CHAPTER 17 PROPERTIES OF SOLUTIONSdevries/chem1C/handouts/... · · 2010-05-11To make sure we...

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CHAPTER 17 PROPERTIES OF SOLUTIONS Solution Composition 12. 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.

13. a. HNO3(l) → H+(aq) + NO3

−(aq) b. Na2SO4(s) → 2 Na+(aq) + SO42−(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) + SO4

2−(aq) i. NaOH(s) → Na+(aq) + OH−(aq)

14. Mol Na2CO3 = 0.0700 L × L

CONamol0.3 32 = 0.21 mol Na2CO3

Na2CO3(s) → 2 Na+(aq) + CO3

2−(aq); mol Na+ = 2(0.21) = 0.42 mol

Mol NaHCO3 = 0.0300 L × LNaHCOmol0.1 3 = 0.030 mol NaHCO3

NaHCO3(s) → Na+(aq) + HCO3

−(aq); mol Na+ = 0.030 mol

L1000.0

mol45.0L030.0L0700.0

mol030.0mol42.0volumetotal

NamoltotalNa ===

+++

+M = 4.5 M Na+

CHAPTER 17 PROPERTIES OF SOLUTIONS 669

15. Molality = g07.62

EGmol1kg

g1000OHg0.60

EGg0.40

2×× = 10.7 mol/kg

Molarity = g07.62

mol1L

cm1000cm

g05.1solutiong0.100

EGg0.40 3

3 ××× = 6.77 mol/L

40.0 g EG × g07.62

mol1 = 0.644 mol EG; 60.0 g H2O ×

g02.18mol1

= 3.33 mol H2O

644.033.3

644.0χ EG += = 0.162 = mole fraction ethylene glycol

16. Hydrochloric acid (HCl):

molarity = g36.5

HClmol1L

cm1000solncmsolng1.19

solng100.HClg38 3

3 ××× = 12 mol/L

molality = g5.36

HClmol1kg

g1000solventg62

HClg38×× = 17 mol/kg

38 g HCl × g5.36

mol1 = 1.0 mol HCl; 62 g H2O ×

g0.18mol1

= 3.4 mol H2O

mole fraction of HCl = 0.14.3

0.1χHCl += = 0.23

Nitric acid (HNO3):

g0.63

HNOmol1L


solng.100HNOg.70 3

3

33 ××× = 16 mol/L

g0.63

HNOmol1kg

g1000solventg.30HNOg.70 33 ×× = 37 mol/kg

70. g HNO3 × g0.63

mol1 = 1.1 mol HNO3; 30. g H2O ×

g0.18mol1

= 1.7 mol H2O

1.17.1

1.1χ3HNO += = 0.39

670 CHAPTER 17 PROPERTIES OF SOLUTIONS Sulfuric acid (H2SO4):

42

423

342

SOHg1.98SOHmol1

Lcm1000

solncmsolng84.1

solng.100SOHg95

××× = 18 mol/L

g1.98

mol1kg

g1000OHg5

SOHg95

2

42 ×× = 194 mol/kg ≈ 200 mol/kg

95 g H2SO4 × g1.98

mol1 = 0.97 mol H2SO4; 5 g H2O ×

g0.18mol1

= 0.3 mol H2O

3.097.0

97.0χ42SOH += = 0.76

Acetic Acid (CH3CO2H):

g05.60

mol1L


solng.100HCOCHg99 3

323 ××× = 17 mol/L

g05.60

mol1kg

g1000OHg1

HCOCHg99

2

23 ×× = 1600 mol/kg ≈ 2000 mol/kg

99 g CH3CO2H × g05.60

mol1 = 1.6 mol CH3CO2H; 1 g H2O ×

g0.18mol1

= 0.06 mol H2O

06.06.1

6.1χ HCOCH 23 += = 0.96

Ammonia (NH3):

g0.17

mol1L

cm1000cm

g90.0solng.100

NHg28 3

33 ××× = 15 mol/L

g0.17

mol1kg

g1000OHg72

NHg28

2

3 ×× = 23 mol/kg

28 g NH3 H g0.17

mol1 = 1.6 mol NH3; 72 g H2O H g0.18

mol1 = 4.0 mol H2O

6.10.4

6.1χ3NH += = 0.29

17. 25 mL C5H12 × mL

g63.0 = 16 g C5H12; 25 mL ×

g15.72mol1

mLg63.0× = 0.22 mol C5H12


45 mL C6H14 × mL

g66.0 = 30. g C6H14; 45 mL × g17.86

mol1mL

g66.0× = 0.34 mol C6H14

Mass % pentane = masstotal

pentanemass × 100 =

g.30g16g16

+ × 100 = 35%

χpentane = moltotal

pentanemol = mol34.0mol22.0

mol22.0+

= 0.39

Molality = hexanekgpentanemol =

kg030.0mol22.0

= 7.3 mol/kg

Molarity = solutionL

pentanemol = L1mL1000

mL45mL25mol22.0

×+

= 3.1 mol/L

18. If we have 100.0 mL of wine:

12.5 mL C2H5OH × mL

g789.0 = 9.86 g C2H5OH and 87.5 mL H2O × mL

g00.1 = 87.5 g H2O

Mass % ethanol = g86.9g5.87

g86.9+

× 100 = 10.1% by mass

Molality = g07.46

mol1OHkg0875.0

OHHCg86.9

2

52 × = 2.45 mol/kg

19. If we have 1.00 L of solution:

1.37 mol citric acid × mol

g1.192 = 263 g citric acid

1.00 × 103 mL solution × mL

g10.1 = 1.10 × 103 g solution

Mass % of citric acid = g1010.1

g2633×

× 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 = OHkg84.0acidcitricmol37.1

2 = 1.6 mol/kg

840 g H2O × g0.18

mol1 = 47 mol H2O;

37.14737.1χ acidcitric +

= = 0.028

672 CHAPTER 17 PROPERTIES OF SOLUTIONS

20. ethanolkg00.1acetonemol00.1 = 1.00 molal; 1.00 × 103 g C2H5OH ×

g07.46mol1

= 21.7 mol C2H5OH

χacetone = 7.2100.1

00.1+

= 0.0441

1 mol CH3COCH3 H g788.0

mL1COCHCHmol

COCHCHg08.58

33

33 × = 73.7 mL CH3COCH3

1.00 × 103 g ethanol × g789.0

mL1 = 1270 mL; total volume = 1270 + 73.7 = 1340 mL

Molarity = L34.1

mol00.1 = 0.746 M

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

Density = mL104

OHg.100POHg0.10volumemass 243 += = 1.06 g/mL = 1.06 g/cm3

Mol H3PO4 = 10.0 g × g99.97

mol1 = 0.102 mol H3PO4

Mol H2O = 100. g × g02.18

mol1 = 5.55 mol H2O

Mole fraction of H3PO4 = mol)55.5102.0(

POHmol102.0 43

+ = 0.0180

OH2

χ = 1.0000 – 0.0180 = 0.9820

Molarity = solnL104.0

POHmol102.0 43 = 0.981 mol/L

Molality = solventkg100.0

POHmol102.0 43 = 1.02 mol/kg

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

0.100 kg H2O × KClmol

g55.74kg

KClmol0.2× = 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. Because 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 H2O). Let x = mass of NaCl.

Mass % = 15 = x

x+.100

× 100, 1500 + 15x = (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 × mL

g79.0 = 79 g CH3OH

Mass % = 25 = x

x+79

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

Dissolve 26 g NaOH in 100. mL CH3OH. d. To make sure we have enough solution, let’s use 100. mL (100. g) of H2O. Let x = mol

C6H12O6.

100. g H2O × g02.18OHmol1 2 = 5.55 mol H2O

6126 OHCχ = 0.10 =

55.5+xx , (0.10)x + 0.56 = x, x = 0.62 mol C6H12O6

0.62 mol C6H12O6 × mol

g2.180 = 110 g C6H12O6

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

6126 OHCχ = 0.10.

Thermodynamics of Solutions and Solubility 23. “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 strengths 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 strengths 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 ΔHsoln is either a small positive number (endothermic) or a


small negative number (exothermic), with a value close to zero. For polar solutes in nonpolar solvents and vice versa, ΔHsoln 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.

24. 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. Because 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.

25. 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

__________________________________________________________________________________________________________ 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.

26. 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 hydrations between Cl and I−. Thus the chloride ion is more strongly hydrated than the iodide ion.

27. Both Al(OH)3 and NaOH are ionic compounds. Because 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.

CHAPTER 17 PROPERTIES OF SOLUTIONS 675 28. a. Mg2+; smaller size, higher charge b. Be2+; smaller size c. Fe3+; smaller size, higher charge d. F−; smaller size e. Cl−; smaller size f. SO4

2−; higher charge 29. Water is a polar molecule capable of hydrogen bonding. Polar molecules, especially

molecules capable of hydrogen bonding, and ions can be hydrated. For covalent compounds, as polarity increases, the attraction to water (hydration) increases. For ionic compounds, as the charge of the ions increase and/or the size of the ions decrease, the attraction to water (hydration) increases.

a. CH3CH2OH; CH3CH2OH is polar, whereas CH3CH2CH3 is nonpolar.

b. CHCl3; CHCl3 is polar, whereas CCl4 is nonpolar.

c. CH3CH2OH; CH3CH2OH is much more polar than CH3(CH2)14CH2OH. 30. 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

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 Kr FS

F F

O C OS

O O

676 CHAPTER 17 PROPERTIES OF SOLUTIONS 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 31. 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 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.

32. The main intermolecular forces are: hexane (C6H14): London dispersion; chloroform (CHCl3): dipole-dipole, London

dispersion; methanol (CH3OH): H-bonding; and 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.

33. 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 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 because the solubility of a gas typically decreases with increasing temperature.

34. Henry’s law is obeyed most accurately for dilute solutions of gases that do not dissociate in

or react with the solvent. O2 will bind to hemoglobin in the blood. Given this reaction in the solvent, O2(g) in blood does not follow Henry’s law.

C

H H

O

C CH

H

H

H


35. Pgas = kC, 0.790 atm = k × L

mol1021.8 4−× , k = 962 L atm/mol

Pgas = kC, 1.10 atm = mol

atmL962 × C, C = 1.14 × 10-3 mol/L

36. 750. mL grape juice × g07.46

OHHCmol1mL

OHHCg79.0juicemL.100

OHHCmL12 525252 ××

OHHCmol2

COmol2

52

2× = 1.54 mol CO2 (carry extra significant figure)

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

L1075

)K298(Kmol

atmL08206.0n

VRTn

P 3

gg

CO2 −×

⎟⎟⎠

⎞⎜⎜⎝

⎛

== = 326ng

L750.0

nmol

atmL32kCP aqCO2

×== = (42.7)naq

2COP = 326ng = (42.7)naq and from above naq = 1.54 − ng; solving: 326ng = 42.7(1.54 − ng), 369ng = 65.8, ng = 0.18 mol

2COP = 326(0.18) = 59 atm in gas phase; 59 atm =

molatmL32 × C, C = 1.8 mol CO2/L in wine

37. 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 a kinetic energy greater than the attractive forces between the gas molecules and the solvent molecules. More gas molecules are able to escape to the vapor phase and the solubility of the gas decreases.

Vapor Pressures of Solution

38. Mol C3H8O3 = 164 g × g09.92

mol1 = 1.78 mol C3H8O3

Mol H2O = 338 mL × g02.18

mol1mL

g992.0× = 18.6 mol H2O

oOHOHsoln 22

PχP = = mol)6.1878.1(

mol6.18+

× 54.74 torr = 0.913 × 54.74 torr = 50.0 torr


39. solutioninmoltotal

solutioninOHHCmolχ;PχP 52OHHC

oOHHCOHHCsoln 525252

==

53.6 g C3H8O3 × g09.92

OHCmol1 383 = 0.582 mol C3H8O3

133.7 g C2H5OH × g07.46

OHHCmol1 52 = 2.90 mol C2H5OH; total mol = 0.582 + 2.90 = 3.48 mol

113 torr = oOHHC

oOHHC 5252

P,Pmol48.3mol90.2

× = 136 torr

40. Compared to H2O, solution d (methanol-water) will have the highest vapor pressure since methanol is more volatile than water ( o

OH2P = 23.8 torr at 25°C). 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. Because there are more solute particles in solution c, the vapor pressure of solution c will be the lowest.

41. 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.

42. Ptotal = totalmol0800.0

ClCHmol0300.0χ;PχP;PP 22LClCH

oLBrCHClCH 222222

==+ = 0.375

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

In the vapor: torr0.57torr9.49

PP

χtotal

ClCHVClCH

22

22== = 0.875; V

BrCH 22χ = 1.000 – 0.875 = 0.125

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

43. o

BBBoBBB P/Pχ,PχP == = 0.900 atm/0.930 atm = 0.968

0.968 = moltotal

benzenemol ; mol benzene = 78.11 g C6H6 × g11.78

mol1 = 1.000 mol

Let x = mol solute; then: χB = 0.968 = x+1.000

mol1.000 , 0.968 + (0.968)x = 1.000, x = 0.033 mol

Molar mass = mol033.0g0.10 = 303 g/mol ≈ 3.0 × 102 g/mol

44. total

VCSCS PχP

22= = 0.855(263 torr) = 225 torr


torr375torr225

PP

χ,PχP oCS

CSLCS

oCS

LCSCS

2

2

2222=== = 0.600

45. a. 25 mL C5H12 × 15.72

mol1mL

g63.0× = 0.22 mol C5H12

45 mL C6H14 × 17.86

mol1mL

g66.0× = 0.34 mol C6H14; total mol = 0.22 + 0.34 = 0.56 mol

mol56.0mol22.0

solutioninmoltotalsolutioninpentanemolχ L

pen == = 0.39, Lhexχ = 1.00 − 0.39 = 0.61

o

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

Ptotal = Ppen + Phex = 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:

torr290

torr100.2PP

vapormoltotalvaporinpentanemolχ

2

total

penVpen

×=== = 0.69

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

46. ;PχP,PχP o

benLbenpen

otol

Ltoltol == for the vapor, V

Aχ = PA/Ptotal. Because the mole fractions of

benzene and toluene are equal in the vapor phase, bentol PP = . torr95)χ00.1()torr28(χ,P)χ00.1(PχPχ L

tolLtol

oben

Ltol

oben

Lben

otol

Ltol −− ===

L

benLtol

Ltol χ;77.0χ,95χ123 == = 1.00 – 0.77 = 0.23

47. Ptotal = Pmeth + Pprop, 174 torr = L

methLprop

Lprop

Lmeth χ000.1χ);torr6.44(χ)torr303(χ −+ =

174 = 500.0χ258129,torr6.44)χ000.1(χ303 L

methLmeth

Lmeth ==−+

500.0000.1χ Lprop −= 500.0=

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

oBB

oAAtotal PχPχP +=


A nonideal liquid-liquid solution does not follow Raoult’s law, giving a total pressure either greater than predicted by Raoult’s law (positive deviation) or less than predicted (negative deviation). In an ideal solution, the strengths of the intermolecular forces in solution are equal to the strengths of the intermolecular forces in pure solute and pure solvent. When this is true, ΔHsoln = 0 and ΔTsoln = 0. For positive deviations from Raoult’s law, the solution has weaker intermolecular forces in solution than in pure solute and pure solvent. Positive deviations have ΔHsoln > 0 (are endothermic) and ΔTsoln < 0. For negative deviations, the solution has stronger intermolecular forces in solution than in pure solute or pure solvent. Negative deviations have ΔHsoln < 0 (are exothermic) and ΔTsoln > 0. Examples of each type of solution are: ideal: benzene-toluene positive deviations: ethanol-hexane negative deviations: acetone-water

49. 50.0 g CH3COCH3 × g08.58

mol1 = 0.861 mol acetone

50.0 g CH3OH × g04.32

mol1 = 1.56 mol methanol

644.0χ000.1χ;356.056.1861.0

861.0χ Lacetone

Lmethanol

Lacetone ==== −

+

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, in the vapor phase:

488.0512.0000.1χ;512.0torr6.188torr5.96

PPχ V

methanoltotal

acetoneVacetone ===== −

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.

50. 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 various solutions are not between these limits, so water and propanol do not form ideal solutions.

b. From the data, the vapor pressures of the various solutions are greater than if the

solutions behaved ideally (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.

CHAPTER 17 PROPERTIES OF SOLUTIONS 681 c. The interactions between propanol and water molecules are weaker than between the pure

substances because the solutions exhibit a positive deviation from Raoult’s law. d. At OH2

χ = 0.54, the vapor pressure is highest as compared to the other solutions. Because a solution boils when the vapor pressure of the solution equals the external pressure, the OH2

χ = 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 as compared to the other solutions.

51. No, the solution is not ideal. For an ideal solution, the strengths of intermolecular forces in

the solution are 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, this solution exhibits negative deviation from Raoult’s law.

52. 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, 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.

53. Solutions of A and B have vapor pressures less than ideal (see Figure 17.11 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. Because χB = 0.6 has a lower vapor pressure at the temperature of the plot than either pure A or pure B, 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 χB = 0.6 will have a higher boiling point than either pure A or pure B. (Note that because o

BP > oAP , B is more volatile than A.)

Colligative Properties 54. 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.

55. 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 π is:

π = MRT where 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 when 1 kg solvent ≈ 1 L solution. This occurs for dilute solutions of water because OH2

d = 1.00 g/cm3.

56. 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 ΔT, the more precise is the measurement, and the more precise is the calculated molar mass. However, if the solute won’t dissolve in camphor, then camphor is no good, and another solvent must be chosen that will dissolve the solute.

57. Molality = m = COHNg60.06

COHNmol1kg

g1000OHg150.0COHNg27.0

solventkgsolutemol

42

42

2

42 ××= = 3.00 molal

ΔTb = Kbm = molal

C51.0 ° × 3.00 molal = 1.5°C

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

58. ΔTf = Kfm, ΔTf = 1.50°C = molal

C86.1 ° × m, m = 0.806 mol/kg

0.200 kg H2O × 383

383

2

383

OHCmolOHCg92.09

OHkgOHCmol0.806

× = 14.8 g C3H8O3

59. Molality = m = g07.62

mol1kg

g1000OHg0.50OHCg0.50

2

262 ×× = 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

60. ΔT = 25.50°C − 24.59°C = 0.91°C = Kfm, m = molal/C1.9C91.0

o

o

= 0.10 mol/kg


Mass H2O = 0.0100 kg t-butanol ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛OHmol

OHg18.02butanol-tkg

OHmol0.10

2

22 = 0.018 g H2O

61. m = kg600.0

g0.58mol1g0.24 ×

= 0.690 mol/kg; ΔTb = Kbm = 0.51°C kg/mol × 0.690 mol/kg = 0.35°C

Tb = 99.725°C + 0.35°C = 100.08°C

62. π = MRT, M = K298molKatmL08206.0

atm00.8RTπ

11 ×−−= = 0.327 mol/L

63. ΔTb = 77.85°C − 76.50°C = 1.35°C; m = mol/kgC03.5

C35.1KTΔ

o

o

b

b = = 0.268 mol/kg

Mol biomolecule = 0.0150 kg solvent × solventkg

nhydrocarbomol268.0 = 4.02 × 10−3 mol

From the problem, 2.00 g biomolecule was used, which must contain 4.02 × 10−3 mol biomolecule. The molar mass of the biomolecule is:

mol1002.4

g00.23−×

= 498 g/mol

64. M = K.300

molKatmL08206.0

torr760atm1torr745.0

RTπ

×

×= = 3.98 × 10−5 mol/L

1.00 L × L

mol1098.3 5−× = 3.98 × 10−5 mol catalase

Molar mass = mol1098.3

g00.105−×

= 2.51 × 105 g/mol

65. ΔTf = Kfm, m = benzenekg

thyroxinemol105.86mol/kgC12.5

C300.0KTΔ 2

f

f−×

°° ==

The moles of thyroxine present are:

0.0100 kg benzene × benzenekg

thyroxinemol105.86 2−× = 5.86 × 410− mol thyroxine


From the problem, 0.455 g thyroxine was used; this must contain 5.86 × 410− mol thyroxine. The molar mass of the thyroxine is:

molar mass = mol10

g455.0486.5 −×

= 776 g/mol

66. m = mol/kgC86.1

C0.30KTΔ

o

o

f

f = = 16.1 mol C2H6O2/kg

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

15.0 L H2O × OHkg

OHCmol1.16OHL

OHkg00.1

2

262

2

2 × = 242 mol C2H6O2

Volume C2H6O2 = 242 mol C2H6O2 × g11.1

cm1OHCmol

g07.62 3

262× = 13,500 cm3 = 13.5 L

ΔTb = Kbm = molal

C51.0 o

× 16.1 molal = 8.2°C; Tb = 100.0°C + 8.2°C = 108.2°C

67. M = g100.9

mol1L

g0.14×

× = 1.1 × 10−5 mol/L; π = MRT

At 298 K: π = atm

torr760K298molK

atmL08206.0L

mol101.1 5

×××× −

, π = 0.20 torr

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

ΔTf = Kfm = molal

C86.1 o

× 1.1 × 10−5 molal = 2.0 × 10−5°C

68. Osmotic pressure is better for determining the molar mass of large molecules. A temperature

change of 10-5°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.

69. π = MRT = K298molK

atmL08206.0Lmol1.0

×× = 2.45 atm ≈ 2 atm

π = 2 atm × atm

Hgmm760 ≈ 2000 mm ≈ 2 m

The osmotic pressure would support a mercury column of approximately 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

70. ΔTf = 5.51 - 2.81 = 2.70°C; C/molal5.12

C2.70KTΔ

o

o

f

f ==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).

2.128

x = moles naphthalene and

2.17860.1 x−

= moles anthracene

)2.178(2.128

)2.128()2.128(60.1)2.178(1005.1,solventkg0200.0

2.17860.1

2.128solventkg

solutemol527.0 2 xxxx

−+×

−+

== −

(50.0)x + 205 = 240., (50.0)x = 240. − 205, (50.0)x = 35, x = 0.70 g naphthalene So the mixture is:

g60.1g70.0 × 100 = 44% naphthalene by mass and 56% anthracene by mass

71. 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 any bacteria present, causing them to die.

Properties of Electrolyte Solutions 72. A strong electrolyte dissociates completely into ions in solution. A weak electrolyte

dissociates only partially into ions in solution. Colligative properties depend on the total number of particles in solution. By measuring a property such as freezing-point depression, boiling-point elevation, or osmotic pressure, we can calculate the van’t Hoff factor (i) to see if an electrolyte is strong or weak.

73. 19.6 torr = )torr8.23(χ OH2

, OH2χ = 0.824; soluteχ = 1.000 - 0.824 = 0.176

0.176 is the mole fraction of all the solute particles present. Because NaCl dissolves to produce two ions in solution (Na+ and Cl−), 0.176 is the mole fraction of Na+ and Cl− ions present. The mole fraction of NaCl is 1/2 (0.176) = 0.0880 = NaClχ .

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

686 CHAPTER 17 PROPERTIES OF SOLUTIONS 74. 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 that contains 1.00 kg of water (solvent).

0.344 mol NaCl × mol

g44.58 = 20.1 g NaCl

Mass % NaCl = g1.20g1000.1

g1.203 +×

× 100 = 1.97%

75. Na3PO4(s) → 3 Na+(aq) + PO4

3−(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 (assuming complete dissociation): 4.0(0.010 molal) = 0.040 molal for the Na3PO4 solution; 3.0(0.020 molal) = 0.060 molal

for the CaBr2 solution; 2.0(0.020 molal) = 0.040 molal for the KCl solution; slightly greater than 0.020 molal for the HF solution because HF only partially dissociates in water (it is a weak acid).

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

because χ increases. Therefore, the 0.020 m HF solution will have the highest vapor pressure because 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). 76. The solutions of glucose, 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 mole of compound.

a. pure water b. CaCl2

solution c. CaCl2 solution

d. pure water e. CaCl2 solution 77. The van’t Hoff factor i is the number of moles of particles (ions) produced for every mole 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.

78. a. As discussed in Figure 17.16 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 will be the same in both directions, and the levels of the liquids in the two arms will stabilize. The height difference between the two arms will be a measure of the osmotic pressure of the NaCl solution.

b. Initially, H2O molecules will have a net migration into the NaCl side. However, NaCl molecules 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 NaCl will be equal in both chambers, and the levels of liquid will be equal.

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

π = iMRT = 2.0 × K293molK

atmL08206.0L

mol10.0×× = 4.8 atm

A pressure greater than 4.8 atm should be applied to ensure purification by reverse osmosis. 80. 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 (Assuming water 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 81. 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


82. ΔTf = iKfm, i = molal0225.0molal/C86.1

C110.0K

TΔo

o

f

f

×=

m = 2.63 for 0.0225 m CaCl2

i =0910.086.1

440.0×

= 2.60 for 0.0910 m CaCl2; i = 278.086.1330.1×

= 2.57 for 0.278 m CaCl2

Note that i is less than the ideal value of 3.0 for CaCl2. This is due to ion pairing in solution.

83. π = iMRT, M = iRTπ =

K298molK

atmL08206.000.2

atm50.2

×× = 5.11 × 210− mol/L

Molar mass of compound =

Lmol1011.5L1000.0

g500.02−×

× = 97.8 g/mol

84. 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:

OHkg

CaClmol71.6CaClg98.110

CaClmol1kg

g1000OHg0.100

CaClg5.74

2

2

2

2

2

2 =××

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 17.82, i ≈ 2.6; ΔTf = iKfm = 2.6 × 1.86 × 6.71 = 32°C; Tf = −32°C.

Assuming i = 2.6, a saturated CaCl2 solution will not melt ice at −34°C (−29°F). Additional Exercises 85. 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 parti-cles compared to a solution. Because of this, colloids will scatter light, whereas solutions will not. The scattering of light by a colloidal suspension is called the Tyndall effect.

86. The micelles form so that the ionic ends of the detergent molecules, the SO4

- ends, are exposed to the polar water molecules on the outside, whereas 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.

CHAPTER 17 PROPERTIES OF SOLUTIONS 689 87. Coagulation is the destruction of a colloid by the aggregation of many suspended particles to

form a large particle that settles out of solution. 88. 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 oppositely 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.

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

mol ethanol = 46 mL C2H5OH × g07.46

OHHCmol1mL

g79.0 52× = 0.79 mol C2H5OH

molarity = L1000.0

mol79.0 = 7.9 M ethanol

90. Mass of H2O = 160. mL mL

g995.0× = 159 g = 0.159 kg

Mol NaDTZ = 0.159 kg kg

mol378.0× = 0.0601 mol

Molar mass of NaDTZ = mol0601.0g4.38 = 639 g/mol

= nonpolar hydrocarbon

= detergent molecule

= SO4-

= dirt


Psoln = oOHOH 22

Pχ ; mol H2O = 159 g g02.18

mol1× = 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.

OH2χ =

mol82.8mol120.0mol82.8+

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

91. ΔT = Kfm, m = C/molal86.1

C79.2K

TΔo

o

f= = 1.50 molal

a. ΔT = Kbm, ΔT = (0.51EC/molal)(1.50 molal) = 0.77EC, Tb = 100.77EC

b. waterowaterwatersoln χ,PχP = =

solutemolOHmolOHmol

2

2

+

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

mol H2O = 1.00 × 103 g H2O × OHg02.18

OHmol1

2

2 = 55.5 mol H2O

χwater = 5.5550.1

mol5.55+

= 0.974; Psoln = (0.974)(23.76 mm Hg) = 23.1 mm Hg

c. We assumed ideal behavior in solution formation, we assumed the solute was nonvolatile, and we assumed i = 1 (no ions formed).

92. ΔT = imKf, i = fK

TΔm

=

molkgC86.1

kg500.0mol250.0

C79.2o

o

× = 3.00

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

38.68 g Cl g45.35

Clmol1× = 1.091 mol Cl

mol M = 1.091 mol Cl Clmol2Mmol1

× = 0.5455 mol M

Molar mass of M = Mmol5455.0

Mg32.61 = 112.4 g/mol; M is Cd, so Q = CdCl2.

CHAPTER 17 PROPERTIES OF SOLUTIONS 691 93.

Benzoic acid is capable of hydrogen-bonding, but a significant part of benzoic acid is the nonpolar benzene ring. In benzene, a hydrogen-bonded dimer forms.

The dimer is relatively nonpolar and thus more soluble in benzene than in water. Because benzoic acid forms dimers in benzene, the effective solute particle concentration will be less than 1.0 molal. Therefore, the freezing-point depression would be less than 5.12°C (ΔTf = Kf m).

94. Benzoic acid (see Exercise 93) would be more soluble in a basic solution 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.

95. a. NH4NO3(s) → NH4

+(aq) + NO3−(aq) ΔHsoln = ?

Heat gain by dissolution process = heat loss by solution; We will keep all quantities

positive in order to avoid sign errors. Because the temperature of the water decreased, the dissolution of NH4NO3 is endothermic (ΔH is positive). Mass of solution = 1.60 + 75.0 = 76.6 g

Heat loss by solution = gCJ18.4

o × 76.6 g × (25.00°C − 23.34°C) = 532 J

ΔHsoln = 34

34

34 NONHmolNONHg05.80

NONHg60.1J532

× = 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/mol

NH4+(aq) + NO3

−(aq) → NH4NO3(s) ΔH = −ΔHsoln = -26.6 kJ/mol

____________________________________________________________________________________________________________ NH4

+(g) + NO3−(g) → NH4NO3(s) ΔH = ΔHhyd − ΔHsoln = -657 kJ/mol

C O H

O

O H O

C

O H O

C

692 CHAPTER 17 PROPERTIES OF SOLUTIONS 96. a. The average values for each ion are: 300. mg Na+, 15.7 mg K+, 5.45 mg Ca2+, 388 mg Cl−, and 246 mg lactate (C3H5O3

−) Note: Because 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− × −

353

353

OHCmg07.89OHNaCmg06.112 = 309.5 mg sodium lactate

The only source of Ca2+ is CaCl2C2H2O.

5.45 mg Ca2+ × +

•2

22

Camg40.08O2HCaClmg147.0 = 19.99 or 20.0 mg CaCl2C2H2O

The only source of K+ is KCl.

15.7 mg K+ × +Kmg10.39KClmg55.74 = 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+ × +Namg99.22NaClmg44.58 = 601.2 mg NaCl

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

20.0 mg CaCl2C2H2O × O2HCaClmg147.0

Clmg70.90

22 •

−

= 9.6 mg Cl−

20.0 mg CaCl2C2H2O = 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− 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 CaCl2C2H2O, 29.9 mg KCl, and 601.2 mg NaCl.

CHAPTER 17 PROPERTIES OF SOLUTIONS 693 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:

mg99.22

mmol1mL.100

Namg285×

+

= 0.124 M; mg10.39

mmol1mL.100

Kmg1.14×

+

= 0.00361 M

mg08.40

mmol1mL.100Camg9.4 2

×+

= 0.0012 M; mg45.35

mmol1mL.100

Clmg368×

−

= 0.104 M

mg07.89

mmol1mL.100

OHCmg231 353 ×−

= 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 = molK

atmL08206.0L

mol259.0× × 310. K = 6.59 atm

Similarly, at maximum concentrations, the concentration for 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 0.287 M.

π = molK

atmL08206.0L

mol287.0× × 310. K = 7.30 atm

Osmotic pressure ranges from 6.59 atm to 7.30 atm.

97. .)150(χ)511(χPPP;PχP;PP

15.0χ Lhex

Lpenhexpentotal

open

Lpenpen

total

penVpen ++ =====

Because L

penLpen

Lpentotal

Lpen

Lhex χ361.150.)150)(χ000.1()511(χP:χ000.1χ +−+− ===

Lpen

LpenL

pen

Lpen

total

penVpen χ511)χ361.150(15.0,

χ361.150)511(χ

15.0,PP

χ === ++

23 + 54 Lpenχ = 511 L

penχ , Lpenχ =

45723

= 0.050

98. 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 temperature. The higher the cooking temperature, the faster is 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.

694 CHAPTER 17 PROPERTIES OF SOLUTIONS c. When water freezes from a solution, it freezes as pure water, leaving behind a more

concentrated salt solution. d. On the CO2 phase diagram, the triple point is above 1 atm, and 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.

99. 14.2 mg CO2 × 2COmg01.44

Cmg01.12 = 3.88 mg C; % C = mg80.4mg88.3

× 100 = 80.8% C

1.65 mg H2O × OHmg02.18

Hmg016.2

2 = 0.185 mg H; % H =

mg80.4mg185.0 × 100 = 3.85% H

Mass % O = 100.00 − (80.8 + 3.85) = 15.4% O Out of 100.00 g:

80.8 g C × g01.12

mol1 = 6.73 mol C;

963.073.6

= 6.99 ≈ 7

3.85 g H × g008.1

mol1 = 3.82 mol H;

963.082.3 = 3.97 ≈ 4

15.4 g O × g00.16

mol1 = 0.963 mol O;

963.0963.0

= 1.00

Therefore, the empirical formula is C7H4O.

ΔTf = Kfm, m =molal/C.40C3.22

KTΔ

o

o

f

f = = 0.56 molal

Mol anthraquinone = 0.0114 kg camphor × camphorkg

oneanthraquinmol56.0 = 6.4 × 10−3 mol

Molar mass = mol104.6

g32.13−×

= 210 g/mol

The empirical mass of C7H4O is 7(12) + 4(1) + 16 ≈ 104 g/mol. Because the molar mass is twice the empirical mass, the molecular formula is C14H8O2.

100. Out of 100.00 g, there are:

31.57 g C × g011.12

Cmol1 = 2.628 mol C;

628.2628.2

= 1.000


5.30 g H × g008.1Hmol1

= 5.26 mol H; 628.226.5

= 2.00

63.13 g O × g999.15

Omol1 = 3.946 mol O;

628.2946.3

= 1.502

Empirical formula: C2H4O3; use the freezing-point data to determine the molar mass.

m =C/molal1.86

C5.20KTΔ

o

o

f

f = = 2.80 molal

Mol solute = 0.0250 kg × kg

solutemol80.2 = 0.0700 mol solute

Molar mass = mol0700.0g56.10

= 151 g/mol

The empirical formula mass of C2H4O3 = 76.051 g/mol. Because the molar mass is about twice the empirical mass, the molecular formula is C4H8O6, which has a molar mass of 152.101 g/mol.

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 atomic masses.

101. a. m = mol/kgC12.5

C32.1KTΔ

o

o

f

f = = 0.258 mol/kg

Mol unknown = 0.01560 kg × kg

unknownmol258.0 = 4.02 × 10−3 mol

Molar mass of unknown = mol1002.4

g22.13−×

= 303 g/mol

Uncertainty in temperature = 32.104.0 × 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 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:

696 CHAPTER 17 PROPERTIES OF SOLUTIONS 1. make the solution 10 times more concentrated (may be solubility problem) 2. use a solvent with a larger Kf value, e.g., camphor

102. MX ⇌ M+ + X− Ksp = [M+][X−]; ΔT = Kfm, m = 86.1028.0

KTΔ

f= = 0.015 mol/kg

g1000

kg1kg

mol015.0× × 250 g = 0.00375 mol total solute particles (carrying extra sig. fig.)

Assume a solution density of 1.0 g/mL so that volume of solution = 250 mL.

[M+] = L0.25

)(0.00375/2 = 7.5 × 10−3 M, [X−] = L0.25

)(0.00375/2 = 7.5 × 10−3 M

Ksp = [M+][X−] = (7.5 × 10−3)(7.5 × 10−3) = 5.6 × 10−5 103. M3X2(s) → 3 M2+(aq) + 2 X3−(aq) Ksp = [M2+]3[X3−]2 Initial s = solubility (mol/L) 0 0 Equil. 3s 2s Ksp = (3s)3(2s)2 = 108s5; total ion concentration = 3s + 2s = 5s.

π = iMRT, iM = total ion concentrationK298molKatmL08206.0

atm1064.2RTπ

11

2

××

−−

−

==

= 1.08 × 10−3 mol/L 5s = 1.08 × 10−3 mol/L, s = 2.16 × 10−4 mol/L Ksp = 108s5 = 108(2.16 × 10−4)5 = 5.08 × 10−17

104. m =molal/C86.1C426.0

KTΔ

o

o

f

f = = 0.229 molal

Assuming a solution density = 1.00 g/mL, then 1.00 L contains 0.229 mol solute. NaCl → Na+ + Cl− i = 2; so: 2(mol NaCl) + mol C12H22O11 = 0.229 mol Mass NaCl + mass C12H22O11 = 20.0 g 2nNaCl +

112212 OHCn = 0.229 and 58.44(nNaCl) + )(3.342112212 OHCn = 20.0

Solving:

112212 OHCn = 0.0425 mol = 14.5 g and nNaCl = 0.0932 mol = 5.45 g


Mass % C12H22O11 = g0.20g5.14 × 100 = 72.5 % and 27.5% NaCl by mass

112212 OHCχ =

mol0932.0mol0425.0mol0425.0

+ = 0.313

105. m = kg5000.0

g0.100mol1g100.0 ×

= 2.00 × 10−3 mol/kg ≈ 2.00 × 10−3 mol/L (dilute solution)

ΔTf = iKfm, 0.0056°C = i(1.86 °C/molal)(2.00 × 10−3 molal), i = 1.5

If i = 1.0, percent dissociation = 0%, and if i = 2.0, percent dissociation = 100%. Because i = 1.5, the weak acid is 50.% dissociated.

HA ⇌ H+ + A− Ka = ]HA[

]A[]H[ −+

Because the weak acid is 50.% dissociated: [H+] = [A−] = [HA]o × 0.50 = 2.00 × 10−3 M × 0.50 = 1.0 × 10−3 M [HA] = [HA]0 − amount HA reacted = 2.00 × 10−3 M − 1.0 × 10−3 M = 1.0 × 10−3 M

Ka = ]HA[

]A[]H[ −+

= 3

33

100.1)100.1)(100.1(

−

−−

××× = 1.0 × 10−3

106. ΔTf = Kfm, m = molal/C86.1C40.5

KTΔ

o

o

f

f = = 2.90 molal

kg0500.0solventkg

solutemol90.2 n= , n = 0.145 mol of ions in solution

Because NaNO3 and Mg(NO3)2 are strong electrolytes: n = 2(x mol of NaNO3) + 3[y mol Mg(NO3)2] = 0.145 mol ions

In addition: 6.50 g = x mol NaNO3 ⎟⎠⎞

⎜⎝⎛

molg00.85

+ y mol Mg(NO3)2 ⎟⎠⎞

⎜⎝⎛

molg3.148

We have two equations: 2x + 3y = 0.145 and (85.00)x + (148.3)y = 6.50 Solving by simultaneous equations:

698 CHAPTER 17 PROPERTIES OF SOLUTIONS −(85.00)x − (127.5)y = −6.16 (85.00)x + (148.3)y = 6.50 ______________________ (20.8)y = 0.34, y = 0.016 mol Mg(NO3)2

Mass of Mg(NO3)2 = 0.016 mol × 148.3 g/mol = 2.4 g Mg(NO3)2, or 37% Mg(NO3)2 by mass Mass of NaNO3 = 6.50 g - 2.4 g = 4.1 g NaNO3, or 63% NaNO3 by mass

107. iM )K2.298(molKatmL08206.0

atm3950.0RTπ

11 −−== = 0.01614 mol/L = total ion concentration

0.01614 mol/L ++−−++ +++ == NaMgClClNaMg 22 2; MMMMMM (charge balance)

Combining: 0.01614 = ++ + NaMg 223 MM

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

443.58

and218.95 NaMg2

yMxM == ++ (Because V = 1.000 L.)

Total ion concentration = 443.58

2218.95

3 yx + = 0.01614 mol/L

Rearranging: 3x + (3.2585)y = 1.537

Solving by simultaneous equations: 3x + (3.2585)y = 1.537 −3(x + y) = −3(0.5000)

________________________________________________ (0.2585)y = 0.037, y = 0.14 g NaCl

Mass MgCl2 = 0.5000 g − 0.14 g = 0.36 g; mass % MgCl2 = g5000.0

g36.0 × 100 = 72%

108. Use the thermodynamic data to calculate the boiling point of the solvent.

At boiling point, ΔG = 0 = ΔH − TΔS, ΔH = TΔS, T = 11

3

molKJ95.95J/mol1090.33

SΔHΔ

−−

×= = 353.3 K

ΔT = Kbm, (355.4 K − 353.3 K) = (2.5 K kg/mol)(m), m = 5.21.2 = 0.84 mol/kg

Mass solvent = 150. mL × g1000

kg1mL

g879.0 × = 0.132 kg


Mass solute = 0.132 kg solvent × mol

g142solventkg

solutemol84.0 × = 15.7 g = 16 g solute

Challenge Problems 109. From the problem, L

CClL

HC 466χχ = = 0.500. We need the pure vapor pressures (Po) in order to

calculate the vapor pressure of the solution. C6H6(l) ⇌ C6H6(g) K o

HCHC 6666PP == at 25°C

== − o

)l(HC,fo

)g(HC,forxn 6666

GΔGΔGΔ 129.66 kJ/mol − 124.50 kJ/mol = 5.16 kJ/mol

ΔG° = −RT ln K, ln K === −−

×−−)K298()molKJ3145.8(

J/mol1016.5RT

GΔ11

3o

−2.08

K = o

HC 66P = e−2.08 = 0.125 atm

For CCl4: o

)l(CCl,fo

)g(CCl,forxn 44

GΔGΔGΔ −= = −60.59 kJ/mol − (−65.21 kJ/mol) = 4.62 kJ/mol

K = oCCl4

P = ⎟⎟⎠

⎞⎜⎜⎝

⎛

×−

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−−=

K298molKJ3145.8J/mol4620exp

RTGΔexp 11

o

= 0.155 atm

== o

HCL

HCHC 666666PχP 0.500(0.125 atm) = 0.0625 atm;

4CClP = 0.500(0.155 atm) = 0.0775 atm

1400.00625.0

atm0775.0atm0625.0atm0625.0

PP

χtotal

HCVHC

66

66===

+ = 0.446; V

CCl4χ = 1.000 − 0.446

= 0.554 110. Let L

Aχ = mole fraction A in solution, so 1.000 − LAχ = L

Bχ . From the problem, VAχ = 2 L

Aχ .

VAχ =

total

A

PP

= )torr0.100)(χ000.1()torr0.350(χ

)torr0.350(χLA

LA

LA

−+

VAχ = 2 L

Aχ = LAL

A

LA χ)0.250(,

0.100χ)0.250(χ)0.350(+

= 75.0, LAχ = 0.300

The mole fraction of A in solution is 0.300.

700 CHAPTER 17 PROPERTIES OF SOLUTIONS 111. For the second vapor collected, V

2,Bχ = 0.714 and V2,Tχ = 0.286. Let L

2,Bχ = mole fraction of benzene in the second solution and L

2,Tχ = mole fraction of toluene in the second solution. L

2,TL

2,B χχ + = 1.000

V2,Bχ = 0.714 =

TB

B

total

B

PPP

PP

+= =

)torr0.300)(χ000.1()torr0.750(χ)torr0.750(χ

L2,B

L2,B

L2,B

−+

Solving: L2,Bχ = 0.500 = L

2,Tχ

This second solution came from the vapor collected from the first (initial) solution, so, V1,Bχ =

V1,Tχ = 0.500. Let L

1,Bχ = mole fraction benzene in the first solution and L1,Tχ = mole fraction

of toluene in first solution. L1,T

L1,B χχ + = 1.000.

V1,Bχ = 0.500 =

TB

B

total

B

PPP

PP

+= =

)torr0.300)(χ000.1()torr0.750(χ)torr0.750(χL

1,BL

1,B

L1,B

−+

Solving: L1,Bχ = 0.286

The original solution had χB = 0.286 and χT = 0.714.

112. a. Freezing-point depression is determined using molality for the concentration units,

whereas molarity units are used to determine osmotic pressure. We need to assume that the molality of the solution equals the molarity of the solution.

b. Molarity =

solutionliterssolventmoles ; molality =

solventkgsolventmoles

When the liters of solution equal the kilograms of solvent present for a solution, then

molarity equals molality. This occurs for an aqueous solution when the density of the solution is equal to the density of water, 1.00 g/cm3. The density of a solution is close to 1.00 g/cm3 when not a lot of solute is dissolved in solution. Therefore, molarity and molality values are close to each other only for dilute solutions.

c. ΔT = Kf m, m = fKTΔ =

kg/molC1.86C0.621

o

o

= 0.334 mol/kg

Assuming 0.334 mol/kg = 0.334 mol/L:

π = MRT = K298molK

atmL08206.0L

mol334.0×× = 8.17 atm


d. m = bKTΔ =

kg/molC0.51C2.0

o

o

= 3.92 mol/kg

This solution is much more concentrated than the isotonic solution in part c. Here, water will leave the plant cells in order to try to equilibrate the ion concentration both inside and outside the cell. Because there is such a large concentration discrepancy, all the water will leave the plant cells, causing them to shrivel and die.

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

g32.84

MgCOmol1L

g10560L

mg560mL

(s)MgCOμg560 33

3 ×× −

==

= 6.6 × 10−3 mol MgCO3/L

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

M L

mol32.0K.300molKatmL08206.0

atm0.8RTπ

11 ===×−−

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 H2O. When V + 45 L of water has been processed, the moles of solute particles will equal:

6.6 × 10−3 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 CO3

2− ions, then the solute concentration increases to 1.3 × 10−2 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 of less than 0.32 mol/L. Salt water has a solute concentration of 2(0.60 M) = 1.2 mol/L ions. The solute concentration of salt water is much too high for this reverse osmosis unit to work.

114. m = C/molal1.86

C0.406K

TΔo

o

f= = 0.218 mol/kg

π = MRT, where M = mol/L; we must assume that molarity = molality so that 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. π = (0.218 M)(0.08206 L atm K−1mol−1)(298 K) = 5.33 atm

115. a. Assuming no ion association between SO4

2−(aq) and Fe3+(aq), then i = 5 for Fe2(SO4)3. π = iMRT = 5(0.0500 mol/L)(0.08206 L atm K−1 mol−1)(298 K) = 6.11 atm b. Fe2(SO4)3(aq) → 2 Fe3+(aq) + 3 SO4

2− (aq)

Under ideal circumstances, 2/5 of π calculated above results from Fe3+ and 3/5 results from SO4

2−. The contribution to π from SO42− is 3/5 × 6.11 atm = 3.67 atm. Because SO4

2− is assumed unchanged in solution, the SO4

2− contribution in the actual solution will also be 3.67 atm. The contribution to the actual π from the Fe(H2O)6

3+ dissociation reaction is 6.73 − 3.67 = 3.06 atm.

The initial concentration of Fe(H2O)6

2+ is 2(0.0500) = 0.100 M. The setup for the weak acid problem is:

Fe(H2O)63+ → H+ + Fe(OH)(H2O)5

2+ Ka = ])OH(Fe[

])OH)(OH(Fe][H[362

252

+

++

Initial 0.100 M ~0 0 x mol/L of Fe(H2O)6

3+ reacts to reach equilibrium Equil. 0.100 − x x x

Total ion concentration = iM === −− )K298(molKatmL08206.0atm06.3

RTπ

11 0.125 M

0.125 M = 0.100 − x + x + x = 0.100 + x, x = 0.025 M

Ka = ])OH(Fe[

])OH)(OH(Fe][H[362

252

+

++

= 075.0

)025.0()025.0100.0(

)025.0(100.0

222

==−− x

x

Ka = 8.3 × 10−3 116. Initial moles VCl4 = 6.6834 g VCl4 × 1 mol VCl4/192.74 g VCl4 = 3.4676 × 10−2 mol VCl4

Total molality of solute particles = im = mol/kgC8.29

C97.5K

TΔo

o

f= , 0.200 mol/kg

Because we have 0.1000 kg CCl4, the total moles of solute particles present is:

0.200 mol/kg (0.1000 kg) = 0.0200 mol


2 VCl4 ⇌ V2Cl8 K = 24

82

]VCl[]ClV[

Initial 3.4676 × 10−2 mol 0 2x mol VCl4 reacts to reach equilibrium Equil. 3.4676 × 10−2 − 2x x Total moles solute particles = 0.0200 mol = mol VCl4 + mol V2Cl8 = 3.4676 × 10−2 − 2x + x 0.0200 = 3.4676 × 10−2 − x, x = 0.0147 mol

At equilibrium, we have 0.0147 mol V2Cl8 and 0.0200 - 0.0147 = 0.0053 mol VCl4. To determine the equilibrium constant, we need the total volume of solution in order to calculate equilibrium concentrations. The total mass of solution is 100.0 g + 6.6834 g = 106.7 g.

Total volume = 106.7 g × 1 cm3/1.696 g = 62.91 cm3 = 0.06291 L The equilibrium concentrations are:

[V2Cl8] = L06291.0

mol0147.0 = 0.234 mol/L; [VCl4] = L06291.0

mol0053.0 = 0.084 mol/L

K === 224

82

)084.0(234.0

]VCl[]ClV[ 33

117. a. π = iMRT, iM ===×−− K298molKatmL08206.0

atm83.7RTπ

11 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 × mL

g071.1 = 1071 g solution

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

mol NaCl added to solution = 10.7 g × g44.58

mol1 = 0.183 mol NaCl

Some of this NaCl dissociates into Na+ and Cl− (two moles of ions per mole of 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 = 183.0046.0 = 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 kilograms of solvent present in 1.000 L of 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 × kg060.1mol320.0 = 0.562°C

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

118. For 30.% A by moles in the vapor, 30. = BA

A

PPP+

× 100:

0.30 = yx

xyx

x)χ00.1(χ

χ30.0,χχ

χ

AA

A

BA

A

−++=

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 = ;30.070.0

30.0yx

y+

χB = 1.00 ! χA

Similarly, if vapor above is 50.% A: yx

yyx

y+

−+

== 00.1χ;χ BA

If vapor above is 80.% A: χA = ;80.020.0

80.0yx

y+

χB = 1.00 ! χA

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

Thus =VAχ

BA

A

PPP+

= yx

xyx

x70.030.0

30.000.1χand70.030.0

30.0 VB +

−+

=

If solution is 50.% A: VA

VB

VA χ00.1χandχ −

+==

yxx

If solution is 80.% A: VA

VB

VA χ00.1χand

20.080.080.0χ −=+

=yx

x


Marathon Problem 119. 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 × L

mol02313.0 = 3.530 × 10−4 mol NanA

Molar mass of NanA = mol10530.3g100.30

4

3

−

−

×× = 85.0 g/mol

To deduce the formula, we will assume various charges and numbers of oxygens present in the oxyanion, and then use the periodic table to see if an element fits the molar mass data. Assuming n = 1 so that 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−.

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 allow determination of the metal M in the formula. See

Exercise 16.47 for a review of relationships in body-centered cubic cells. In a body-centered cubic unit cell and there are two atoms per unit cell, and the body diagonal of the cubic cell is related to the radius of the metal by the equation 4r = 3l where l = cubic edge length.

l = 3

)cm10984.1(43r4 8−×= = 4.582 × 10−8 cm

Volume of unit cell = l3 = (4.582 × 10−8)3 = 9.620 × 10−23 cm3

Mass of M in a unit cell = 9.620 × 10−23 cm3 × 3cmg243.5 = 5.044 × 10−22 g M

Mol M in a unit cell = 2 atoms × 2310022.6mol1×

= 3.321 × 10−24 mol M

Molar mass of M = Mmol10321.3

Mg10044.524

22

−

−

×× = 151.9 g/mol

706 CHAPTER 17 PROPERTIES OF SOLUTIONS From the periodic table, M is europium (Eu). Given that the charge of Eu is +3, then the

formula of the salt is Eu(NO3)3•zH2O. c. Part c data allow determination of the molar mass of Eu(NO3)3•zH2O, from which we can

determine z, the number of waters of hydration.

B = iMRT, iM = )K298(molKatmL08206.0

torr760atm1torr558

RTπ

11 −−

⎟⎟⎠

⎞⎜⎜⎝

⎛

= = 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 mole of

Eu(NO3)3•zH2O dissolves to form four ions (Eu3+ + 3 NO3−), the molarity of

Eu(NO3)3•zH2O is 0.0300/4 = 0.00750 M.

Mol Eu(NO3)3•zH2O = 0.01000 L × L

mol00750.0 = 7.50 × 10−5 mol

Molar mass of Eu(NO3)3•zH2O = mol1050.7

g1045.335

3

−

−

×× = 446 g/mol

446 g/mol = 152.0 + 3(62.0) + z(18.0), z(18.0) = 108, z = 6.00 The formula for the strong electrolyte is Eu(NO3)3•6H2O.

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