the freezing point or boiling point constant for the solvent and m is the molality of the solute. Table 11.5 lists the K values for several solvents. The solvent which shows the largest change in freezing point for a certain concentration of solute is camphor; it has the largest Kf value. Water, with the smallest Kb value, will show the smallest increase in boiling point for a certain concentration of solute.

To calculate molar mass, you need to know the mass of the unknown solute, the mass and identity of solvent used, and the change in temperature (T) of the freezing or boiling point of the solution. Since the mass of unknown solute is known, one manipulates the freezing point data to determine the number of moles of solute present. Once the mass and moles of solute are known, one can determine the molar mass of the solute. To determine the moles of solute present, one determines the molality of the solution from the freezing point data and multiplies this by the kilograms of solvent present; this equals the moles of solute present.

8. Osmotic pressure: the pressure that must be applied to a solution to stop osmosis; osmosis is the flow of solvent into the solution through a semipermeable membrane. The equation to calculate osmotic pressure, , is:

= MRTwhere M is the molarity of the solution, R is the gas constant, and T is the Kelvin temperature. The molarity of a solution approximately equals the molality of the solution

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375 CHAPTER 11 PROPERTIES OF SOLUTIONS

when 1 kg solvent 1 L solution. This occurs for dilute solutions of water since = 1.00 g/cm3.

With addition of salt or sugar, the osmotic pressure inside the fruit cells (and bacteria) is less than outside the cell. Water will leave the cells which will dehydrate bacteria present, causing them to die.

Dialysis allows the transfer of solvent and small solute molecules through a membrane. To purify blood, the blood is passed through a cellophane tube (the semipermeable membrane); this cellophane tube is immersed in a dialyzing solution which contains the same concen-trations of ions and small molecules as in blood, but has none of the waste products normally removed by the kidney. As blood is passed through the dialysis machine, the unwanted waste products pass through the cellophane membrane, cleansing the blood.

Desalination is the removal of dissolved salts from an aqueous solution. Here, a solution is subjected to a pressure greater than the osmotic pressure and reverse osmosis occurs, i.e., water passes from the solution through the semipermeable membrane back into pure water. Desalination plants can turn sea-water with its high salt content into drinkable water.

9. A strong electrolyte completely dissociates into ions in solution, a weak electrolyte only partially dissociates into ions in solution, and a nonelectrolyte does not dissociate into ions when dissolved in solution. Colligative properties depend on the total number of solute particles in solution. By measuring a property such as freezing point depression, boiling point elevation, or osmotic pressure, we can determine the number of solute particles present from the solute and thus characterize the solute as a strong, weak, or nonelectrolyte.

The van’t Hoff factor, i, is the number of moles of particles (ions) produced for every mol of solute dissolved. For NaCl, i = 2 since Na+ and Cl are produced in water; for Al(NO3)3, i = 4 since Al3+ and 3 NO3

ions are produced when Al(NO3)3 dissolves in water. In real life, the van’t Hoff factor is rarely the value predicted by the number of ions a salt dissolves into; i is generally something less than the predicted number of ions. This is due to a phenomenon called ion pairing where at any instant a small percentage of oppositely charged ions pair up and act like a single solute particle. Ion pairing occurs most when the concentration of ions is large. Therefore, dilute solutions behave most ideally; here i is close to that determined by the number of ions in a salt.

10. A colloidal dispersion is a suspension of particles in a dispersing medium. See Table 11.7 for some examples of different types of colloids.

Both solutions and colloids have suspended particles in some medium. The major difference between the two is the size of the particles. A colloid is a suspension of relatively large particles as compared to a solution. Because of this, colloids will scatter light while solutions will not. The scattering of light by a colloidal suspension is called the Tyndall effect.

Coagulation is the destruction of a colloid by the aggregation of many suspended particles to form a large particle that settles out of solution.

Solution Review

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CHAPTER 11 PROPERTIES OF SOLUTIONS 376

9. = 9.74 M

10. 1.28 g CaCl2 × = 19.9 mL

11. mol Na2CO3 = 0.0700 L × = 0.21 mol Na2CO3

Na2CO3(s) → 2 Na+(aq) + CO32(aq); mol Na+ = 2(0.21) = 0.42 mol

mol NaHCO3 = 0.0300 L × = 0.030 mol NaHCO3

NaHCO3(s) → Na+(aq) + HCO3(aq); mol Na+ = 0.030 mol

= 4.5 M Na+

12. a. HNO3(l) → H+(aq) + NO3(aq) b. Na2SO4(s) → 2 Na+(aq) + SO4

2(aq)

c. Al(NO3)3(s) → Al3+(aq) + 3 NO3(aq) d. SrBr2(s) → Sr2+(aq) + 2 Br(aq)

e. KClO4(s) → K+(aq) + ClO4(aq) f. NH4Br(s) → NH4

+(aq) + Br(aq)

g. NH4NO3(s) → NH4+(aq) + NO3

(aq) h. CuSO4(s) → Cu2+(aq) + SO42(aq)

i. NaOH(s) → Na+(aq) + OH(aq)

Questions

13. As the temperature increases, the gas molecules will have a greater average kinetic energy. A greater fraction of the gas molecules in solution will have kinetic energy greater than the attractive forces between the gas molecules and the solvent molecules. More gas molecules will escape to the vapor phase and the solubility of the gas will decrease.

14. Henry’s law is obeyed most accurately for dilute solutions of gases that do not dissociate in or react with the solvent. NH3 is a weak base and reacts with water by the following reaction:

NH3(aq) + H2O(l) → NH4+(aq) + OH(aq)

O2 will bind to hemoglobin in the blood. Due to these reactions in the solvent, NH3(g) in water and O2(g) in blood do not follow Henry’s law.

15 Because the solute is volatile, both the water and solute will transfer back and forth between the two beakers. The volume in each beaker will become constant when the concentrations of solute in the beakers are equal to each other. Because the solute is less volatile than water,

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377 CHAPTER 11 PROPERTIES OF SOLUTIONS

one would expect there to be a larger net transfer of water molecules into the right beaker than the net transfer of solute molecules into the left beaker. This results in a larger solution volume in the right beaker when equilibrium is reached, i.e., when the solute concentration is identical in each beaker.

16. Solutions of A and B have vapor pressures less than ideal (see Figure 11.13 of the text), so this plot shows negative deviations from Rault’s law. Negative deviations occur when the intermolecular forces are stronger in solution than in pure solvent and solute. This results in an exothermic enthalpy of solution. The only statement that is false is e. A substance boils when the vapor pressure equals the external pressure. Since B = 0.6 has a lower vapor pressure at the temperature of the plot than either pure A or pure B, then one would expect this solution to require the highest temperature in order for the vapor pressure to reach the external pressure. Therefore, the solution with B = 0.6 will have a higher boiling point than either pure A or pure B. (Note that because P°B > P°A, B is more volatile than A, and B will have a lower boiling point temperature than A).

17. No, the solution is not ideal. For an ideal solution, the strength of intermolecular forces in solution is the same as in pure solute and pure solvent. This results in ΔHsoln = 0 for an ideal solution. ΔHsoln for methanol/water is not zero. Because ΔHsoln < 0 (heat is released), this solution shows a negative deviation from Raoult’s law.

18. The micelles form so the ionic ends of the detergent molecules, the SO4 ends, are exposed to the polar water molecules on the outside, while the nonpolar hydrocarbon chains from the detergent molecules are hidden from the water by pointing toward the inside of the micelle. Dirt, which is basically nonpolar, is stabilized in the nonpolar interior of the micelle and is washed away.

19. Normality is the number of equivalents per liter of solution. For an acid or a base, an equivalent is the mass of acid or base that can furnish 1 mol of protons (if an acid) or accept 1 mol of protons (if a base). A proton is an H+ ion. Molarity is defined as the moles of solute per liter of solution. When the number of equivalents equals the number of moles of solute, then normality = molarity. This is true for acids which only have one acidic proton in them

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CHAPTER 11 PROPERTIES OF SOLUTIONS 378

and for bases that accept only one proton per formula unit. Examples of acids where equivalents = moles solute are HCl, HNO3, HF, and HC2H3O2. Examples of bases where equivalents = moles solute are NaOH, KOH, and NH3. When equivalents moles solute, then normality molarity. This is true for acids that donate more than one proton (H2SO4, H3PO4, H2CO3, etc.) and for bases that react with more than one proton per formula unit [Ca(OH)2, Ba(OH)2, Sr(OH)2, etc.].

20. It is true that the sodium chloride lattice must be broken in order to dissolve in water, but a lot of energy is released when the water molecules hydrate the Na+ and Cl ions. These two processes have relatively large values for the amount of energy associated with them, but they are opposite in sign. The end result is they basically cancel each other out resulting in a Hsoln

0. So energy is not the reason why ionic solids like NaCl are so soluble in water. The answer lies in nature’s tendency toward the higher probability of the mixed state. Processes, in general, are favored that result in an increase in disorder because the disordered state is the easiest (most probable) state to achieve. The tendency of processes to increase disorder will be discussed in Chapter 16 when entropy, S, is introduced.

21. Only statement b is true. A substance freezes when the vapor pressure of the liquid and solid are the same. When a solute is added to water, the vapor pressure of the solution at 0 C is less than the vapor pressure of the solid and the net result is for any ice present to convert to liquid in order to try to equalize the vapor pressures (which never can occur at 0C). A lower temperature is needed to equalize the vapor pressure of water and ice, hence the freezing point is depressed.

For statement a, the vapor pressure of a solution is directly related to the mole fraction of solvent (not solute) by Raoult’s law. For statement c, colligative properties depend on the number of solute particles present and not on the identity of the solute. For statement d, the boiling point of water is increased because the sugar solute decreases the vapor pressure of the water; a higher temperature is required for the vapor pressure of the solution to equal the external pressure so boiling can occur.

22. This is true if the solute will dissolve in camphor. Camphor has the largest Kb and Kf

constants. This means that camphor shows the largest change in boiling point and melting point as a solute is added. The larger the change in T, the more precise the measurement and the more precise the calculated molar mass. However, if the solute won’t dissolve in camphor, then camphor is no good and another solvent must be chosen which will dissolve the solute.

23. Isotonic solutions are those which have identical osmotic pressures. Crenation and hemolysis refer to phenomena that occur when red blood cells are bathed in solutions having a mismatch in osmotic pressures inside and outside the cell. When red blood cells are in a solution having a higher osmotic pressure than that of the cells, the cells shrivel as there is a net transfer of water out of the cells. This is called crenation. Hemolysis occurs when the red blood cells are bathed in a solution having lower osmotic pressure than that inside the cell. Here, the cells rupture as there is a net transfer of water to into the red blood cells.

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379 CHAPTER 11 PROPERTIES OF SOLUTIONS

24. Ion pairing is a phenomenon that occurs in solution when oppositely charged ions aggregate and behave as a single particle. For example, when NaCl is dissolved in water, one would expect sodium chloride to exist as separate hydrated Na+ ions and Cl ions. A few ions, however, stay together as NaCl and behave as just one particle. Ion pairing increases in a solution as the ion concentration increases (as the molality increases).

Exercises

Concentration of Solutions

25. Because the density of water is 1.00 g/mL, 100.0 mL of water has a mass of 100. g.

density = = 1.06 g/mL = 1.06 g/cm3

mol H3PO4 = 10.0 g × = 0.102 mol H3PO4

mol H2O = 100. g × = 5.55 mol H2O

mole fraction of H3PO4 = = 0.0180

= 1.000 – 0.0180 = 0.9820

molarity = = 0.981 mol/L

molality = = 1.02 mol/kg

26. molality = = 10.7 mol/kg

where EG = ethylene glycol (C2H6O2)

molarity = = 6.77 mol/L

40.0 g EG × = 0.644 mol EG; 60.0 g H2O × = 3.33 mol H2O

= 0.162 = mole fraction ethylene glycol

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CHAPTER 11 PROPERTIES OF SOLUTIONS 380

27. Hydrochloric acid:

molarity = = 12 mol/L

molality = = 17 mol/kg

38 g HCl × = 1.0 mol HCl; 62 g H2O × = 3.4 mol H2O

mole fraction of HCl = = 0.23

Nitric acid:

= 16 mol/L

= 37 mol/kg

70. g HNO3 × = 1.1 mol HNO3; 30. g H2O × = 1.7 mol H2O

= 0.39

Sulfuric acid:

= 18 mol/L

= 194 mol/kg ≈ 200 mol/kg

95 g H2SO4 × = 0.97 mol H2SO4; 5 g H2O × = 0.3 mol H2O

= 0.76

Acetic Acid:

= 17 mol/L

= 1600 mol/kg ≈ 2000 mol/kg

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381 CHAPTER 11 PROPERTIES OF SOLUTIONS

99 g HC2H3O2 × = 1.6 mol HC2H3O2; 1 g H2O × = 0.06 mol H2O

= 0.96

Ammonia:

= 15 mol/L

= 23 mol/kg

28 g NH3 × = 1.6 mol NH3; 72 g H2O × = 4.0 mol H2O

= 0.29

28. a. If we use 100. mL (100. g) of H2O, we need:

0.100 kg H2O × = 14.9 g = 15 g KCl

Dissolve 15 g KCl in 100. mL H2O to prepare a 2.0 m KCl solution. This will give us slightly more than 100 mL, but this will be the easiest way to make the solution. Since we don’t know the density of the solution, we can’t calculate the molarity and use a volumetric flask to make exactly 100 mL of solution.

b. If we took 15 g NaOH and 85 g H2O, the volume would probably be less than 100 mL. To make sure we have enough solution, let’s use 100. mL H2O (100. g). Let x = mass of NaCl.

mass % NaOH = 15 = × 100, 1500 + 15 x = 100. x, x = 17.6 g ≈ 18 g

Dissolve 18 g NaOH in 100. mL H2O to make a 15% NaOH solution by mass.

c. In a fashion similar to part b, let’s use 100. mL CH3OH. Let x = mass of NaOH.

100. mL CH3OH × = 79 g CH3OH

mass % NaOH = 25 = × 100, 25(79) + 25 x = 100. x, x = 26.3 g ≈ 26 g

Dissolve 26 g NaOH in 100. mL CH3OH.

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CHAPTER 11 PROPERTIES OF SOLUTIONS 382

d. To make sure we have enough solution, let’s use 100. mL (100. g) of H 2O. Let x = mol C6H12O6.

100. g H2O × = 5.55 mol H2O

; 0.10 x + 0.56 = x, x = 0.62 mol C6H12O6

0.62 mol C6H12O6 × = 110 g C6H12O6

Dissolve 110 g C6H12O6 in 100. mL of H2O to prepare a solution with = 0.10.

29. 25 mL C5H12 × = 16 g C5H12; 25 mL × = 0.22 mol C5H12

45 mL C6H14 × = 30. g C6H14; 45 mL × = 0.34 mol C6H14

mass % pentane = × 100 = = 35%

χpentane = = = 0.39

molality = = = = 7.3 mol/kg

molarity = = = 3.1 mol/L

30. If there are 100.0 mL of wine:

12.5 mL C2H5OH × = 9.86 g C2H5OH and 87.5 mL H2O × = 87.5 g

H2O

mass % ethanol = × 100 = 10.1% by mass

molality = = 2.45 mol/kg

31. If we have 1.00 L of solution:

1.37 mol citric acid × = 263 g citric acid (H3C6H5O7)

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383 CHAPTER 11 PROPERTIES OF SOLUTIONS

1.00 × 103 mL solution × = 1.10 × 103 g solution

mass % of citric acid = × 100 = 23.9%

In 1.00 L of solution, we have 263 g citric acid and (1.10 × 103 263) = 840 g of H2O.

molality = = 1.6 mol/kg

840 g H2O × = 47 mol H2O; = 0.028

Since citric acid is a triprotic acid, the number of protons citric acid can provide is three times the molarity. Therefore, normality = 3 × molarity:

normality = 3 × 1.37 M = 4.11 N

32. = 1.00 molal; 1.00 × 103 g C2H5OH × = 21.7 mol C2H5OH

χacetone = = 0.0441

1 mol CH3COCH3 × = 73.7 mL CH3COCH3

1.00 × 103 g ethanol × = 1270 mL; Total volume = 1270 + 73.7 = 1340 mL

molarity = = 0.746 M

Energetics of Solutions and Solubility

33. Using Hess’s law:

NaI(s) → Na+(g) + I(g) ΔH = ΔHLE = (686 kJ/mol) Na+(g) + I(g) → Na+(aq) + I(aq) ΔH = ΔHhyd = 694 kJ/mol

ΔH = ΔHhyd = -694 kJ/mol__________________________________________________________________

NaI(s) → Na+(aq) + I(aq) ΔHsoln = 8 kJ/mol

ΔHsoln refers to the heat released or gained when a solute dissolves in a solvent. Here, an ionic compound dissolves in water.

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CHAPTER 11 PROPERTIES OF SOLUTIONS 384

34. a. CaCl2(s) → Ca2+(g) + 2 Cl(g) ΔH = ΔHLE = (2247 kJ) Ca2+(g) + 2 Cl(g) → Ca2+(aq) + 2 Cl(aq) ΔH = ΔHhyd

_______________________________________________________________ CaCl2(s) → Ca2+(aq) + 2 Cl(aq) ΔHsoln = 46 kJ

46 kJ = 2247 kJ + ΔHhyd, ΔHhyd = 2293 kJ

CaI2(s) → Ca2+(g) + 2 I(g) ΔH = ΔHLE = (2059 kJ) Ca2+(g) + 2 I(g) → Ca2+(aq) + 2 I(aq) ΔH = ΔHhyd

______________________________________________________________ CaI2(s) → Ca2+(aq) + 2 I(aq) ΔHsoln = 104 kJ

104 kJ = 2059 kJ + ΔHhyd, ΔHhyd = 2163 kJ

b. The enthalpy of hydration for CaCl2 is more exothermic than for CaI2. Any differences must be due to differences in hydration between Cl and I. Thus, the chloride ion is more strongly hydrated as compared to the iodide ion.

35. Both Al(OH)3 and NaOH are ionic compounds. Since the lattice energy is proportional to the charge of the ions, the lattice energy of aluminum hydroxide is greater than that of sodium hydroxide. The attraction of water molecules for Al3+ and OH cannot overcome the larger lattice energy and Al(OH)3 is insoluble. For NaOH, the favorable hydration energy is large enough to overcome the smaller lattice energy and NaOH is soluble.

36. The dissolving of an ionic solute in water can be thought of as taking place in two steps. The first step, called the lattice energy term, refers to breaking apart the ionic compound into gaseous ions. This step, as indicated in the problem requires a lot of energy and is unfavorable. The second step, called the hydration energy term, refers to the energy released when the separated gaseous ions are stabilized as water molecules surround the ions. Since the interactions between water molecules and ions are strong, a lot of energy is released when ions are hydrated. Thus, the dissolution process for ionic compounds can be thought of as consisting of an unfavorable and a favorable energy term. These two processes basically cancel each other out; so when ionic solids dissolve in water, the heat released or gained is minimal, and the temperature change is minimal.

37. Water is a polar solvent and dissolves polar solutes and ionic solutes. Carbon tetrachloride (CCl4) is a nonpolar solvent and dissolves nonpolar solutes (like dissolves like). To predict the polarity of the following molecules, draw the correct Lewis structure and then determine if the individual bond dipoles cancel or not. If the bond dipoles are arranged in such a manner that they cancel each other out, then the molecule is nonpolar. If the bond dipoles do not cancel each other out, then the molecule is polar.

a. KrF2, 8 + 2(7) = 22 e b. SF2, 6 + 2(7) = 20 e

nonpolar; soluble in CCl4 polar; soluble in H2O

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385 CHAPTER 11 PROPERTIES OF SOLUTIONS

c. SO2, 6 + 2(6) = 18 e d. CO2, 4 + 2(6) = 16 e

+ 1 more

polar; soluble in H2O nonpolar; soluble in CCl4

e. MgF2 is an ionic compound so it is soluble in water.

f. CH2O, 4 + 2(1) + 6 = 12 e g. C2H4, 2(4) + 4(1) = 12 e

polar; soluble in H2O nonpolar (like all compounds made up of only carbon and hydrogen); soluble in CCl4

38. a. water b. water c. hexane d. water

39. Water is a polar molecule capable of hydrogen bonding. Polar molecules, especially molecules capable of hydrogen bonding, and ions are all attracted to water. For covalent compounds, as polarity increases, the attraction to water increases. For ionic compounds, as the charge of the ions increases and/or the size of the ions decreases, the attraction to water increases.

a. CH3CH2OH; CH3CH2OH is polar while CH3CH2CH3 is nonpolar.

b. CHCl3; CHCl3 is polar while CCl4 is nonpolar.

c. CH3CH2OH; CH3CH2OH is much more polar than CH3(CH2)14CH2OH.40. For ionic compounds, as the charge of the ions increases and/or the size of the ions decreases,

the attraction to water (hydration) increases.

a. Mg2+; smaller size, higher charge b. Be2+; smaller

c. Fe3+; smaller size, higher charge d. F; smaller

e.  Cl; smaller f. SO42-; higher charge

41. As the length of the hydrocarbon chain increases, the solubility decreases. The ‒OH end of the alcohols can hydrogen bond with water. The hydrocarbon chain, however, is basically nonpolar and interacts poorly with water. As the hydrocarbon chain gets longer, a greater

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CHAPTER 11 PROPERTIES OF SOLUTIONS 386

portion of the molecule cannot interact with the water molecules and the solubility decreases, i.e., the effect of the ‒OH group decreases as the alcohols get larger.

42. Benzoic acid is capable of hydrogen bonding, but a significant part of benzoic acid is the non-polar benzene ring which is composed of only carbon and hydrogen. In benzene, a hydrogen bonded dimer forms:

The dimer is relatively nonpolar since the polar part of benzoic acid is hidden in the dimer formation. Thus, benzoic acid is more soluble in benzene than in water due to the dimer formation.

Benzoic acid would be more soluble in 0.1 M NaOH because of the reaction:

C6H5CO2H + OH → C6H5CO2 + H2O

By removing the proton from benzoic acid, an anion forms, and like all anions, the species becomes more soluble in water.

43. C = kP, = k × 0.790 atm, k = 1.04 × mol/Latm

C = kP, C = × 1.10 atm = 1.14 × mol/L

44. 750. mL grape juice ×

= 1.54 mol CO2 (carry extra significant

figure)

1.54 mol CO2 = total mol CO2 = mol CO2(g) + mol CO2(aq) = ng + naq

= = = 326 ng

= = = 43.0 naq = 43.0 naq

= 326 ng = 43.0 naq and from above naq = 1.54 ng; Solving:

O H OC

O H OC

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387 CHAPTER 11 PROPERTIES OF SOLUTIONS

326 ng = 43.0(1.54 ng), 369 ng = 66.2, ng = 0.18 mol

= 326(0.18) = 59 atm in gas phase

C = = × 59 atm, C = 1.8 mol CO2/L in wine

Vapor Pressures of Solutions

45. mol C3H8O3 = 164 g × = 1.78 mol C3H8O3

mol H2O = 338 mL × = 18.6 mol H2O

= × 54.74 torr = 0.913 × 54.74 torr = 50.0 torr

46. = =

53.6 g C3H8O3 × = 0.582 mol C3H8O3

133.7 g C2H5OH × = 2.90 mol C2H5OH; total mol = 0.582 + 2.90

= 3.48 mol

113 torr = = 136 torr

47. = 0.900 atm/0.930 atm = 0.968

0.968 = ; mol benzene = 78.11 g C6H6 × = 1.000 mol

Let x = mol solute, then: χB = 0.968 = , 0.968 + 0.968 x = 1.000, x = 0.033 mol

molar mass = = 303 g/mol ≈ 3.0 × 102 g/mol

48. 19.6 torr = (23.8 torr), = 0.824; = 1.000 0.824 = 0.176

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CHAPTER 11 PROPERTIES OF SOLUTIONS 388

0.176 is the mol fraction of all the solute particles present. Since NaCl dissolves to produce two ions in solution (Na+ and Cl), 0.176 is the mole fraction of Na+ and Cl ions present (assuming complete dissociation of NaCl).

At 45°C, = 0.824 (71.9 torr) = 59.2 torr

49. a. 25 mL C5H12 × = 0.22 mol C5H12

45 mL C6H14 × = 0.34 mol C6H14; total mol = 0.22 + 0.34 = 0.56

mol

= 0.39, = 1.00 - 0.39 = 0.61

= 0.39(511 torr) = 2.0 × 102 torr; = 0.61(150. torr) = 92 torr

= 2.0 × 102 + 92 = 292 torr = 290 torr

b. From Chapter 5 on gases, the partial pressure of a gas is proportional to the number of moles of gas present. For the vapor phase:

= 0.69

Note: In the Solutions Guide, we have added V or L to the mole fraction symbol to empha-size the value for which we are solving. If the L or V is omitted, then the liquid phase is assumed.

50. = = 0.375

Ptotal = 0.375 (133 torr) + (1.000 0.375) (11.4 torr) = 49.9 + 7.13 = 57.0 torr

In the vapor: = 0.875; = 1.000 0.875 = 0.125

51. 174 torr = (303 torr) + (44.6 torr); = 1.000

174 = 303 + (1.000 ) 44.6, = 0.500; = 1.000 0.500 = 0.500

52. For the vapor, Since the mole fractions of benzene and toluene are equal in the vapor phase, then

= (1.00 (28 torr) = (1.00 ) 95 torr

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389 CHAPTER 11 PROPERTIES OF SOLUTIONS

123 = 95, = 0.77; = 1.00 0.77 = 0.23

53. Compared to H2O, solution d (methanol/water) will have the highest vapor pressure because methanol is more volatile than water. Both solution b (glucose/water) and solution c (NaCl/water) will have a lower vapor pressure than water by Raoult's law. NaCl dissolves to give Na+ ions and Cl ions; glucose is a nonelectrolyte. Since there are more solute particles in solution c, the vapor pressure of solution c will be the lowest.

54. Solution d (methanol/water); Methanol is more volatile than water, which will increase the total vapor pressure to a value greater than the vapor pressure of pure water at this temperature.

55. 50.0 g CH3COCH3 × = 0.861 mol acetone

50.0 g CH3OH × = 1.56 mol methanol

= = 0.356; = 1.000 = 0.644

Ptotal = Pmethanol + Pacetone = 0.644(143 torr) + 0.356(271 torr) = 92.1 torr + 96.5 torr

= 188.6 torr

Because partial pressures are proportional to the moles of gas present, then in the vapor phase:

= 0.512; = 1.000 - 0.512 = 0.488

The actual vapor pressure of the solution (161 torr) is less than the calculated pressure assuming ideal behavior (188.6 torr). Therefore, the solution exhibits negative deviations from Raoult’s law. This occurs when the solute-solvent interactions are stronger than in pure solute and pure solvent.

56. a. An ideal solution would have a vapor pressure at any mole fraction of H2O between that of pure propanol and pure water (between 74.0 torr and 71.9 torr). The vapor pressures of the solutions are not between these limits, so water and propanol do not make ideal solutions.

b. From the data, the vapor pressures of the various solutions are greater than in the ideal solution (positive deviation from Raoult’s law). This occurs when the intermolecular forces in solution are weaker than the intermolecular forces in pure solvent and pure solute. This gives rise to endothermic (positive) ΔHsoln values.

c. The interactions between propanol and water molecules are weaker than between the pure substances since this solution exhibits a positive deviation from Raoult’s law.

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CHAPTER 11 PROPERTIES OF SOLUTIONS 390

d. At = 0.54, the vapor pressure is highest as compared to the other solutions. Since a solution boils when the vapor pressure of the solution equals the external pressure, the

= 0.54 solution should have the lowest normal boiling point; this solution will have a vapor pressure equal to 1 atm at a lower temperature than the other solutions.

Colligative Properties

57. molality = m = = 3.00 molal

ΔTb = Kbm = × 3.00 molal = 1.5°C

The boiling point is raised from 100.0°C to 101.5°C (assuming P = 1 atm).

58. ΔTb = 77.85°C 76.50°C = 1.35°C; m = = 0.268 mol/kg

mol biomolecule = 0.0150 kg solvent × = 4.02 × mol

From the problem, 2.00 g biomolecule was used that must contain 4.02 × mol biomolecule. The molar mass of the biomolecule is:

= 498 g/mol

59. ΔTf = Kfm, ΔTf = 1.50°C = × m, m = 0.806 mol/kg

0.200 kg H2O × = 14.8 g C3H8O3

60. ΔTf = 25.50°C 24.59°C = 0.91°C = Kfm, m = = 0.10 mol/kg

mass H2O = 0.0100 kg t-butanol = 0.018 g H2O

61. molality = m = = 16.1 mol/kg

ΔTf = Kfm = 1.86°C/molal × 16.1 molal = 29.9°C; Tf = 0.0°C 29.9°C = 29.9°C

ΔTb = Kbm = 0.51°C/molal × 16.1 molal = 8.2°C; Tb = 100.0°C + 8.2°C = 108.2°C

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391 CHAPTER 11 PROPERTIES OF SOLUTIONS

62. m = = 13.4 mol C2H6O2/kg

Since the density of water is 1.00 g/cm3, the moles of C2H6O2 needed are:

15.0 L H2O × = 201 mol C2H6O2

Volume C2H6O2 = 201 mol C2H6O2 × = 11,200 cm3 = 11.2 L

ΔTb = Kbm = × 13.4 molal = 6.8°C , Tb = 100.0°C + 6.8°C = 106.8°C

63. ΔTf = Kfm, m =

The mol of thyroxine present is:

0.0100 kg benzene × = 5.86 × mol thyroxine

From the problem, 0.455 g thyroxine were used; this must contain 5.86 × mol thyroxine. The molar mass of the thyroxine is:

molar mass = = 776 g/mol

64. empirical formula mass ≈ 7(12) + 4(1) + 16 = 104 g/mol

ΔTf = Kfm, m = = 0.56 molal

mol anthraquinone = 0.0114 kg solvent × = 6.4 × mol

molar mass = = 210 g/mol

= 2.0; molecular formula = C14H8O2

65. a. M = = 1.1 × mol/L; π = MRT

At 298 K: π = × 298 K × , π = 0.20 torr

Because d = 1.0 g/cm3, 1.0 L solution has a mass of 1.0 kg. Because only 1.0 g of protein is present per liter of solution, 1.0 kg of H2O is present and molality equals molarity.

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CHAPTER 11 PROPERTIES OF SOLUTIONS 392

ΔTf = Kfm = × 1.1 × molal = 2.0 × °C

b. Osmotic pressure is better for determining the molar mass of large molecules. A temperature change of °C is very difficult to measure. A change in height of a column of mercury by 0.2 mm (0.2 torr) is not as hard to measure precisely.

66. M = = 4.0 × mol/L

1.00 L × = 4.0 × mol catalase

molar mass = = 2.5 × 105 g/mol

67. π = MRT, M = = 0.327 mol/L

68. M = = 0.62 M solute particles

This represents the total molarity of the solute particles. NaCl is a soluble ionic compound that breaks up into two ions, Na+ and Cl. Therefore, the concentration of NaCl needed is 0.62/2 = 0.31 M; this NaCl concentration will produce a 0.62 M solute particle solution assuming complete dissociation.

1.0 L × = 18.1 ≈ 18 g NaCl

Dissolve 18 g of NaCl in some water and dilute to 1.0 L in a volumetric flask. To get 0.31 ± 0.01 mol/L, we need 18.1 g ± 0.6 g NaCl in 1.00 L solution.

Properties of Electrolyte Solutions

69. Na3PO4(s) → 3 Na+(aq) + i = 4.0; CaBr2(s) → Ca2+(aq) + 2 Br(aq), i = 3.0

KCl(s) → K+(aq) + Cl(aq), i = 2.0.

The effective particle concentrations of the solutions are:

4.0(0.010 molal) = 0.040 molal for Na3PO4 solution; 3.0(0.020 molal) = 0.060 molal for CaBr2 solution; 2.0(0.020 molal) = 0.040 molal for KCl solution; slightly greater than 0.020 molal for HF solution since HF only partially dissociates in water (it is a weak acid).

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393 CHAPTER 11 PROPERTIES OF SOLUTIONS

a. The 0.010 m Na3PO4 solution and the 0.020 m KCl solution both have effective particle concentrations of 0.040 m (assuming complete dissociation), so both of these solutions should have the same boiling point as the 0.040 m C6H12O6 solution (a nonelectrolyte).

b. P = χP°; As the solute concentration decreases, the solvent’s vapor pressure increases since χ increases. Therefore, the 0.020 m HF solution will have the highest vapor pressure since it has the smallest effective particle concentration.

c. ΔT = Kfm; The 0.020 m CaBr2 solution has the largest effective particle concentration so it will have the largest freezing point depression (largest ΔT).

70. The solutions of C12H22O11, NaCl and CaCl2 will all have lower freezing points, higher boiling points and higher osmotic pressures than pure water. The solution with the largest particle concentration will have the lowest freezing point, the highest boiling point and the highest osmotic pressure. The CaCl2 solution will have the largest effective particle concentration because it produces three ions per mol of compound.

a. pure water b. CaCl2 solution c. CaCl2 solution

d. pure water e. CaCl2 solution

71. a. MgCl2(s) → Mg2+(aq) + 2 Cl(aq), i = 3.0 mol ions/mol solute

ΔTf = iKfm = 3.0 × 1.86°C/molal × 0.050 molal = 0.28°C; Tf = -0.28°C (Assuming water freezes at 0.00°C.)

ΔTb = iKbm = 3.0 × 0.51°C/molal × 0.050 molal = 0.077°C; Tb = 100.077°C (Assumingwater boils at 100.000°C.)

b. FeCl3(s) → Fe3+(aq) + 3 Cl(aq), i = 4.0 mol ions/mol solute

ΔTf = iKfm = 4.0 × 1.86°C/molal × 0.050 molal = 0.37°C; Tf = 0.37°C

ΔTb = iKbm = 4.0 × 0.51°C/molal × 0.050 molal = 0.10°C; Tb = 100.10°C

72. NaCl(s) → Na+(aq) + Cl(aq), i = 2.0

π = iMRT = 2.0 × × 293 K = 4.8 atm

A pressure greater than 4.8 atm should be applied to insure purification by reverse osmosis.

73. ΔTf = iKfm, i = = 2.63 for 0.0225 m CaCl2

i = = 2.60 for 0.0910 m CaCl2; i = = 2.57 for 0.278 m

CaCl2

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CHAPTER 11 PROPERTIES OF SOLUTIONS 394

iave = (2.63 + 2.60 + 2.57)/3 = 2.60

Note that i is less than the ideal value of 3.0 for CaCl2. This is due to ion pairing in solution. Also note that as molality increases, i decreases. More ion pairing occurs as the solute con-centration increases.

74. a. MgCl2, i(observed) = 2.7

ΔTf = iKfm = 2.7 × 1.86°C/molal × 0.050 molal = 0.25°C; Tf = 0.25°C

ΔTb = iKbm = 2.7 × 0.51°C/molal × 0.050 molal = 0.069°C; Tb = 100.069°C

b. FeCl3, i(observed) = 3.4

ΔTf = iKfm = 3.4 × 1.86 °C/molal × 0.050 molal = 0.32°C; Tf = 0.32°C

ΔTb = iKbm = 3.4 × 0.51°C/molal × 0.050 molal = 0.087°C; Tb = 100.087°C

75. a. TC = 5(TF 32)/9 = 5(29 32)/9 = 34°C; Assuming the solubility of CaCl2 is temperature independent, the molality of a saturated CaCl2 solution is:

ΔTf = iKfm = 3.00 × 1.86°C kg/mol × 6.71 mol/kg = 37.4°C

Assuming i = 3.00, a saturated solution of CaCl2 can lower the freezing point of water to 37.4°C. Assuming these conditions, a saturated CaCl2 solution should melt ice at 34°C (29°F).

b. From Exercise 11.73, iave = 2.60; ΔTf = iKfm = 2.60 × 1.86 × 6.71 = 32.4°C

Tf = 32.4°C

Assuming i = 2.60, a saturated CaCl2 solution will not melt ice at 34°C(29°F).

76. = iMRT, M = = = 5.11 × mol/L

molar mass of compound = = 97.8 g/mol

Additional Exercises

77 a. NH4NO3(s) → NH4+(aq) + NO3

(aq) ΔHsoln = ?

Heat gain by dissolution process = heat loss by solution; We will keep all quantities posi-tive in order to avoid sign errors. Since the temperature of the water decreased, the dis-solution of NH4NO3 is endothermic (ΔH is positive). Mass of solution = 1.60 + 75.0 = 76.6 g.

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395 CHAPTER 11 PROPERTIES OF SOLUTIONS

heat loss by solution = × 76.6 g × (25.00°C 23.34°C) = 532 J

ΔHsoln = = 2.66 × 104 J/mol = 26.6 kJ/mol

b. We will use Hess’s law to solve for the lattice energy. The lattice energy equation is:

NH4+(g) + NO3

(g) → NH4NO3(s) ΔH = lattice energy

NH4+(g) + NO3

(g) → NH4+(aq) + NO3

(aq) ΔH = ΔHhyd = 630. kJ/molNH4

+(aq) + NO3(aq) → NH4NO3(s) ΔH = ΔHsoln = 26.6 kJ/mol

____________________________________________________________________ NH4

+(g) + NO3(g) → NH4NO3(s) ΔH = ΔHhyd ΔHsoln

= 657 kJ/mol

78. The main intermolecular forces are: hexane (C6H14): London dispersion; chloroform (CHCl3): dipole-dipole; London dispersion; methanol (CH3OH): H bonding; H2O: H bonding (two places)

There is a gradual change in the nature of the intermolecular forces (weaker to stronger). Each preceding solvent is miscible in its predecessor because there is not a great change in the strengths of the intermolecular forces from one solvent to the next.

79. a. Water boils when the vapor pressure equals the pressure above the water. In an open pan,Patm ≈ 1.0 atm. In a pressure cooker, Pinside > 1.0 atm, and water boils at a higher temper-ature. The higher the cooking temperature, the faster the cooking time.

b. Salt dissolves in water forming a solution with a melting point lower than that of pure water (ΔTf = Kfm). This happens in water on the surface of ice. If it is not too cold, the ice melts. This won't work if the ambient temperature is lower than the depressed freezing point of the salt solution.

c. When water freezes from a solution, it freezes as pure water, leaving behind a more concentrated salt solution. Therefore, the melt of frozen sea ice is pure water.

d. On the CO2 phase diagram in chapter 10, the triple point is above 1 atm, so CO2(g) is the stable phase at 1 atm and room temperature. CO2(l) can't exist at normal atmospheric pressures. Therefore, dry ice sublimes instead of boils. In a fire extinguisher, P > 1 atm and CO2(l) can exist. When CO2 is released from the fire extinguisher, CO2(g) forms as predicted from the phase diagram.

e. Adding a solute to a solvent increases the boiling point and decreases the freezing point of the solvent. Thus, the solvent is a liquid over a wider range of temperatures when a solute is dissolved.

80. A 92 proof ethanol solution is 46% C2H5OH by volume. Assuming 100.0 mL of solution:

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CHAPTER 11 PROPERTIES OF SOLUTIONS 396

mol ethanol = 46 mL C2H5OH × = 0.79 mol C2H5OH

molarity = = 7.9 M ethanol

81. Because partial pressures are proportional to the moles of gas present, then

= 0.855 (263 torr) = 225 torr

= 0.600

82. π = MRT = × 298 K = 2.45 atm ≈ 2 atm

π = 2 atm × ≈ 2000 mm ≈ 2 m

The osmotic pressure would support a mercury column of ≈ 2 m. The height of a fluid column in a tree will be higher because Hg is more dense than the fluid in a tree. If we assume the fluid in a tree is mostly H2O, then the fluid has a density of 1.0 g/cm3. The density of Hg is 13.6 g/cm3.

Height of fluid ≈ 2 m × 13.6 ≈ 30 m

83. Out of 100.00 g, there are:

31.57 g C × = 2.629 mol C; = 1.000

5.30 g H × = 5.26 mol H; = 2.00

63.13 g O × = 3.946 mol O; = 1.501

empirical formula: C2H4O3; Use the freezing point data to determine the molar mass.

m = = 2.80 molal

mol solute = 0.0250 kg × = 0.0700 mol solute

molar mass = = 151 g/mol

The empirical formula mass of C2H4O3 = 76.05 g/mol. Since the molar mass is about twice the empirical mass, the molecular formula is C4H8O6, which has a molar mass of 152.10 g/mol.

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397 CHAPTER 11 PROPERTIES OF SOLUTIONS

Note:  We use the experimental molar mass to determine the molecular formula. Knowing this, we calculate the molar mass precisely from the molecular formula using the atomic masses in the periodic table.

84. a. As discussed in Figure 11.18 of the text, the water would migrate from right to left. Initially, the level of liquid in the right arm would go down and the level in the left arm would go up. At some point, the rate of solvent transfer would be the same in both directions and the levels of the liquids in the two arms would stabilize. The height difference between the two arms is a measure of the osmotic pressure of the NaCl solution.

b. Initially, H2O molecules will have a net migration into the NaCl side. However, Na+ and Cl ions can now migrate into the H2O side. Because solute and solvent transfer are both possible, the levels of the liquids will be equal once the rate of solute and solvent transfer is equal in both directions. At this point, the concentration of Na+ and Cl ions will be equal in both chambers and the levels of liquid will be equal.

85. If ideal, NaCl dissociates completely and i = 2.00. ΔTf = iKfm; Assuming water freezes at 0.00°C:

1.28°C = 2 × 1.86°C kg/mol × m, m = 0.344 mol NaCl/kg H2O

Assume an amount of solution which contains 1.00 kg of water (solvent).

0.344 mol NaCl × = 20.1 g NaCl; mass % NaCl = × 100

= 1.97%

86. The main factor for stabilization seems to be electrostatic repulsion. The center of a colloid particle is surrounded by a layer of same charged ions, with opposite charged ions forming another charged layer on the outside. Overall, there are equal numbers of charged and oppositely charged ions, so the colloidal particles are electrically neutral. However, since the outer layers are the same charge, the particles repel each other and do not easily aggregate for precipitation to occur.

Heating increases the velocities of the colloidal particles. This causes the particles to collide with enough energy to break the ion barriers, allowing the colloids to aggregate and eventually precipitate out. Adding an electrolyte neutralizes the adsorbed ion layers which allows colloidal particles to aggregate and then precipitate out.

87. T = Kfm, m = = 1.50 molal

a. T = Kbm, T = (0.51C/molal)(1.50 molal) = 0.77C, Tb = 100.77C

b. =

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CHAPTER 11 PROPERTIES OF SOLUTIONS 398

Assuming 1.00 kg of water, we have 1.50 mol solute and:

mol H2O = 1.00 × 103 g H2O × = 55.5 mol H2O

water = = 0.974; Pwater = (0.974)(23.76 mm Hg) = 23.1 mm Hg

c. We assumed ideal behavior in solution formation and assumed i = 1 (no ions form).

Challenge Problems

88. For the second vapor collected, = 0.714 and = 0.286, where = mole fraction

of benzene in the second solution and = mole fraction of toluene in the second solution.

= 1.000

= 0.714 = =

Solving: = 0.500 =

This second solution came from the vapor collected from the first (initial) solution. So,

= = 0.500. Let = mole fraction benzene in the first solution and = mole frac-

tion of toluene in first solution. = 1.000

= 0.500 = =

Solving: = 0.286

The original solution had B = 0.286 and T = 0.714.

89. For 30.% A by moles in the vapor, 30. = × 100:

0.30 =

χA x = 0.30(χA x) + 0.30 y 0.30 χA y, χA x 0.30 χA x + 0.30 χA y = 0.30 y

χA(x 0.30 x + 0.30 y) = 0.30 y, χA = χB = 1.00 χA

Similarly, if vapor above is 50.% A:

If vapor above is 80%A: χA = χB = 1.00 χA

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399 CHAPTER 11 PROPERTIES OF SOLUTIONS

If the liquid solution is 30.%A by moles, χA = 0.30.

Thus, =

If solution is 50.%A:

If solution is 80.%A:

90. m = = 0.218 mol/kg

= MRT where M = mol/L; We must assume that molarity = molality so we can calculate the osmotic pressure. This is a reasonable assumption for dilute solutions when 1.00 kg of water 1.00 L of solution. Assuming complete dissociation of NaCl, a 0.218 m solution corresponds to 6.37 g NaCl dissolved in 1.00 kg of water. The volume of solution may be a little larger than 1.00 L, but not by much (to three sig figs). The assumption that molarity = molality will be good here.

= (0.218 M)(0.08206 L atm/Kmol)(298 K) = 5.33 atm

91. m = = 0.229 molal

Let x = mol NaCl in mixture and y = mol C12H22O11.

NaCl(aq) → Na+(aq) + Cl(aq); In solution, NaCl exists as separate Na+ and Cl ions.

0.229 mol = mol Na+ + mol Cl + mol C12H22O11 = x + x + y = 2x + y

The molar mass of NaCl is 58.44 g/mol and the molar mass of C12H22O11 = 342.3 g/mol. Setting up another equation for the mass of the mixture:

20.0 g = x(58.44) + y(342.3)

Substituting: 20.0 = 58.44 x + (0.229 – 2x) 342.3

Solving: x = mol NaCl = 0.0932 mol and

y = mol C12H22O11 = 0.229 – 2(0.0932) = 0.043 mol

mass NaCl = 0.0932 mol × 58.44 g/mol = 5.45 g NaCl

mass% NaCl = × 100 = 27.3%, mass % C12H22O11 = 100.0 – 27.3 = 72.7%

sucrose = = 0.32

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CHAPTER 11 PROPERTIES OF SOLUTIONS 400

92. a. π = iMRT, iM = = 0.320 mol/L

Assuming 1.000 L of solution:

total mol solute particles = mol Na+ + mol Cl + mol NaCl = 0.320 mol

mass solution = 1000. mL × = 1071 g solution

mass NaCl in solution = 0.0100 × 1071 g = 10.7 g NaCl

mol NaCl added to solution = 10.7 g × = 0.183 mol NaCl

Some of this NaCl dissociates into Na+ and Cl (two mol ions per mol NaCl) and some remains undissociated. Let x = mol undissociated NaCl = mol ion pairs.

mol solute particles = 0.320 mol = 2(0.183 x) + x

0.320 = 0.366 x, x = 0.046 mol ion pairs

fraction of ion pairs = = 0.25, or 25%

b. ΔT = Kfm where Kf = 1.86 °C kg/mol; From part a, 1.000 L of solution contains 0.320 mol of solute particles. To calculate the molality of the solution, we need the kg of solvent present in 1.000 L solution.

mass of 1.000 L solution = 1071 g; mass of NaCl = 10.7 g

mass of solvent in 1.000 L solution = 1071 g 10.7 g = 1060. g

ΔT = 1.86 °C kg/mol × = 0.562°C

Assuming water freezes at 0.000°C, then Tf = 0.562°C.

93. = 0.15 = ; Ppen = = (511 torr); Ptotal = Ppen + Phex = (511) + (150.)

Since = 1.000 , then: Ptotal = (511 torr) + (1.000 )(150.) = 150. + 361

= , 0.15 = , 0.15 (150. + 361 ) = 511

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401 CHAPTER 11 PROPERTIES OF SOLUTIONS

23 + 54 = 511 , = = 0.050

94. a. m = = 0.258 mol/kg

mol unknown = 0.01560 kg × = 4.02 × mol

molar mass of unknown = = 303 g/mol

Uncertainty in temperature = × 100 = 3%; A 3% uncertainty in 303 g/mol

= 9 g/mol.

So, molar mass = 303 ± 9 g/mol.

b. No, codeine could not be eliminated since its molar mass is in the possible range including the uncertainty.

c. We would really like the uncertainty to be ± 1 g/mol. We need the freezing point depression to be about 10 times what it was in this problem. Two possibilities are:1. make the solution ten times more concentrated (may be a solubility problem) or2. use a solvent with a larger Kf value, e.g., camphor

95. ΔTf = 5.51C 2.81C = 2.70°C; m = = 0.527 molal

Let x = mass of naphthalene (molar mass = 128.2 g/mol). Then, 1.60 x = mass of anthracene (molar mass = 178.2 g/mol).

= moles napthalene and = moles anthracene

= , 1.05 × =

50.0 x + 205 = 240., 50.0 x = 35, x = 0.70 g naphthalene

So mixture is: × 100 = 44% naphthalene by mass and 56% anthracene by mass

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CHAPTER 11 PROPERTIES OF SOLUTIONS 402

96. iM = = 0.01614 mol/L = total ion concentration

0.01614 mol/L = + + ; = + (charge balance)

Combining: 0.01614 = +

Let x = mass MgCl2 and y = mass NaCl, then x + y = 0.5000 g.

= and = (Because V = 1.000 L.)

Total ion concentration = = 0.01614 mol/L; Rearranging: 3 x + 3.258 y

= 1.537Solving by simultaneous equations:

3 x + 3.258 y = 1.5373 (x + y) = 3(0.5000)__________________________________________

0.258 y = 0.037, y = 0.14 g NaCl

mass MgCl2 = 0.5000 g 0.14 g = 0.36 g; mass % MgCl2 = × 100 = 72%

97. HCO2H → H+ + HCO2; Only 4.2% of HCO2H ionizes. The amount of H+ or HCO2

produced is:

0.042 × 0.10 M = 0.0042 M

The amount of HCO2H remaining in solution after ionization is:

0.10 M 0.0042 M = 0.10 M

The total molarity of species present = = 0.10 + 0.0042 + 0.0042 = 0.11 M

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CHAPTER 11 PROPERTIES OF SOLUTIONS

Assuming 0.11 M = 0.11 molal and assuming ample significant figures in the freezing point and boiling point of water at P = 1 atm:

ΔT = Kfm = 1.86°C/molal × 0.11 molal = 0.20°C; freezing point = 0.20°C

ΔT = Kbm = 0.51°C/molal × 0.11 molal = 0.056°C; boiling point = 100.056°C

98. a. The average values for each ion are:

300. mg Na+; 15.7 mg K+; 5.45 mg Ca2+; 388 mg Cl; 246 mg lactate, C3H5O3

Note: Since we can precisely weigh to ± 0.1 mg on an analytical balance, we'll carry extra significant figures and calculate results to ± 0.1 mg.

The only source of lactate is NaC3H5O3.

246 mg C3H5O3 × = 309.5 mg sodium lactate

The only source of Ca2+ is CaCl22H2O.

5.45 mg Ca2+ × = 19.99 or 20.0 mg CaCl22H2O

The only source of K+ is KCl.

15.7 mg K+ × = 29.9 mg KCl

From what we have used already, let's calculate the mass of Na+ added.

309.5 mg sodium lactate = 246.0 mg lactate + 63.5 mg Na+

Thus, we need to add an additional 236.5 mg Na+ to get the desired 300. mg.

236.5 mg Na+ × = 601.2 mg NaCl

Now, let's check the mass of Cl added:

20.0 mg CaCl22H2O × = 9.6 mg Cl

20.0 mg CaCl22H2O = 9.6 mg Cl

29.9 mg KCl 15.7 mg K+ = 14.2 mg Cl

601.2 mg NaCl 236.5 mg Na+ = 364.7 mg Cl _______________________________________

Total Cl = 388.5 mg Cl

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CHAPTER 11 PROPERTIES OF SOLUTIONS 404

This is the quantity of Cl we want (the average amount of Cl).

An analytical balance can weigh to the nearest 0.1 mg. We would use 309.5 mg sodium lactate, 20.0 mg CaCl22H2O, 29.9 mg KCl and 601.2 mg NaCl.

b. To get the range of osmotic pressure, we need to calculate the molar concentration of each ion at its minimum and maximum values. At minimum concentrations, we have:

= 0.124 M; = 0.00361 M

= 0.0012 M; = 0.104 M

= 0.0259 M (Note: molarity = mol/L = mmol/mL.)

Total = 0.124 + 0.00361 + 0.0012 + 0.104 + 0.0259 = 0.259 M

π = MRT = × 310. K = 6.59 atm

Similarly at maximum concentrations, the concentration of each ion is:

Na+: 0.137 M; K+: 0.00442 M; Ca2+: 0.0015 M; Cl: 0.115 M; C3H5O3: 0.0293 M

The total concentration of all ions is the sum, 0.287 M.

π = × 310. K = 7.30 atm

Osmotic pressure ranges from 6.59 atm to 7.30 atm.

99. a. Assuming MgCO3(s) does not dissociate, the solute concentration in water is:

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CHAPTER 11 PROPERTIES OF SOLUTIONS

= 6.6 × mol MgCO3/L

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CHAPTER 11 PROPERTIES OF SOLUTIONS

An applied pressure of 8.0 atm will purify water up to a solute concentration of:

M =

When the concentration of MgCO3(s) reaches 0.32 mol/L, the reverse osmosis unit can no longer purify the water. Let V = volume (L) of water remaining after purifying 45 L of H 2O. When V + 45 L of water have been processed, the moles of solute particles will equal:

6.6 × mol/L × (45 L + V) = 0.32 mol/L × V

Solving: 0.30 = (0.32 0.0066) × V, V = 0.96 L

The minimum total volume of water that must be processed is 45 L + 0.96 L = 46 L.

Note: If MgCO3 does dissociate into Mg2+ and CO32 ions, the solute concentration will

increase to 1.3 × M and at least 47 L of water must be processed.

b. No; A reverse osmosis system that applies 8.0 atm can only purify water with a solute concentration less than 0.32 mol/L. Salt water has a solute concentration of 2(0.60 M) = 1.20 M ions. The solute concentration of salt water is much too high for this reverse osmosis unit to work.

Integrative Problems

100. 10.0 mL = 8.8 × mol

C4H7N3O

mass of blood = 10.0 mL = 10.3 g

molality = = 8.5 × mol/kg

= MRT, M = = 8.8 × mol/L

= 8.8 × mol/L × 298 K = 2.2 × atm

101. T = imKf, i = = = 3.00

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407 CHAPTER 11 PROPERTIES OF SOLUTIONS

We have 3 ions in solutions and we have twice as many anions as cations. Therefore, the formula of Q is MCl2. Assuming 100.00 g compound:

38.68 g Cl = 1.091 mol Cl

mol M = 1.091 mol Cl = 0.5455 mol M

molar mass of M = = 112.4 g/mol; M is Cd so Q = CdCl2.

102. mass of H2O = 160. mL = 159 g = 0.159 kg

mol NaDTZ = 0.159 kg = 0.0601 mol

molar mass of NaDTZ = = 639 g/mol

Psoln = ; mol H2O = 159 g = 8.82 mol

Sodium diatrizoate is a salt because there is a metal (sodium) in the compound. From the short-hand notation for sodium diatrizoate, NaDTZ, we can assume this salt breaks up into Na+ and DTZ ions. So the moles of solute particles are 2(0.0601) = 0.120 mol solute particles.

= = 0.987; Psoln = 0.987 × 34.1 torr = 33.7 torr

Marathon Problem

103. a. From part a information, we can calculate the molar mass of NanA and deduce the formula.

mol NanA = mol reducing agent = 0.01526 L = 3.530 × mol NanA

molar mass of NanA = = 85.0 g/mol

To deduce the formula, we will assume various charges and numbers of oxygens present in the oxyanion, then use the periodic table to see if an element fits the molar mass data.

Assuming n = 1 so the formula is NaA. The molar mass of the oxyanion, A , is 85.0 23.0 = 62.0 g/mol. The oxyanion part of the formula could be EO or EO2

or EO3 where E is some

element. If EO, then the molar mass of E is 62.0 16.0 = 46.0 g/mol; no element has this molar mass. If EO2

, molar mass of E = 62.0 32.0 = 30.0 g/mol. Phosphorus is close, but PO2

anions are not common. If EO3, molar mass of E = 62.0 48.0 = 14.0. Nitrogen has this

molar mass and NO3 anions are very common. Therefore, NO3

is a possible formula for A.

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CHAPTER 11 PROPERTIES OF SOLUTIONS 408

Next, we assume Na2A and Na3A formulas and go through the same procedure as above. In all cases, no element in the periodic table fits the data. Therefore, we assume the oxyanion is NO3

= A.

b. The crystal data in part b allows determination of the metal, M, in the formula. See Exercise 10.49 for a review of relationships in body-centered cubic cells. In a bcc unit cell, there are 2 atoms per unit cell and the body diagonal of the cubic cell is related to the radius of the metal by the equation 4r = ℓ where ℓ = cubic edge length.

ℓ = = 4.582 × cm

volume of unit cell = ℓ 3 = (4.582 × )3 = 9.620 × cm3

mass of M in a unit cell = 9.620 × cm3 × = 5.044 × g M

mol M in a unit cell = 2 atoms × = 3.321 × mol M

molar mass of M = = 151.9 g/mol

From the periodic table, M is europium, Eu. Given the charge of Eu is +3, then the formula of the salt is Eu(NO3)3zH2O.

c. Part c data allows determination of the molar mass of Eu(NO3)3zH2O, from which we can determine z, the number of waters of hydration.

π = iMRT, iM = = = 0.0300 mol/L

The total molarity of solute particles present is 0.0300 M. The solute particles are Eu3+ and NO3

ions (the waters of hydration are not solute particles). Because each mol of Eu(NO3)3zH2O dissolves to form 4 ions (Eu3+ + 3 NO3

), the molarity of Eu(NO3)3zH2O is 0.0300/4 = 0.00750 M.

mol Eu(NO3)3zH2O = 0.01000 L × = 7.50 × mol

molar mass of Eu(NO3)3zH2O = = 446 g/mol

446 g/mol = 152.0 + 3(62.0) + z(18.0), z(18.0) = 108, z = 6.00

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409 CHAPTER 11 PROPERTIES OF SOLUTIONS

The formula for the strong electrolyte is Eu(NO3)36H2O.