solutions - gordon state collegeptfaculty.gordonstate.edu/lgoodroad/summer 2011/chem 1212...12.2 a...

Solutions

Review Questions

As seawater moves through the intestine, it flows past cells that line the digestive tract, whichconsist of largely fluid interiors surrounded by membranes. Although cellular fluids them-selves contain dissolved ions, including sodium and chloride, the fluids are more dilute thanseawater. Nature's tendency towards mixing (which tends to produce solutions of uniformconcentration), together with the selective permeability of the cell membranes (which allowwater to flow in and out, but restrict the flow of dissolved solids), cause a flow of solvent outof the body's cells and into the seawater.

12.2 A solution is a homogeneous mixture of two or more substances. A solution has at least twocomponents. The majority component is usually called the solvent and the minority compo-nent is usually called the solute.

12.3 A substance is soluble in another substance if they can form a homogeneous mixture. Thesolubility of a substance is the amount of the substance that will dissolve in a given amountof solvent. Many different units can be used to express solubility, including grams of soluteper 100 grams of solvent, grams of solute per liter of solvent, moles of solute per liter ofsolution, and moles of solute per kilogram of solvent.

12.4 Ideal gases do not interact with each other in any way (that is, there are no significant forcesbetween their constituent particles). When the two gases mix their potential energy remainsunchanged, so this does not drive the mixing. The tendency to mix is related, rather, to a con-cept called entropy. Entropy is a measure of energy randomization or energy dispersal in asystem. Recall that a gas at any temperature above 0 K has kinetic energy due to the motionof its atoms. When the gases are separated, their kinetic energies are also confined to thoseregions. However, when the gases mix the kinetic energy of each gas becomes spread out ordispersed over a larger volume. Therefore, the mixture of the two gases has greater energydispersal, or greater entropy, than the separated components. The pervasive tendency for allkinds of energy to spread out, or disperse, whenever they are not restrained from doing so isthe reason that two ideal gases mix.

12.5 Entropy is a measure of energy randomization or energy dispersal in a system. When twosubstances mix to form a solution there is an increase in randomness, due to the fact that thecomponents are no longer segregated to separate regions. This makes the formation of a solu-tion energetically favorable, even when it is endothermic.

12.6 Whether two substances will spontaneously mix to form a homogeneous solution is depend-ent on a number of different types of intermolecular forces including dispersion forces,dipole-dipole forces, hydrogen bonding, and ion-dipole forces.

12.7 A solution always forms if the solvent-solute interactions are comparable to, or stronger than,the solvent-solvent interactions and the solute-solute interactions.

12.8 The statement "like dissolves like" means that similar kinds of solvents dissolve similar kindsof solutes. Polar solvents, such as water, dissolve many polar or ionic solutes, and nonpolarsolvents, such as hexane, dissolve many nonpolar solutes.

457

Chapter 12 Solutions 461

12.30 (a) water, methanol, ethanol; dispersion, dipole-dipole, hydrogen bonding

(b) water, acetone, methanol, ethanol; dispersion, ion-dipole

(c) hexane, toluene, or CCl^; dispersion forces

(d) water, acetone, methanol, ethanol; dispersion, ion-dipole

HOCH2CH2CH2OH would be more soluble in water because it has -OH groups on both ends of the mole-cule, so it can hydrogen bond on both ends.

12.32 CH2C12 would be more soluble in water because it is a polar molecule and can exhibit dipole-dipole inter-actions with the water molecules. CCLj is a nonpolar molecule.

12.34

12.33 (a^ water; dispersion, dipole-dipole, hydrogen bonding

hexane; dispersion forces

water; dispersion, dipole-dipole

water; dispersion, dipole-dipole, hydrogen bonding





Energetics of Solution Formation

12.35 (a) endothermici

(b) The lattice energy is greater in magnitude than the heat of hydration.

(c) *

(b)

(c)

(d)

(a)

(b)

(c)

(d)

AH= -lattice energy

NH4Cl(s)

AH,

rNH/(a<j) -

I A

12.36

(d) The solution forms because chemical systems tend towards greater entropy.

(a) exothermic

(b) The lattice energy is smaller in magnitude than the heat of hydration.


12.40 Given: KNO3: lattice energy = - 163.8 kcal/mol, AHhydration = - 155.5 kcal/mol; 1.00 x 102 kj absorbedFind: AHsoInand m (KNO3)Conceptual Plan: Lattice Energy, AHhydration -» AHsoln (kcal) -» AHsoin (kcal) then

AHsoin = AHsou,,,, + AHhydraHon where AHMiute = - AHlatlke

mol, q — > AHsoin then mol — » g101. 11 g

Solution: AH^ = AHsoIute + AHhydration where AHsolute = -AHlattice so AHsoln = AHhydration - AHlattice

AHsoln = [-155.5 kcal/mol - (-163.8 kcal/mol)](4. 184 kj/kcal) = 34. 7 kj/ mol then g = n AHsoIn.

<? !.OOX102fcf 101. 11 gRearrange to solve for n. n = - - = - - = 2.8818 mol then 2.8818 frtel x - - = 2.9 x 102 g.

AHsoln _k£ Ifttelo4./

molCheck: The units (kj/mol and g) are correct. The magnitude of the answer (+ 35) makes physical sensebecause the lattice energy and the heat of hydration are about the same, with the lattice energy dominating;therefore the answer is positive. The problem hints at a positive heat of solution by saying that heat isabsorbed. The magnitude of the mass (290) makes physical sense since 100 kj will require over 2 moles of salt.

Solution Equilibrium and Factors Affecting Solubility

12.41 The solution is unsaturated since we are dissolving 25 g of NaCl per 100 g of water and the solubility fromthe figure is - 35 g NaCl per 100 g of water at 25° C.

12.42 The solution is almost saturated since we are dissolving 32 g of KNO3 per 100 g of water and the solubilityfrom the figure is ~ 36 g KNO3 per 100 g of water at 25° C.

12.43 At 40 °C the solution has 45 g of KNO3 per 100 g of water and it can contain up to 63 g of KNO3 per 100 gof water. At 0 °C the solubility from the figure is ~ 14 g KNO3 per 100 g of water, so ~ 31 g KNO3 per 100 gof water will precipitate out of solution.

12.44 At 60 °C the solution has 42 g of KC1 per 100 g of water and it can contain up to 45 g of KC1 per 100 g ofwater. At 0 °C the solubility from the figure is - 26 g KC1 per 100 g of water, so - 16 g KC1 per 100 g of waterwill precipitate out of solution.

Since the solubility of gases decreases as the temperature increases boiling will cause dissolved oxygen tobe removed from the solution.

Since the solubility of gases decreases as the temperature increases, dissolved oxygen was removed from thesolution and there was no oxygen in the water for the fish to breathe.

12.47 Henry's law says that as pressure increases, nitrogen will more easily dissolve in blood. To reverse thisprocess, divers should ascend to lower pressures.

12.48 Henry's law says that as pressure increases, oxygen will more easily dissolve in blood. To reverse thisprocess, divers should ascend to lower pressures or breathe special gas mixtures with lower oxygen levels.

12.49 Given: room temperature, 80.0 L aquarium, PTotal = 1-0 arm; XN2 = °-78 Find: m (N2)Other: A:H(N2) = 6.1 x 10"4 M/L at 25 °CConceptual Plan: PTotat/ XN2 -"

PN2 then pN2r kH(N2) -* SNz then L -» mol -» gamount solute (moles) 28 .01 g N2

IVfc = *NA«l SN2 = *H(N2)PN2 M = voiume solution (L) 1 mol N2

Solution: PNz = *N2PTotai = 0.78 x 1.0 arm = 0.78 atm then

SN2 = ^H(N2)PN2 = 6 . 1 x l O " 4 - x 0 . 7 8 a t m = 4.758x KT4M then

„ 28.01 g"4„80.0 k x 4.758 x 10"4 — x = 1.1 gCheck: The units (g) are correct. The magnitude of the answer (1) seems reasonable since we have 80 L ofwater and expect much less than a mole of nitrogen.

464 Chapter 12 Solution

12.50 Given: Helium, 25 °C, PHe = 1.0 atm Find: SHe (M) Other: JtH(He) = 3.7 x HT4 M/L at 25 °CConceptual Plan: PHe, fcH(He) -> SHe

SHe = fcH(He)PHe

Solution: SHe = ArH(He)PHe = 3.7 x 10~4 x 1.0 atm = 3.7 x 10"4Matm

Check: The units (M) are correct. The magnitude of the answer (10"4) seems reasonable since this is the valu

oncentrations of SolutionsGiven: NaCl and water; 112 g NaCl in 1.00 L solution Find: M, m, and mass percentOther: d = 1.08 g/mLConceptual Plan: gNaC, -» mol and L -> mL -> gsoin and gsoln gNaC1 -> gH2o -» kgHzo

12.52

1 mol NaCl 1000 mL l-58.44 g NaCl 1L

mol, V —> M and mol, kgHzO ~* ™ andamount solute (moles) amount solute (moles)\*

volume solution (L)

Solution: 112 g~NaCl x

mass solvent (kg)

1 mol NaCl

ImL

gsoln gNaCl

mass percent =

i - &soin ~

mass percentmass solute

1kg1000 g

1 .00 t> x1000 mi

58.44g-Na€ll.OSg

mass solution

= 1.9164956 mol NaCl and

X 100%

IrrtL= 1080 g soln and

SH20 = gsoln - gNaCl = 1080 g - 112 g = 968 g-ffjO X1000 %= 0.968 kg H2O then

M =amount solute (moles) 1.9164956 mol NaCl

m =

volume solution (L)amount solute (moles)

mass solvent (kg)

mass solute

1.00 L soln1.9164956 mol NaCl

= 1.92Mand

mass percent =

0.968 kg H2O112 g NaCl

= 2.0 m and

X 100% = X 100% = 10.4% by mass.mass solution 1080 g soln

Check: The units (M, m, and percent by mass) are correct. The magnitude of the answer (2 M) seemreasonable since we have 112 g NaCl, which is a couple of moles and we have 1 L. The magnitude of thanswer (2 m) seems reasonable since it is a little higher than the molarity, which we expect since we only usthe solvent weight in the denominator. The magnitude of the answer (10%) seems reasonable since we hav112 g NaCl and just over 1000 g of solution.

Given: KNO3 and water; 72.5 g KNO3 in 2.00 L solution Find: M, m, and mass percentOther: d = 1.05 g/mLConceptual Plan: gKNO3 -" n™1 and L -» mL -> gsoln and gsoln gKNO3 ~* gH2O -" kgH2o

then

lmolKNO3 1000 mL l.OSg 1kglOl.llgKNOj 1L ImL ^^ ~8sota ~

mol, V -> M and mol, kgH2o ~* m and gsoln gKNO3 ~* mass percentamount solute (moles) amount solute (moles) mass solute

M =volume solution (L)

Solution: 72.5 g~Na€l x

mass solvent (kg)

1 mol KNO3

1000 g

mass solutionX 100%

= 0.71704085 mol KNO3 and

£H2c? =

lOOOTnL L05g-x— - = 2100gsoln and

It. 1 mL

gsoln - SKNO, = 2100 g - 72.5 g = 2027.,

M =

m =

amount solute (moles) 0.71704085 mol KNO3

volume solution (L)amount solute (moles)

mass solvent (kg)

2.00 L soln0.71704085 mol KNO3

2.0275 kg H2O

xTo^ = 2^75ksH20then

= 0.358 M and

= 0.354 m and

'hapter 12 Solutions 465

mass solute 72 .5 g KNO3mass percent = - — X 100% = - - X 100% = 3.45% by mass.

mass solution 2100 g solnCheck: The units (M, m, and percent by mass) are correct. The magnitude of the answer (0.358 M) seems rea-sonable since we have 72.5 g KNO3/ which is less than a mole and we have 2 L. The magnitude of the answer(0.345 m) seems reasonable since it is a little higher than the molarity, which we expect since we only use thesolvent weight in the denominator. The magnitude of the answer (3.45%) seems reasonable since we have72.5 g KNO3 and ~ 2010 g of solution.

iven: initial solution: 50.0 mL of 5.00 M KI; final solution contains: 3.05 g KI in 25.0 mLFind: final volume to dilute initial solution toConceptual Plan: final solution: gj<i — * mol and mL — > L then mol, V — > M2 then Mj, Vj, M2 — * V2

1 mol KI 1 L _ amount solute (moles)

166.006gKI 1000 mL M " volume solution (L) MlV) = ̂ ^

Solution: 3.05 g-KI x *"*" ̂ = 0.01837283 mol KI and 25.0 mi x * L T = 0.0250 mL

166.006g~KI 1000 mLamount solute (moles) 0.01837283 mol KI

then M = - — = - - = 0.7349132 M then MjVj = M2V2.volume solution (L) 0 .0250L soln

Rearrange to solve for V2. V2 = - x Vj = ,_ - — x 50.0 mL = 340. mL diluted volume.

Check: The units (mL) are correct. The magnitude of the answer (340 mL) seems reasonable since we arestarting with a concentration of 5 M and ending with a concentration of less than 1 M.

2.54 Given: initial solution: 125 mL of 8.00 M CuCl2; final solution contains: 4.67 g CuCl2 in 50.0 mLFind: final volume to dilute initial solution toConceptual Plan: final Solution: gcuci2 ~~* m°l and mL — » L then mol, V — > M2 then Mj, Vj, M2 — > V2

I mol CuQ2 1 L amount solute (moles)

134.45gCuCl2 1000 mL M = volume solution (L) M'Vl = M*V*

Solution: 4 .67 g^uGfe x — m° " * = 0.03473410 mol CuCl2 and 50 .0 mi x * L T = 0 .0500 mL

134.45 gctitlj 1000 mLamount solute (moles) 0.03473410 mol CuCl2

then M = - — = - -- 0.6946820 M then MlVl = M2V2.volume solution (L) 0 .0500 L soln

Rearrange to solve for V2.

V2 = ̂ * V> = 0.6894682^M X ™ mL = 144° mL ***** ̂ ^Check: The units (mL) are correct. The magnitude of the answer (1440 mL) seems reasonable since we arestarting with a concentration of 8 M and ending with a concentration of less than 1 M.

2.55 Given: AgNO3 and water; 3.4% Ag by mass, 4.8 L solution Find: m (Ag) Other: d = 1.01 g/mLConceptual Plan: L -» mL -» gsoln -> gAg

1000 mL 1-Olg 3-4 gAg1 L 1 mL 100 g soln

Solution: 4.8 L x - x — - = 4848 g soln then1 k 1 mL

3-4 gAg 9= 160gAg = 1.6xl02gAg.

Check: The units (g) are correct. The magnitude of the answer (160 g) seems reasonable since we havealmost 5000 g solution.

2.56 Given: dioxin and water; 0.085% dioxin by mass, 2.5 L solution Find: m (dioxin) Othen d = 1.00 g/mLConceptual Plan: L -» mL -> gsoin -» gdioxin

1000 mL 1-Olg 0.085 gdioxinlT ImL 100 gsoln

1000 fnL 1 -00 g 0 .085 g dioxinSolution: 2.5Kx- - X - - = 2500 g soln then 2500 g^olft x - - = 2.1 g dioxin.

1 L 1 IrtL 100 g~seta,Check: The units (g) are correct. The magnitude of the answer (2 g) seems reasonable since we have 2500 gsolution and a low concentration.

468 Chapter 12 Solutioi

Prepare the solution by carefully adding 1.06 g NaNO3 to a container. Add -100 mL of distilled waland agitate the solution until the salt dissolves completely. Finally add enough water to generatetotal volume of solution (125 mL).Check: The units (g) are correct. The magnitude of the answer (7 g) seems reasonable since we amaking a small volume of solution and the formula weight of KC1 is ~ 75 g/mol.

(b) Given: 125 g of 0.100 m NaNOs Find: describe final solution preparationConceptual Plan: m — > mol NaNO3 /I kg solvent — > g NaNO3/l kg solvent then

amount solute (moles) 85 .00 g NaNO3

mass solvent (kg) 1 mol NaNOg

g NaNO/1 kg solvent, gso]n -> gNaNCV gH2o then describe methodgsoln = gNaNO, +

amount solute (moles) 0.500 mol NaNO3Solution: m = - : - — — so 0.100 m = - — — — — - - so

mass solvent (kg) 1 kg H2O0.100 ffroWaNOg 85.00gNaN03 8.50gNaNO3

S°lkgH20 IffioWaNQs 1000 g H2O0.00850 g NaNO3 xgNaNO3

substitute into raticlgHz0 125gsoln-xgNaN03-

Rearrange and solve for x g NaNO3 0.00850(125 g soln - x g NaNO3) = x g NaNO3 -»

1 .0625 - 0.00850 (x g NaNO3) = x g NaNO3 -» 1 .0625 = 1.00850 (x g NaNO3) -»1 .0625, " n = ^ g NaN03 = 1.05 g NaN03 then gH2o = gsoln - SNaNO3 = 125 g - 1.05 g = 124 g H2Cl.UUoouPrepare the solution by carefully adding 1.05 g NaNO3 to a container with 124 g of distilled water aragitate the solution until the salt dissolves completely.Check: The units (g) are correct. The magnitude of the answer (1 g) seems reasonable since we amaking a small volume of solution and the formula weight of NaNO3 is 85 g/mol.

(c) Given: 125 g of 1.0 % NaNO3 by mass Find: describe final solution preparationConceptual Plan: gsojn -» g NaNo3 then g NaNO3/ gsoln ~* gH2o

LOgNaNQj100 gsoln so

then describe method1.0gNaNO3

Solution: 125 g-Bebi x - — = 1.25 g NaNO3 then gso]n = £NaNO3 + £H2O So100 g~sotfi

= gsoln - £NaN03 = 125 g - 1-25 g = 124 g H2O.

Prepare the solution by carefully adding 1.3 g NaNO3 to a container with 124 g of distilled water aragitate the solution until the salt dissolves completely.Check: The units (g) are correct. The magnitude of the answer (1 g) seems reasonable since we amaking a small volume of solution and the solution is 1% by mass NaNO3.

(a) Given: 28.4 g of glucose (C6Hi2O6) in 355 g water; final volume = 378 mL Find: molarityConceptual Plan: mL —» L and g c6Hi2O6 ~~* m°l Cfft^Ot then mol,

1 mol QHjjQs amount solute (moles)M = -

1000 mL 180 .16 g C6H]2O6 volume solution (L)

Solution: 378 mi x —— — - = 0.378 L and1UOO mk

1 mol C,H19Ofi= 0.157638 mol C6H12O6

amount solute (moles) 0 .157638 mol C6Hi2O6M = — — = — - = 0.417 Mvolume solution (L) 0 .378 L

Check: The units (M) are correct. The magnitude of the answer (0.4 M) seems reasonable since vhave 1/8 mole in about 1/3 L.

(b) Given: 28.4 g of glucose (C6H12O6) in 355 g water; final volume = 378 mL Find: molaliryConceptual Plan: g H2o — » kg H2o and g c6H12O6 -*

m°l C6HnO6 then mol QHi2O6/ kg H2o ~* m

1 kg 1 mol C6H12O6 _ amount solute (mole

lOOOg ISO.iegC^jOe mass solvent (kg)


1kgSolution: 355 % x = 0 .355 kg and

1000 ̂

x - . Q - = 0.157638 molC6H1206sU.logcgi-trTtifcamount solute (moles) 0 .157638 mol C6H12O6

•yti — - — - — C\ A.A.A. 1*7

mass solvent (kg) 0 .355 kgCheck: The units (m) are correct. The magnitude of the answer (0.4 m) seems reasonable since we have1 /8 mole in about 1 /3 kg.

(c) Given: 28.4 g of glucose (CgH^Og) in 355 g water; final volume = 378 mL Find: percent by massConceptual Plan: g c6H12CV gH2O ~* gsoln then g QHuO,, Ssoln ~» percent by mass

mass soluteSsom = «C6H,206 + «H2o mass percent = masssolution x 100%

Solution: g^ = gc6H12o6 + gH2o = 28.4 g + 355 g = 383.4 g soln thenmass solute 28 -4 g C6H12O6

mass percent = - - X 100% = - - X 100% = 7.41 percent by mass.mass solution 383 .4 g soln

Check: The units (percent by mass) are correct. The magnitude of the answer (7%) seems reasonablesince we are dissolving 28 g in 355 g.

(d) Given: 28.4 g of glucose (CeH12O6) in 355 g water; final volume = 378 mL Find: mole fractionConceptual Plan:

m°l C6H1206 and 8H2o -" molH2o then mol c6H12o6 / molH2o -" Xc,,H12o6

1 mol H2O amount solute (in moles)

180.16 gC6H12O6 18.02gH2O total amount of solute and solvent (in moles)

Solution: 28.4 gCgH^ x = 0.157638 mol C6Hj2O6 andloU.l -

= 19-7003 mol H2O then

amount solute (in moles) 0.157638 frtelv = - - = - - = 0.00794

total amount of solute and solvent (in moles) 0.157638 fttel + 19.7003 melCheck: The units (none) are correct. The magnitude of the answer (0.008) seems reasonable since wehave many more grams of water and water has a much lower molecular weight.

(e) Given: 28.4 g of glucose (C6H12O6) in 355 g water; final volume = 378 mL Find: mole percentConceptual Plan: use answer from part d) then \ c6H12o6 ~~* mole percent

X x 100%

Solution: mole percent = x x 100% = 0.00794 x 100% = 0.794 mole percentCheck: The units (%) are correct. The magnitude of the answer (0.8) seems reasonable since we havemany more grams of water, water has a much lower molecular weight, than glucose, and we areincreasing the answer from part d) by a factor of 100.

12.64 (a) Given: 20.2 mL of methanol (CH3OH) in 100.0 mL water; final volume = 118 mL Find: molarityOther d (CH3OH) = 0.782 g/mL; d (H2O) = 1.00 g/mLConceptual Plan: mL — » L and mLCH3OH ~* g CH3OH ~* m°l CH3OH then mol CH3OH/ V — » M

1L 0.782g lmolCH3OH amount solute (moles)

lOOOmL ImL 32.04gCH3OH " volume solution (L)

Solution: 118 fttL x - - = 0.118 L and1000 fri

- «**- -«**amount solute (moles) 0.493021 mol CH3OH

M = - = 4.18 Mvolume solution (L) 0 . 1 18 L

Check: The units (M) are correct. The magnitude of the answer (4 M) seems reasonable since we have

1/2 mole in about 1/8 L.

hapter 12 Solutions 471

'.65 Given: 3.0 % H2O2 by mass, d = 1.01 g/mL Find: molarityConceptual Plan:Assume exactly 100 g of solution; gsoiution ~» gH2o2 -* molH2o2 and gsoiutim -»• mLsolution -» Laolution

3.0gH202 lmolH202 1 mL 1L1 00 g solution 34 .02 g H2C>2 1 .01 g 1000 mL

then molH2o2/ Lsolution -*• M

amount solute (moles)

volume solution (L)

lmolH2O2Solution: 100 rsTjhrtJen x - - x - r^r- = 0 .0881834 mol H2O2 and100 gsoitrtien 34 . -

1 L solution= 0 .0990099 L solution then

1.01 g-strttrtien 1000 mtsohrtienamount solute (moles) 0 .0881834 mol H2O2

M = - — = - — = 0.89 M H2O2.volume solution (L) 0 .0990099 L solutionCheck: The units (M) are correct. The magnitude of the answer (1) seems reasonable since we are startingwith a low concentration solution and pure water is ~ 55.5 M.

2.66 Given: 4.55 % NaOCl by mass, d = 1.02 g/mL Find: molarityConceptual Plan: Assume exactly 100 g of solution; gsoiution -> gNaoci ~* molNaOCi and g^non ~*

4. 55 g NaOCl l mol NaOCl 1 mL100 gsoiution 74. 44 g NaOCl 1.02g

mLsoiution — * Lsolution then molNa0ci' Lsoiution — » M1 L amount solute (moles)

M = •1000 mL volume solution (L)

4 .55 g-NaGCl l mol NaOClSolution: 100 gôhrtien x - - x - - = 0.06112305 mol NaOCl and

100 g-5ohrfk» 74 .44 -1 fntrsohrtien 1 L solution

= 0.0980392 L solution then100 g~5ortrtien x — x1.02 g-stJhrtien 1000 mtôhrtien

amount solute (moles) 0.06112305 mol NaOClM = - — = - - = 0.623 M NaOCl.

volume solution (L) 0.0980392 L solutionCheck: The units (M) are correct. The magnitude of the answer (1) seems reasonable since we are startingwith a low concentration solution and pure water is ~ 55.5 M.

Given: 36 % HC1 by mass Find: molality and mole fractionConceptual Plan: Assume exactly 100 g of solution; gsoiution ~* gnci ~* n»olHci and gHcir gsoiution -"

36 g HC1 1 mol HC1100 g solution 36.46 g HC1 ?sota " *HCI + gttf>

gsolvent ~* kgsolvent then molHCl/ kgsolvent ~* m a"d gsolvent ~~* m°lsolvent then molHa/ molsolvent ~» X HQ1 kg amount solute (moles) 1 mol H2O amount solute (in moles)

1000 g mass solvent (kg) 18.02gH2O total amount of solute and solvent (in moles)

36 g HC1 l mol HC1Solution: 100 gôrtrtiGn x - — = 36 g~HQ x - - = 0 .987383 mol HC1 and

100g-5oitrtien -+ SH20- Rearrange to solve for gsolvent- gHf> = ^soln ~ ^HCI = 100 g - 36 g = 64 g H2O

1 kg

amount solute (moles) 0 .987383 mol HC1m = - - = — = - = 15 m HC1 and

mass solvent (kg) 0.064kg

= 3.55161 mol H2O then

amount solvent (in moles) 0 .987383 frtely = - - = - = 0.22.

total amount of solute and solvent (in moles) 0 .987383 frtel + 3.55161 frtelCheck: The units (m and unitless) are correct. The magnitudes of the answers (15 and 0.2) seem reasonablesince we are starting with a high concentration solution and the molar mass of water is much less than thatof HC1.

:hapter 12 Solutions 473

molc6H]4 then mol CloH8/ mol QH,, -» x c6H14 then x c6H14, PC^HM -* PQH^1 mol C6H]4 amount solute (in moles)

X ~ . | ^^l.,rt«i ^™ ™rtl™.\ ^solution ~~ A?solvent*solvent86.17 g CsHu total amount of solute and solvent (in moles)

12.35gC10H8Solution: J° means 12.35 g C10H8 and (100 g - 12.35 g) = 87.65 g C6H14 then

>0 g (C10H8 + C6H14)

12.35 gtrroHs x - = 0.096363920 mol C10H8 and

87.65 g C6H14 x a _ = 1.0171754 mol C6H14 then

amount solvent (in moles) 1.0171754 fnelX = ~ - = - - = 0.91346159 then

total amount of solute and solvent (in moles) 0.096363920 irtel + 1.0171754 irrel^solution = A-soivent Solvent = 0.91346159 x 151 torr = 138 torr C6H14Check: The units (torr) are correct. The magnitude of the answer (140 torr) seems reasonable since it is adrop from the pure vapor pressure. Only a fraction of a mole of naphthalene is added, so the pressure will

/-v— -~^^ n°t drop much.

I .2.73 ) Given: 50.0 g of heptane (C7H16) and 50.0 g of octane (C8H18) at 25 °C; Pc7H16 = 45.8 torr; Pc8H18 = 10.9 torr

(a) Find: PC7Hl6, Pc8HI8

Conceptual Plan: g c7H16 ~* mol C?H16 and g QHl8 -» mol CgHl8 then mol C?H16/1 mol C7H]6 1 mol C8Hi8 amount C7H]6 (in moles)

100.20gC7H]6 114.22gC8H18 Xc?Hti " total amount (in moles)

CgHl8 -» X c7H1(/ X C8H18 then x c7H,«/ Pc7H16 ~* P C7H]6 and x c8HI8/ ^C8H18 -»c7H16

PC8Hle =

Solution: 50.0 ̂ 7^^ x = 0.499002 mol C7Hi6 and

50.0 g-CgHtt, x u! = 0'43^752 mo1

amount C7Hi6 (in moles) _ _ 0.499002 irtel~

amount (m moies) 0.499002 rrtel + 0.437752 irtel

XCsflw = 1 - A-C7H16 = 1 - 0.532693 = 0.467307 thenPc7Hu = Arc7H16Pc7H16 = 0.532693x45 .8 torr = 24 .4 torr andPcsHu = Xc^Pc^ = 0.467307 x 10.9 torr = 5.09 torrCheck: The units (torr) are correct. The magnitude of the answer (24 and 5 torr) seems reasonablebecause we expect a drop in half from the pure vapor pressures since we have roughly a 50:50 moleratio of the two components.

(b) Find: PTotal

Conceptual Plan: Pc7H16/J°c8H18 -> P Total

Solution: PTotal = PC7H16 + Pc8H18 = 24.4 torr + 5.09 torr = 29.5 torrCheck: The units (torr) are correct. The magnitude of the answer (30 torr) seems reasonable consider-ing the two pressures.

(c) Find: mass percent composition of the gas phaseConceptual Plan: since n a P and we are calculating a mass percent, which is a ratio of masses, wecan simply convert 1 torr to 1 mole soPc7H]6, Pc8H18 ^ «c7H16, »c8H]8then molC7H1(i -» gc7H16and molC8H18 -* gc8H18

100.20gC7H16 114.22 gCsH18

1 mol C7H,6 1 mol C8H18

then gc7H16/ g c8H18 ~* mass percentsmass solute

mass percent = — - — x 100%mass solution

Solution: so «c7H,6 = 24-4 m°l ancl "C8H18 = 5-09 mol then

474 Chapter 12 Solutioi

24,

5.i

100.20gC7H16x - - = 2444.88 g C7H16 and

114.22gC8H,8x . _ .„„ = 581.380 g C8H18 then

2444.mass solutemass percent = - —— x 100% =

2444. + 581X 100% =

mass solution= 80.8 percent by mass C/H^then 100% - 80.8% = 19.2 percent by mass C8H18

Check: The units (%) are correct. The magnitudes of the answers (81% and 19%) seem reasonable ccsidering the two pressures.

(d) The two mass percents are different because the vapor is richer in the more volatile component (tlighter molecule).

Given: pentane(C5H12) and hexane(C6H14) PTotaJ = 258 torr; Pc5Hi2 = 425 torr; Pcy^ = 151 torr at 25°C

Find:xC5H12'XQH14

Conceptual Plan: PTotai = PCsH12 + Pc6H14 where PCsH12 = Xc&J'csHu and PCtnlt = *C6H14 PQH14

but XCfflu = 1 ~ *c5H12 so PTotai = XCsHuPcjHu + (1 - Xc&^PCfHu substitute in values and solve

Solution: PTota] = AfC5H12 Pc5H12 + (! ~ XC&u)Pc<Pu so258 toK = A>C5H12425lDK + (1 - *c5H12)

151 to* -* 258 - 151 = *C5H12(425 - 151) -»

*C5H12 = H = 0.391 and *c6HI4 = 1 - Xc^ = 1 - 0.391 = 0.609Check: The units (none) are correct. The magnitudes of the answers (0.4 and 0.6) seem reasonable since the tovapor pressure is closer to the vapor pressure of hexane, so we expect there to be more hexane in the liquid.

Given: 4.08 g of chloroform (CHC13) and 9.29 g of acetone (CH3COCH3); at 35 °C PCHCU = 295 toPcH3coCH3 = 332 torr; assume ideal behavior; PTotai measured = 312 ton-Find: PCHCV PcH3cocH3' Piotai/ an<^ ^ me so^n *s ideal.Conceptual Plan: g CHci3 ~* m°l cna3

and g cH3cocH3 ~* mol CH3COCH3 then1 mol CHC13 lmolCH3COCH3

119.38 gCHC!3 58.08 g CH3COCH3

molCHci3/ molcH3cocH3 ~* XcHO3/ XcH3cocH3 then XCHCV PcHQ3 ~* PcHCi3 and

amount CHC13 (in moles)

total amount (in moles)

XcHjCOCiv PcH3cocH3 -" PcH3cocH3 then PcHOy PcH3cocH3 -» PTOUI then compare valuesP = P° P = ""* ' "

Solution:

4.08 g-CHG^ x 11lQ

m°l^tT"

31 = 0.0341766 mol CHC13 and

= 0.159952 mol CH3COCH3 then

AfCHCl, =amount CHC13 (in moles) 0.0341766 frtel

= 0.176052 andtotal amount (in moles) 0.0341766 irtel + 0.159952 fuel

ATCH3COCH3 = 1 - *CHC13 = 1 - 0.176052 = 0.823948 thenPCHCIJ = ATCHC1/CHC13 = 0.176052 x 295 torr = 51.9 torr andPCH3cocH3 = *CH3cocH3PcH3COCH3 = 0.823948 x 332 torr = 274 torr thenPTotai = PCHCI, + PCH3COCH3 = 51.9 torr + 274 torr = 326 torr.Since 326 torr + 312 torr, the solution is not behaving ideally. The chloroform-acetone interactions astronger than the chloroform-chloroform and acetone-acetone interactions.Check: The units (torr) are correct. The magnitude of the answer seems reasonable since each is a fracticof the pure vapor pressure. We are not surprised that the solution is not ideal, since the types of bondsthe two molecules are very different.


12.76 Given: methanol (CH3OH) and water; XH2o = 0.312; at 39.9 °C PTotal meaSured = 211 torr; PcH3OH = 256 ton;?H2o = 55.3 ton Find: is soln ideal?

Conceptual Plan: xn2o -" X CH3OH then XH2o/ PH.O ~* Pn2o and \ CH}OH, P°cn3OHXCHjOH = 1 - XHf> Ptifl =

thenPH20/PCH3oH -» PTotal then compare values

Solution:*CH3OH = 1 - *H2o = 1 - 0.312 = 0.688 then PH2o = XH2O ^H2O = 0.312 x 55.3 ton = 17.2536 ton and

^CH3OH = *CH3OH-PcH3OH = 0.688 x 256 ton = 176.128 ton then

^Totai = ^H2o + f"cH3OH = 17.2536 ton + 176.128 ton = 193 torr.

Since 193 ton + 211 ton, the solution is not behaving ideally. The methanol-water interactions are weakerthan the methanol-methanol and water-water interactions.Check: The units (torr) are correct. The magnitude of the answer seems reasonable since each is a fractionof the pure vapor pressures. We are not surprised that the solution is not ideal, since the types of bonds inthe two molecules are very different.

jiig Point Depression, Boiling Point Elevation, and Osmosisiven: 55.8 g of glucose (C6H12O6) in 455 g water Find: Tf and Tb Other. K{ = 1.86 °C/m; Kb = 0.512 °C/m;

Conceptual Plan: gH2o -> kgH2o and g QH12o6 ~" mo!C6H12o6 then mol cy^cv k§H2o ~* m1 kg 1 mol QHi2O6 amount solute (moles)

1000 g 180.16 gQHi2O6 mass solvent (kg)m, Kf -> A Tf -> Tf and m, Kb -> A Tb -> Tb

AT, = K,m T( = r;-Ar, ATb = Kb m ATb = Tb - T'b

Solution: 455 % x - = 0.455 kg and 55.8 g^gH^A, x = 0.309725 mol C6H12O6 then

amount solute (moles) 0.309725 mol C6H12O6m = - - = - = 0.680714 m then

mass solvent (kg) 0 .455 kg

ATf = K(m = 1.86 — x 0.680714 to = 1.27 °C then Tf = T°f - ATf = 0.00 °C - 1.27 °C = - 1.27 °C and)7t°c

ATb = Kbm = 0.512 — x 0.680714 )» = 0.349 °C and ATb = Tb - T°b soJn

Tb = T°b+ ATb = 100.000 °C + 0.349 °C = 100.349 °CCheck: The units (°C) are conect. The magnitudes of the answers seem reasonable since the molality is ~ 2/3.The shift in boiling point is less than the shift in freezing point because the constant for boiling is smaller thanthe constant for freezing is smaller than the con.

12.78 Given: .21.2 g of ethylene glycol ̂ HgCy in 85.4 g water Find: Tf and Tb

Other: Kf = 1.86 °C/m; Kb = 0.512 °C/m;Conceptual Plan: gH2o ~» kgHzo and g c2H6o2 -" mol C2H6o2 then mol c2H6o2/ kgH2o ~* «

1 kg 1 mol C2HjO2 amount solute (moles)1000 g 62 .07 g C2H6O2 mass solvent (kg)

m, Kf — > A Tf — > Tf and m, Kb — > A Tb — * Tb

ATf = K f m Tf = T;-AT( ATb = Kbm ATb = Tb - Tb

1 kg 1Solution: 85.4 % x — -~ = 0.0854 kg and 21.2 gGzH^ x ,-^- = 0.341561 mol C2H6O2 then

1000 gamount solute (moles) 0.341561 mol C2H6O2

m = -- = - = - = 3.99941 m thenmass solvent (kg) 0 .0854 kg

ATf = Kfm = 1.86 — x 3.99941 )» = 7.43 °C then T{ = T°f - ATf = 0.00 °C - 7.43 °C = - 7.43 °C and7M

ATb = Kbm = 0.512 — x 3.99941 Jn = 2.04 °C and ATb = Tb - T^ so

Tb = Tb + ATb = 100.00 °C + 2.04 °C = 102.04 °CCheck: The units (°C) are correct. The magnitudes of the answers seem reasonable since the molality is ~ 4.The shift in boiling point is less than the shift in freezing point because the constant for boiling is smallerthan the constant for freezing.

476

12.79

12.80

Chapter 12 Solutioi

Given: 17.5 g of unknown nonelectrolyte in 100.0 g water, Tf = - 1.8 °C Find: MOther: Kf = 1.86°C/mConceptual Plan: gHzo *̂ kgH2o and Tf -> ATf then ATf, Kf -» m then m, kgH2O

Tf = ij - ATf AT, = K,m1 kg

mol Unkamount solute (moles)

1000 g

then g Unk, mol unk -* -M.?Unk

mass solvent (kg)

Solution: 100 .Og-x1kg

= 0.1000 kg and Tf = Tf - ATf so1000 £ATf = Tf° - Tf = 0.00 °C - (- 1.8 °C) = + 1.8 °C ATf = K{m. Rearrange to solve for m.

m =1,

= 0.96774 m then m =amount solute (moles)

mass solvent (kg)so

"' 1.86 —m

mo/Unk = mUnkxitgH2o = 0.96774- *——x 0.1000 kg = 0.096774 mol Unk then

M =17.5 g

= 180 = 1.8m°hnk 0.096774 mol " mol mol"Check: The units (g/mol) are correct. The magnitude of the answer (180 g/mol) seems reasonable since tlmolality is - 0.1 and we have -18 g. It is a reasonable molecular weight for a solid or liquid.

Given: 35.9 g of unknown nonelectrolyte in 150.0 g water, Tf = - 1.3 °C Find: MOther. Kf = 1.86 °C/m;Conceptual Plan: gHzO -* kgH2o

and Tf —* ATf then ATf, Kf —> m then m, kgH2o

T, = T, - AT, ATf = Kftti1kg amount solute (moles)

1000 g mass solvent (kg)

then gunk/mol Unk -* MgUnk

M = :mo'Unk

1kgSolution: 150.0 g> x —r^- = 0.1500 kg and Tf = Tf - ATf so

1000 g-ATf = Tf - Tf = 0.00 °C 1.3 °C = + 1.3 °C ATf = Kfm. Rearrange to solve for m.

m =ATf

= 0.69892 m then m =1.86-

m

amount solute (moles)

mass solvent (kg)so

= 0.69892 — ^— — x 0.1500 kg = 0.104839 mol Unk then

&Unk 35.9 g gM = m°lunk 0.104839 mol molCheck: The units (g/mol) are correct. The magnitude of the answer (340 g/mol) seems reasonable since trmolality is ~ 0.7 and we have -36 g. It is a reasonable molecular weight for a solid or liquid.

Given: 24.6 g of glycerin (C3H8O3) in 250.0 mL of solution at 298 K Find: IIConceptual Plan: mLsoin -> Lsoin and g c3H8o3 ~* mol c3H8o3 then mol c3H8cv Lsoin

1L

M, T -» nn = MRT

Solution:

250.0fnLx

1000 mL

= 0.2500 L and

M =

Mthenamount solute (moles)

volume solution (L)

M =

1000 M, 92.09g-e3HB09

amount solute (moles) 0.267130 mol C3H8O3

= 0.267130 mol then

volume solution (L) 0.2500L= 1.06852 M then

= M RT = 1.06852 — x0.08206 ^'atin_ x298K = 26.1 armK K- frtel


Check: The units (atm) are correct. The magnitude of the answer (26 atm) seems reasonable since themolarity is ~ 1.

12.82 Given: sucrose (C^HÔn) in 5.00 x 102 g water; II = 8.55 atm at 298 K Find: mOther: d = 1.0 g/mLConceptual Plan: II, T -» M then g H2o d, M

amount solute (moles) I L 342.30 g C]2H22Oi] 1.0 mLM * volume solution (L) * 1000 mL' 1 mol C^HÔn ' a"d l.Og

Solution: U = M RT for M. M = — = - 8.55atm - = Q ̂ _mol

0.08206 - x 2 9 8 KK- mol

Substitute quantities into the definition of M./ 1 mol C12H22Oii \

/ /̂ " T T f~\ \ I ^ £~£- 1 1

amount solute (moles) _ \342.30gC12H22Oii/ _ mol

volume solution (L) 2 u ,-, , r- u /-> \ l-0mt IL(5.00xl02gH20 -

Rearrange to solve for x g Ci2H22On. x g C^HÔn = 0.119681 x (5 .00 x 102 g H2O +59.8405

0. 880312 x(*gC12H22Qu) = 59.8405 -* xgC^HÔn = Q = 68.0gC12H22On.

Check: The units (g) are correct. The magnitude of the answer (68 g) seems reasonable since the molarity is~ 1/3 and we have 0.5 L of water.

Given: 27.55 mg unknown protein in 25.0 mL solution; II = 3.22 torr at 25 °C Find: Alunknown protein

Conceptual Plan: °C — » K and torr — > atm then II, T — » M then mLsoin — » Lsoin then

760 torr 1000 mL

'-•soln/ "I * mol unknown protein 3nO mg •* g tnen g unknown protein/ HlOl unknown protein * ^ûnknown proteinamount solute (moles) I g gunknown protein

M = M = —:volume solution (L) 1000 mg moluntno,,,, proteil,

Solution: 25 °C + 273.15 = 298 K and 3.22 to« x — - = 0.00423684 atm O = M RT for M.760 fore

0 0.00423684 atm .mol I LM = -- = - - = 1.73258xlO~4- -then 25.0 rrtLx——- - = 0.0250Lthen

RT 0.08206 -^x 298 K L

K- molamount solute (moles)

M = - . Rearrange to solve for molunkno^ protem.volume solution (L)

unknown protein = M x L = 1.73258 x 10~4 — x 0.0250 K = 4.33146 x 10~6 mol and

1 g gunknown protein 0.02755 g g27.55frtgx- - = 0.02755 g then M = - - = 6.36xl03

1000 rng molunknown protein 4.33146 x 10"6 molCheck: The units (g/mol) are correct. The magnitude of the answer (6400 g/mol) seems reasonable for alarge biological molecule. A small amount of material is put into 0.025 L, so the concentration is very smalland the molecular weight is large.

12.84 Given: 18.75 mg of hemoglobin in 15.0 mL of solution at 25 °C, Xhemoglobin = 6-5 x 104 g/mol Find: IIConceptual Plan: mLsoin —* Lsoin and mgH —* gH ~* molH then mo!H, Lsoin —» M then

1L l g imolH amount solute (moles)M = -

1000 mL 1000 mg 6 5 x l 0 4 g H volume solution (L)

M, T -» nn = MRT

Solution:15.0 mi x 1L = 0.0150 L and 18.75 itig« x - —|— x - - = 2.88465 x l<T7mol H then

1000 mt 1000 frtg 6.5xl04g«

amount solute (moles) 2.88465 x 10~7 mol HM = = — - = 1.9231 x 10 5M then

volume solution (L) 0.0150 L

478 Chapter 12 Solutior

= MRT = 1.9231 x 10~5 ̂ x 0.08206 Vatm x 298K = 4.7 x 10~4atm = 0.36 torr

K K- melCheck: The units (atm) are correct. The magnitude of the answer (10"4 atm) seems reasonable since trmolarity is so small.

(b)

(c)

ATb then ATb -im,=3 Tb = Tb

Tb

ATb

Given: 0.100 m of K2S, completely dissociated Find: Tf, Tb

Other: Kf = 1.86 °C/m; Kb = 0.512 °C/m;Conceptual Plan: m, i, Kf -» ATf then ATf -> Tf and m, i, Kb

AT, = Kf im,,3 T, = Tj - ATf ATb =

°cSolution: ATf = Xf zm = 1.86 — x 3 x 0.100 te = 0.558 °C then

mTf = Tf - ATf = 0.000 °C - 0.558 °C = - 0.558 °C and

ATb = Kb im = 0.512 — x 3 x 0.100 Jn = 0.154 °C then

Tb = Tb - ATb = 100.000 °C + 0.154 °C = 100.154 °C.Check: The units (°C) are correct. The magnitude of the answer (- 0.6 °C and 100.2 °C) seems reasoiable since the molality of the particles is 0.3. The shift in boiling point is less than the shift in freezinpoint because the constant for boiling is larger than the constant for freezing.

Given: 21.5 g CuCl2 in 4.50 x 102 g water, completely dissociated Find: Tf, Tb

Other: Kf = 1.86 °C/m; Kb = 0.512 °C/m;Conceptual Plan: gH2o ~* kgH2o

and 8 CuCi2 -» m°lcuCi2 then mol cud,, kgH2o ~* m1 kg 1 mol CuCl2 amount solute (moles)

1000 g 134.46 gCuCl2

m,i,Kf -» ATf -> Tfandm,i,Kb -» ATb -» Tb

AT, = K, im, - 3 T, = T°f- AT, ATb = Kb im, = 3 Tb = Tb + ATb

Solution:

4 '5 x 102 6 x = ° -450

m =amount solute (moles)

.21.!

0.159904 mol CuCl2

0.450kg

mass solvent (kg)

= 0.159904 mol CuCl2 then

= 0.355341 m thenmass solvent (kg)

°CATf = K( im = 1.86 — x 3 x 0.355341 te = 1.98 °C then

>»Tf = Tf - ATf = 0.000 °C - 1.98 °C = - 1.98 °C and

ATb = Kb im = 0.512 — x 3 x 0.355341 )H = 0.546 °C then~tn

Tb = Tb - ATb = 100.000 °C + 0.546 °C = 100.546 °C.Check: The units (°C) are correct. The magnitude of the answer (- 2 °C and 100.5 °C) seems reasorable since the molality of the particles is ~ 1. The shift in boiling point is less than the shift in freesing point because the constant for boiling is larger than the constant for freezing.

Given: 5.5 % by mass NaNO3, completely dissociated Find: Tf, Tb

Other. Kf = 1.86 °C/m; Kb = 0.512 °C/m;Conceptual Plan: percent by mass -» gNaNO3/ gH2O

tnen gH2O "" kgH2o and g NaNO3mass solute 1 kg

mass percent = :—:— X 100%mass solution 1000 g

then mol NaNo3, kgH2o —* nt then m, i, Kf —> ATf -> T{andm,i,Kb -> ATbamount solute (moles) „

AT, = Kf im,- = 2 Tf = Tf — ATf ATb = JCb im,- = 2 T\

•mol1 mol Na

NaNO3

84.99gNaNO3

Solution: mass percent =

mass solvent (kg)

mass solute

= T°b ATb

mass solutionX 100% so 5.5 % by mass NaNO3 means 5.5 g NaNO3 and

100.0 g - 5.5 g = 94.5 g water. Then 94.5 g x

5.5

1kg = 0.0945 kg and1000 g-

= 0.064713 mol NaNO3 then


mol FeClamolFeCl3 = m x kgH2o = 1-149 — - — - x 0.250 fegRje = 0.2872 mol FeCl3 then

162.21gFeCl30.2872 BTDhFeQg x - - = 47 g FeCl3 = 50 g FeCl3.1 rrtorrevSlgCheck: The units (g) are correct. The magnitude of the answer (50 g) seems reasonable since the tenperature change is significant and we are making 0.25 L of solution.

Given: 0.100 M of ionic solution, IT = 8.3 arm at 25 °C Find: /measuredConceptual Plan: °C -> K then II, M, T -> i

K = °C + 273.15 n = iM RT

Solution: 25 °C + 273.15 = 298 K then U = iM RT. Rearrange to solve for i.. = n _ _ 8.3 atm _" ~ ~

0.100 x 0.08206 298 K

Check: The units (none) are correct. The magnitude of the answer (3) seems reasonable for an ionic solutiowith a high osmotic pressure.

12.90 Given: 8.92 g of KBr in 500.0 mL solution, U = 6.97 arm at 25 °C Find: zmeasuredConceptual Plan: °C — > K and mLsoln — » Lsoln and gKBr — » mo!KBr then molKBr, Lsoln -> M then

1 L 1 mol KBr amount solute (moles)1000 mL 119 .01 gKBr " volume solution (L)

n, M, T -> in = iMRi

Solution: 25 °C + 273.15 = 298 K and 500.0 frtL x — — — - = 0.5000 L and1000 iTrL

8.93 g-K& x - = 0-0750357 mol KBr then

_ amount solute (moles) = 0.0750357 mo! KBr = ^olKI* =

volume solution (L) 0.5000 L Ln 6.97 atm

Rearrange to solve for z. z = = — 1.90.0.150071 x 0.08206 -atm x 298 K

L K- rrreiCheck: The units (none) are correct. The magnitude of the answer (1.9) seems reasonable for KBr since wexpect z to be 2 if it completely dissociates. Since both ions are large and have only one charge each, wexpect i to be close to the theoretical value.

12.91 Given: 5.50 % NaCl by mass in water at 25 °C Find: PH2o Other: Pn2o = 23.78 torrConceptual Plan: % NaCl by mass —» gNaci/ g H2o then g NaC1 -> mol Naa and

5.50 g NaCl 1 mol NaCl100 g (NaCl + H2O) 58.44 g NaCl

gH2o -" molH2o then mol NaCi, mol H2o ~» X H2o then x H2o P°H2o1 mol H2O amount solute (in moles)

18.01 g H2O total amount of solute and solvent (in moles) solution

5.50 g NaClSolution: - - means 5.50 g NaCl and (100 g - 5.50 g) = 94.5 g H2O then

100 g (NaCl + H2O)

5.50 g-NaCl x 1 mo1 NaC1 = Q.0941136 mol NaCl and 94.5 g^O x m°_J = 5.24708 mol H2O the

58.44 g~Nad LS.Olg^Hjt!amount solvent (in moles)

number of moles of solute = /Nad x nNaClso Xsolv ~ , .total amount solute and solvent particles (in moles)5.24708 Tnel

= 0.96536 then Psom = Xsolv^solv = 0.96536 X 23.78 torr = 23.0 torr5.24708 Trtel + 2(0.0941136 Tnel)Check: The units (torr) are correct. The magnitude of the answer (23 torr) seems reasonable since it isdrop from the pure vapor pressure. Only a fraction of a mole of NaCl is added, so the pressure will ncdrop much.

484 Chapter 12 Solution,

Conceptual Plan: lattice energy, AHhydration -> AHsoln and g -» mol then mol, AHsoln -» ^(kj) -> q(JImol 1000 J

AH«oln = AH^^ + AHhydraaon where AHsoiufc, = - AHlattice q = n AH,,,],,00 ,,,],,

and Tj, Tf -> AT then 4, AT, Cs -> gsoln then gsoln -» mLsoin

s

Solution: AH^ = AHsoiute + AHhydration where AH^,^ = - AHlattice so AHsoln = AHhydration - AHlattice

AHsoln = - 932 kj/mol - (-887 kj/mol) = - 45 kj/mol and 25.0 g. x p = 0.625 mol then

/ kj \ 1000 Ja = nAHsoln =0.625frt©lx -45— '— }= - 28.125 k jx— — ̂ = - 28125 J released then

V frtel/ 1 Kj.AT = Tf - T; = 100.0 °C - 25.0 °C = 75.0 °C. Since heat is released when NaOH dissolves the

temperature will rise or q = + 28125 J and q = m Cs AT. Rearrange to solve for m.a + 28125 J ImL

m = — — = - — z^ - = 93.526 g soln then 93.526 g. x = 89 mL soln.4.01^— x 75.0 «G ° S

g-«6Check: The units (kj/mol and mL) are correct. The magnitude of the answer (- 45 kj/mol) makes physicasense because the lattice energy is smaller than the heat of hydration. The magnitude of the solution volumi(90 mL) makes physical sense since we have 2/3 mole of NaOH and large temperature change. NaOH is ;

-- "\ strong base and so we expect heat to be released and need to take precautions in the lab.

12.97 yGiven: Argon, 0.0537 L; 25 °C, PAr = 1.0 arm to make 1.0 L saturated solution Find: fcH(Ar)V ./ Conceptual Plan: °C -> K and PAr/ V, T -> mol^ then molArr Vsohv PAr -> fcH(Ar)

K = °C + 273.15 PV = nRT SA, = JtH(Ar)PAr with SA, =

Solution: 25 °C + 273.15 = 298 K and PV = nRT. Rearrange to solve for n.PV 1.0atmx0.0537K

RT 0.08206 ̂ ^Lx 298 K Lsoln

K- molSubstitute in values and rearrange to solve for kj-j-

mo/A, 0.00219597 mol , MjtH(Ar) = -- — = - - = 2.2 x 10~3 -- .. .

1 .0 Lsoln x 1 .0 arm atmCheck: The units (M/atm) are correct. The magnitude of the answer (10"3) seems reasonable since it is consistent with other values in the text.

12.98 Given: gas: 1.65 L; 25 °C, P = 725 torr; and ktt = 0.112 M/atm Find: volume of saturated solutionConceptual Plan: °C — » K and torr -> atm P, V, T — > molgas then molHe/ kH, p ~* vao\n

1 atm m<V>K = °C + 273.15 PV = nRT Sa s = k^as)? with S = -—

Solution: 25 °C + 273.15 = 298 K and 725 forc x — ̂ - = 0.953947 atm then PV = nRT. Rearrange to760 tore

PV 0.953947 atm xl. 65 Ksolve fom.n = — = - '—- - — = 0.0643666 mol then Sgas = ̂ H(gas)Pgas

0.08206 -^^x 298 KK- mol

mo/gaswith Sgas = - — . Substitute in values and rearrange to solve for Vso]n.

^solnmo/Kas 0 .0643666 frtel

- = 0.602 L soln.*H(gas)Pgas

Check: The units (L) are correct. The magnitude of the answer (0.6 L) seems reasonable since the Henry'slaw constant is so large.

12.99 Given: 0.0020 ppm by mass Hg = legal limit; 0.0040 ppm by mass Hg = contaminated water; 50.0 mg H£ingested Find: volume of contaminated water


1 mol H2OSolution: 19 .5 fi^Q x x = 0.00108273 mol H2O and

55 °C + 273.15 = 328 K then PV = nRT. Rearrange to solve for P.

0.00108273Tnel x 0.08206 x 328 K"

0 .0291424 atm x — - = 22.1482 torr and PH2o = A/H2o

PH2O- Rearrange to solve for x H2o.1

PH,O 22.1482*H20 = T2- = 23 78 forr. = °-93!381 then *solute = : ~ *H'° = J ~ °-93i381 " 0-0686192 then

^H2Omole percent solute = ^solute x 100% = 0.0686192 x 100% = 6.86 mole percent.Check: The units (mole percent) are correct. The magnitude of the answer (7 mole percent) seems reasorable since we expect there to be more water than solute. »

Given: Tb = 106.5 °C aqueous solution Find: Tf Other. Kf = 1.86 °C/m; Kb = 0.512 °C/mConceptual Plan: Tb -» ATb then ATb, Kb -> m then m, Xf -* ATf -> Tf

Tb = Tb + ATb ATb = Kbm ATf = K,m T, = Tj - AT,

Solution: Tb = Tb + ATb so ATb = Tb - Tb = 106.5 °C - 100.0 °C = 6.5 °C then ATb = Kbm.A T* ^ £ *yT"*

Rearrange to solve for m.m = — — = - — —7 = 12.695 m thenXb 0.512 —

m

ATf = Kf ro = 1.86 — x 12.695 >» = 23.6 °C then Tf = T°f - ATf = 0.000 °C - 23.6 °C °C = - 24 °C.

Check: The units (°C) are correct. The magnitude of the answer (- 24 °C) seems reasonable since the shift iboiling point is less than the shift in freezing point because the constant for boiling is smaller than the corstant for freezing.

12.104 Given: PH2o = 20.5 torr at 25 °C aqueous solution Find: Tb Other: P°Hf> = 23.78 torr; Kb = 0.512 °C/m

Conceptual Plan: PH,o/ PH2O -» X H2o assume 1 kg water kg H2o -» mo!H2o then1 mol H2O

i8.01gH2O

mo!H2o/XH2o -» mol solute then mol soiute, kg H2o ~* Wgoiute then m, Kb -» ATb -> Tbmoles H2O amount solute (moles)

^b = Kbm Tb = Tb + ATbmoles H20+ moles solute mass solvent (kg)

PH2O 20.5 towSolution: PH2o = *H2o PH&- Rearrange to solve for x H2o. ATH2O = ̂ — = 23 78!o« = 0'86?069 then

1000 g-HjQ x = 55.52470 mol H2O then18.01

moles HbO 55.52470 frtelVH n = - - - = - - = 0.862069. Solve for x moles of solute.

moles H2O + moles solute 55.52470 rrtdl + x Itiel(55.52470 - 47.8661) mol

55.52470 mol = 0.862069 (55.52470 mol + x mol) -> x = - 862069 = 8-8395 mo1 ̂ ^

amount solute (moles) 8.88395 mol °Cm = dmuuzu humre ^"""g*; = _ _ = 8 88395 m then ATfa = Kbm = 0.512 — x 8.88395 >H = 4.5 °-mass solvent (kg) 1 kg m

then Tb = Tb + ATb = 100.0 °C + 4.5 °C = 104.5 °C.Check: The units (°C) are correct. The magnitude of the answer (4.5 °C) seems reasonable since there issignificant lowering of the vapor pressure.

12.105 (a) Given: 0.90 % NaCl by mass per volume; isotonic aqueous solution at 25 °C; KC1; i - 1.9Find: % KC1 by mass per volumeConceptual Plan: Isotonic solutions will have the same number of particles. Since i is the same,

1 mol KC11 mol NaCl

the new % mass per volume will be the mass ratio of the two salts.mass solute , 1 mol NaCl J 74.56gKCl

percent by mass per volume = — - -- xlOO% 58_44gNaC, ™* imolKC,

solutions - gordon state collegeptfaculty.gordonstate.edu/lgoodroad/summer 2011/chem 1212...12.2 a...

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