CHAPTER 1 PHYSICAL PROPERTIES OF SOLUTIONS 1.9 CsF is an inic slid; the in in attractins are t strng t be verce in the disslving prcess in benzene. The in induced diple interactin is t weak t stabilize the in. Nnplar naphthalene lecules fr a lecular slid in which the nly interparticle frces are f the weak dispersin type. The sae frces perate in liquid benzene causing naphthalene t disslve with relative ease. Like disslves like. 1.10 Strategy: In predicting slubility, reeber the saying: Like disslves like. A nnplar slute will disslve in a nnplar slvent; inic cpunds will generally disslve in plar slvents due t favrable in-diple interactins; slutes that can fr hydrgen bnds with a slvent will have high slubility in the slvent. Slutin: Strng hydrgen bnding (diple-diple attractin) is the principal interlecular attractin in liquid ethanl, but in liquid cyclhexane the interlecular frces are dispersin frces because cyclhexane is nnplar. Cyclhexane cannt fr hydrgen bnds with ethanl, and therefre cannt attract ethanl lecules strngly enugh t fr a slutin. 1.11 The rder f increasing slubility is: O < Br < LiCl < CH 3 OH. Methanl is iscible with water because f strng hydrgen bnding. LiCl is an inic slid and is very sluble because f the high plarity f the water lecules. Bth xygen and brine are nnplar and exert nly weak dispersin frces. Brine is a larger lecule and is therefre re plarizable and susceptible t diple induced diple attractins. 1.1 The lnger the C C chain, the re the lecule "lks like" a hydrcarbn and the less iprtant the OH grup beces. Hence, as the C C chain length increases, the lecule beces less plar. Since like disslves like, as the lecules bece re nnplar, the slubility in plar water decreases. The OH grup f the alchls can fr strng hydrgen bnds with water lecules, but this prperty decreases as the chain length increases. 1.15 Percent ass equals the ass f slute divided by the ass f the slutin (that is, slute plus slvent) ties 100 (t cnvert t percentage). (a) (c) 5.50 g NaBr 100% 78. g sln 7.03% 31.0 g KCl 100% (31.0 + 15)g sln 16.9% 4.5 g tluene 100% (4.5 + 9)g sln 13% 1.16 Strategy: We are given the percent by ass f the slute and the ass f the slute. We can use Equatin (1.1) f the text t slve fr the ass f the slvent (water).
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 305 Slutin: (a) The percent by ass is defined as ass f slute percent by ass f slute 100% ass f slute + ass f slvent Substituting in the percent by ass f slute and the ass f slute, we can slve fr the ass f slvent (water). 5.00 g urea 16.% 100% 5.00 g urea + ass f water (0.16)(ass f water) 5.00 g (0.16)(5.00g) ass f water 5.9 g Siilar t part (a), 6. g MgCl 1.5% 100% 6. g MgCl + ass f water ass f water 1.7 10 3 g 1.17 (a) The lality is the nuber f les f sucrse (lar ass 34.3 g/l) divided by the ass f the slvent (water) in kg. 1l l sucrse 14.3 g sucrse 0.0418 l 34.3 g sucrse Mlality 0.0418 l sucrse 0.676 kg H O 0.0618 Mlality 7.0 l ethylene glycl 3.546 kg H O.03 1.18 lality les f slute ass f slvent (kg) (a) 1.08 g ass f 1 L sln 1000 L 1080 g 1L 58.44 g NaCl ass f water 1080 g.50 l NaCl 934 g 0.934 kg 1lNaCl.50 l NaCl.68 0.934 kg HO 100 g f the slutin cntains 48. g KBr and 51.8 g H O. 1lKBr l f KBr 48. g KBr 0.405 l KBr 119.0 g KBr
306 CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS ass f H 1kg O (in kg) 51.8 g HO 0.0518 kg HO 1000 g 0.405 l KBr 7.8 0.0518 kg H O 1.19 In each case we cnsider ne liter f slutin. ass f slutin vlue density (a) 34.3 g sugar 1 kg ass f sugar 1. l sugar 418 g sugar 0.418 kg sugar 1 l sugar 1000 g 1.1 g 1 kg ass f sln 1000 L 110 g 1.10 kg 1L 1000g lality 1. l sugar (1.10 0.418) kg H O 1.74 40.00 g NaOH ass f NaOH 0.87 l NaOH 35 g NaOH 1lNaOH ass slvent (H O) 1040 g 35 g 1005 g 1.005 kg lality 0.87 l NaOH 1.005 kg H O 0.87 (c) 84.01 g NaHCO ass f NaHCO 5.4 l NaHCO 440 g NaHCO 3 3 3 3 1lNaHCO3 ass slvent (H O) 1190 g 440 g 750 g 0.750 kg lality 5.4 l NaHCO3 0.750 kg H O 6.99 1.0 Let s assue that we have 1.0 L f a 0.010 M slutin. Assuing a slutin density f 1.0 g/l, the ass f 1.0 L (1000 L) f the slutin is 1000 g r 1.0 10 3 g. The ass f 0.010 le f urea is: 60.06 g urea 0.010 l urea 0.60 g urea 1lurea The ass f the slvent is: (slutin ass) (slute ass) (1.0 10 3 g) (0.60 g) 1.0 10 3 g 1.0 kg les slute 0.010 l ass slvent 1.0 kg 0.010
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 307 1.1 We find the vlue f ethanl in 1.00 L f 75 prf gin. Nte that 75 prf eans 75 %. 75 Vlue 1.00 L % 0.38 L 3.8 10 L 0.798 g Ethanl ass (3.8 10 L) 3.0 10 g 1L 1. (a) Cnverting ass percent t lality. Strategy: In slving this type f prble, it is cnvenient t assue that we start with 100.0 gras f the slutin. If the ass f sulfuric acid is 98.0% f 100.0 g, r 98.0 g, the percent by ass f water ust be 100.0% 98.0%.0%. The ass f water in 100.0 g f slutin wuld be.0 g. Fr the definitin f lality, we need t find les f slute (sulfuric acid) and kilgras f slvent (water). Slutin: Since the definitin f lality is lality les f slute ass f slvent (kg) we first cnvert 98.0 g H SO 4 t les f H SO 4 using its lar ass, then we cnvert.0 g f H O t units f kilgras. 1lHSO 98.0 g H 4 SO4 0.999 l HSO4 98.09 g H SO 4 1kg 3.0 g HO.0 10 kg HO 1000 g Lastly, we divide les f slute by ass f slvent in kg t calculate the lality f the slutin. l f slute 0.999 l 5.0 10 kg f slvent 3.0 10 kg Cnverting lality t larity. Strategy: Fr part (a), we knw the les f slute (0.999 le H SO 4 ) and the ass f the slutin (100.0 g). T slve fr larity, we need the vlue f the slutin, which we can calculate fr its ass and density. Slutin: First, we use the slutin density as a cnversin factr t cnvert t vlue f slutin. 1L? vlue f slutin 100.0 g 54.6 L 0.0546 L 1.83 g Since we already knw les f slute fr part (a), 0.999 le H SO 4, we divide les f slute by liters f slutin t calculate the larity f the slutin. M l f slute 0.999 l 18.3 M Lfsln 0.0546L
308 CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 1.3 1lNH3 l NH3 30.0 g NH3 1.76 l NH3 17.03 g NH3 1L 1L Vlue f the slutin 100.0 g sln 0.10 L 0.98 g 1000 L larity 1.76 l NH 3 0.10 L sln 17.3 M kg f slvent (H 1kg O) 70.0 g HO 0.0700 kg HO 1000 g lality 1.76 l NH3 0.0700 kg HO 5.1 1.4 Assue 100.0 g f slutin. (a) The ass f ethanl in the slutin is 0.100 100.0 g 10.0 g. The ass f the water is 100.0 g 10.0 g 90.0 g 0.0900 kg. The aunt f ethanl in les is: 1l 10.0 g ethanl 0.17 l ethanl 46.07 g l slute 0.17 l kg slvent 0.0900 kg.41 The vlue f the slutin is: 1L 100.0 g 10 L 0.10 L 0.984 g The aunt f ethanl in les is 0.17 le [part (a)]. M l slute 0.17 l liters f sln 0.10 L.13 M (c) 1L Slutin vlue 0.15 l 0.0587 L 58.7 L.13 l 1.7 The aunt f salt disslved in 100 g f water is: 3.0 g salt 100 g HO 35. g salt 9.10 g HO Therefre, the slubility f the salt is 35. g salt/100 g H O.
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 309 1.8 At 75 C, 155 g f KNO 3 disslves in 100 g f water t fr 55 g f slutin. When cled t 5 C, nly 38.0 g f KNO 3 reain disslved. This eans that (155 38.0) g 117 g f KNO 3 will crystallize. The aunt f KNO 3 fred when 100 g f saturated slutin at 75 C is cled t 5 C can be fund by a siple unit cnversin. 117 g KNO3 crystallized 100 g saturated sln 45.9 g KNO 3 55 g saturated sln 1.9 The ass f KCl is 10% f the ass f the whle saple r 5.0 g. The KClO 3 ass is 45 g. If 100 g f water will disslve 5.5 g f KCl, then the aunt f water t disslve 5.0 g KCl is: The 0 g f water will disslve: The KClO 3 reaining undisslved will be: 100 g HO 5.0 g KCl 0 g HO 5.5 g KCl 7.1 g KClO3 0gHO 1.4gKClO3 100 g HO (45 1.4) g KClO 3 44 g KClO 3 1.35 When a disslved gas is in dynaic equilibriu with its surrundings, the nuber f gas lecules entering the slutin (disslving) is equal t the nuber f disslved gas lecules leaving and entering the gas phase. When the surrunding air is replaced by heliu, the nuber f air lecules leaving the slutin is greater than the nuber disslving. As tie passes the cncentratin f disslved air beces very sall r zer, and the cncentratin f disslved heliu increases t a axiu. 1.36 Accrding t Henry s law, the slubility f a gas in a liquid increases as the pressure increases (c kp). The sft drink tastes flat at the btt f the ine because the carbn dixide pressure is greater and the disslved gas is nt released fr the slutin. As the iner ges up in the elevatr, the atspheric carbn dixide pressure decreases and disslved gas is released fr his stach. 1.37 We first find the value f k fr Henry's law k c 0.034 l/l 0.034 l/l at P 1at Fr atspheric cnditins we write: c kp (0.034 l/l at)(0.00030 at) 1.0 10 5 l/l 1.38 Strategy: The given slubility allws us t calculate Henry's law cnstant (k), which can then be used t deterine the cncentratin f N at 4.0 at. We can then cpare the slubilities f N in bld under nral pressure (0.80 at) and under a greater pressure that a deep-sea diver ight experience (4.0 at) t deterine the les f N released when the diver returns t the surface. Fr the les f N released, we can calculate the vlue f N released.
310 CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS Slutin: First, calculate the Henry's law cnstant, k, using the cncentratin f N in bld at 0.80 at. k k c P 4 5.6 10 l/l 4 7.0 10 l/l at 0.80 at Next, we can calculate the cncentratin f N in bld at 4.0 at using k calculated abve. c kp c (7.0 10 4 l/l at)(4.0 at).8 10 3 l/l Fr each f the cncentratins f N in bld, we can calculate the nuber f les f N disslved by ultiplying by the ttal bld vlue f 5.0 L. Then, we can calculate the nuber f les f N released when the diver returns t the surface. The nuber f les f N in 5.0 L f bld at 0.80 at is: (5.6 10 4 l/l )(5.0 L).8 10 3 l The nuber f les f N in 5.0 L f bld at 4.0 at is: (.8 10 3 l/l)(5.0 L) 1.4 10 l The aunt f N released in les when the diver returns t the surface is: (1.4 10 l) (.8 10 3 l) 1.1 10 l Finally, we can nw calculate the vlue f N released using the ideal gas equatin. The ttal pressure pushing n the N that is released is atspheric pressure (1 at). The vlue f N released is: V N nrt P V N (1.1 10 l)(73 + 37)K 0.081 L at (1.0 at) l K 0.8 L 1.51 The first step is t find the nuber f les f sucrse and f water. The le fractin f water is: 1l Mles sucrse 396 g 1.16 l sucrse 34.3 g 1l Mles water 64 g 34.6 l water 18.0 g HO 34.6 l 34.6 l + 1.16 l 0.968
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 311 The vapr pressure f the slutin is fund as fllws: P slutin HO HO (0.968)(31.8 Hg) P 30.8 Hg 1.5 Strategy: Fr the vapr pressure f water at 0 C and the change in vapr pressure fr the slutin (.0 Hg), we can slve fr the le fractin f sucrse using Equatin (1.5) f the text. Fr the le fractin f sucrse, we can slve fr les f sucrse. Lastly, we cnvert fr les t gras f sucrse. Slutin: Using Equatin (1.5) f the text, we can calculate the le fractin f sucrse that causes a.0 Hg drp in vapr pressure. Δ P P1 Δ P sucrsepwater sucrse ΔP.0 Hg P 17.5 Hg water 0.11 Fr the definitin f le fractin, we can calculate les f sucrse. sucrse nsucrse nwater + nsucrse 1l les f water 55 g 30.6 l HO 18.0 g sucrse nsucrse 0.11 30.6 + nsucrse n sucrse 3.8 l sucrse Using the lar ass f sucrse as a cnversin factr, we can calculate the ass f sucrse. 34.3 g sucrse 3 ass f sucrse 3.8 l sucrse 1.3 10 g sucrse 1 l sucrse 1.53 Let us call benzene cpnent 1 and caphr cpnent. n1 P1 1P1 P1 n1 + n 1l n 1 98.5 g benzene 1.6 l benzene 78.11 g n 1l 4.6 g caphr 0.16 l caphr 15. g 1.6 l P 1 100.0 Hg 88.6 Hg (1.6 + 0.16) l
31 CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 1.54 Fr any slutin the su f the le fractins f the cpnents is always 1.00, s the le fractin f 1 prpanl is 0.700. The partial pressures are: Pethanl ethanl Pethanl (0.300)(100 Hg) 30.0 Hg P1 prpanl 1 prpanl P1 prpanl (0.700)(37.6 Hg) 6.3 Hg Is the vapr phase richer in ne f the cpnents than the slutin? Which cpnent? Shuld this always be true fr ideal slutins? 1.55 (a) First find the le fractins f the slutin cpnents. 1l Mles ethanl 30.0 g 0.936 l CH3OH 3.04 g 1l Mles ethanl 45.0 g 0.977 l CH5OH 46.07 g ethanl 0.936 l 0.936 l + 0.977 l 0.489 ethanl 1 ethanl 0.511 The vapr pressures f the ethanl and ethanl are: P ethanl (0.489)(94 Hg) 46 Hg P ethanl (0.511)(44 Hg) Hg Since n PV/RT and V and T are the sae fr bth vaprs, the nuber f les f each substance is prprtinal t the partial pressure. We can then write fr the le fractins: ethanl 46 Hg P ethanl 0.68 Pethanl + Pethanl 46 Hg + Hg X ethanl 1 ethanl 0.3 (c) The tw cpnents culd be separated by fractinal distillatin. See Sectin 1.6 f the text. 1.56 This prble is very siilar t Prble 1.5. Δ P urea Pwater.50 Hg X urea (31.8 Hg) X urea 0.0786 The nuber f les f water is: 1lHO n water 450 g HO 5.0 l HO 18.0 g HO
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 313 nurea urea nwater + nurea nurea 0.0786 5.0 + nurea n urea.13 l ass f urea 60.06 g urea.13 l urea 1lurea 18 g f urea 1.57 ΔT b K b (.53 C/)(.47 ) 6.5 C The new biling pint is 80.1 C + 6.5 C 86.4 C ΔT f K f (5.1 C/)(.47 ) 1.6 C The new freezing pint is 5.5 C 1.6 C 7.1 C Δ T 1.58 f 1.1 C K 1.86 C/ f 0.59 1.59 METHOD 1: The epirical frula can be fund fr the percent by ass data assuing a 100.0 g saple. 1l Mles C 80.78 g 6.76 l C 1.01 g 1l Mles H 13.56 g 13.45 l H 1.008 g 1l Mles O 5.66 g 0.354 l O 16.00 g This gives the frula: C 6.76 H 13.45 O 0.354. Dividing thrugh by the sallest subscript (0.354) gives the epirical frula, C 19 H 38 O. The freezing pint depressin is ΔT f 5.5 C 3.37 C.1 C. This iplies a slutin lality f: Δ Tf.1 C Kf 5.1 C/ 0.41 Since the slvent ass is 8.50 g r 0.00850 kg, the aunt f slute is: 0.41 l 3 0.00850 kg benzene 3.5 10 l 1 kg benzene Since 1.00 g f the saple represents 3.5 10 3 l, the lar ass is: lar ass 1.00 g 3 3.5 10 l 86 g/l
314 CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS The ass f the epirical frula is 8 g/l, s the lecular frula is the sae as the epirical frula, C 19 H 38 O. METHOD : Use the freezing pint data as abve t deterine the lar ass. lar ass 86 g/l Multiply the ass % (cnverted t a decial) f each eleent by the lar ass t cnvert t gras f each eleent. Then, use the lar ass t cnvert t les f each eleent. 1lC n C (0.8078) (86 g) 19. l C 1.01 g C 1lH n H (0.1356) (86 g) 38.5 l H 1.008 g H 1lO n O (0.0566) (86 g) 1.01 l O 16.00 g O Since we used the lar ass t calculate the les f each eleent present in the cpund, this ethd directly gives the lecular frula. The frula is C 19 H 38 O. 1.60 METHOD 1: Strategy: First, we can deterine the epirical frula fr ass percent data. Then, we can deterine the lar ass fr the freezing-pint depressin. Finally, fr the epirical frula and the lar ass, we can find the lecular frula. Slutin: If we assue that we have 100 g f the cpund, then each percentage can be cnverted directly t gras. In this saple, there will be 40.0 g f C, 6.7 g f H, and 53.3 g f O. Because the subscripts in the frula represent a le rati, we need t cnvert the gras f each eleent t les. The cnversin factr needed is the lar ass f each eleent. Let n represent the nuber f les f each eleent s that n C 1lC 40.0 g C 1.01 g C 3.33 l C 1lH n H 6.7gH 1.008 g H 6.6lH 1lO n O 53.3 g O 16.00 g O 3.33 l O Thus, we arrive at the frula C 3.33 H 6.6 O 3.3, which gives the identity and the ratis f ats present. Hwever, cheical frulas are written with whle nubers. Try t cnvert t whle nubers by dividing all the subscripts by the sallest subscript. 3.33 C: 1.00 3.33 This gives us the epirical, CH O. 6.6 H:.0 3.33 3.33 O : 1.00 3.33 Nw, we can use the freezing pint data t deterine the lar ass. First, calculate the lality f the slutin. Δ Tf 1.56 C 0.195 K 8.00 C/ f
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 315 Multiplying the lality by the ass f slvent (in kg) gives les f unknwn slute. Then, dividing the ass f slute (in g) by the les f slute, gives the lar ass f the unknwn slute. 0.195 l slute? l f unknwn slute 0.078 kg diphenyl 1 kg diphenyl 0.0054 l slute 0.650 g lar ass f unknwn 1.0 10 g/l 0.0054 l Finally, we cpare the epirical lar ass t the lar ass abve. epirical lar ass 1.01 g + (1.008 g) + 16.00 g 30.03 g/l The nuber f (CH O) units present in the lecular frula is: lar ass 1.0 10 g epirical lar ass 30.03 g 4.00 Thus, there are fur CH O units in each lecule f the cpund, s the lecular frula is (CH O) 4, r C 4 H 8 O 4. METHOD : Strategy: As in Methd 1, we deterine the lar ass f the unknwn fr the freezing pint data. Once the lar ass is knwn, we can ultiply the ass % f each eleent (cnverted t a decial) by the lar ass t cnvert t gras f each eleent. Fr the gras f each eleent, the les f each eleent can be deterined and hence the le rati in which the eleents cbine. Slutin: We use the freezing pint data t deterine the lar ass. First, calculate the lality f the slutin. Δ Tf 1.56 C 0.195 K 8.00 C/ f Multiplying the lality by the ass f slvent (in kg) gives les f unknwn slute. Then, dividing the ass f slute (in g) by the les f slute, gives the lar ass f the unknwn slute. 0.195 l slute? l f unknwn slute 0.078 kg diphenyl 1 kg diphenyl 0.0054 l slute 0.650 g lar ass f unknwn 1.0 10 g/l 0.0054 l Next, we ultiply the ass % (cnverted t a decial) f each eleent by the lar ass t cnvert t gras f each eleent. Then, we use the lar ass t cnvert t les f each eleent. 1lC n C (0.400) (1.0 10 g) 4.00 l C 1.01 g C 1lH n H (0.067) (1.0 10 g) 7.98 l H 1.008 g H
316 CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS n O 1lO (0.533) (1.0 10 g) 4.00 l O 16.00 g O Since we used the lar ass t calculate the les f each eleent present in the cpund, this ethd directly gives the lecular frula. The frula is C 4 H 8 O 4. 1.61 We want a freezing pint depressin f 0 C. Δ Tf 0 C Kf 1.86 C/ 10.8 The ass f ethylene glycl (EG) in 6.5 L r 6.5 kg f water is: 10.8 l EG 6.07 g EG 3 ass EG 6.50 kg HO 4.36 10 g EG 1kgHO 1lEG The vlue f EG needed is: V 3 1LEG 1L (4.36 10 g EG) 3.93 L 1.11 g EG 1000 L Finally, we calculate the biling pint: ΔT b K b (10.8 )(0.5 C/) 5.6 C The biling pint f the slutin will be 100.0 C + 5.6 C 105.6 C. 1.6 We first find the nuber f les f gas using the ideal gas equatin. n 1at 748 Hg (4.00 L) PV 760 Hg l K RT (7 + 73) K 0.081 L at 0.160 l 0.160 l lality.76 0.0580 kg benzene ΔT f K f (5.1 C/)(.76 ) 14.1 C freezing pint 5.5 C 14.1 C 8.6 C 1.63 The experiental data indicate that the benzic acid lecules are assciated tgether in pairs in slutin due t hydrgen bnding. O H O C C O H O
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 317 1.64 First, fr the freezing pint depressin we can calculate the lality f the slutin. See Table 1. f the text fr the nral freezing pint and K f value fr benzene. ΔT f (5.5 4.3) C 1. C Δ Tf 1. C Kf 5.1 C/ 0.3 Multiplying the lality by the ass f slvent (in kg) gives les f unknwn slute. Then, dividing the ass f slute (in g) by the les f slute, gives the lar ass f the unknwn slute. 0.3 l slute? l f unknwn slute 0.050 kg benzene 1 kg benzene 0.0058 l slute.50 g lar ass f unknwn 4.3 10 g/l 0.0058 l The epirical lar ass f C 6 H 5 P is 108.1 g/l. Therefre, the lecular frula is (C 6 H 5 P) 4 r C 4 H 0 P 4. 1.65 π MRT (1.36 l/l)(0.081 L at/k l)(.0 + 73) K 3.9 at 1.66 Strategy: We are asked t calculate the lar ass f the plyer. Gras f the plyer are given in the prble, s we need t slve fr les f plyer. want t calculate given lar ass f plyer gras f plyer les f plyer need t find Fr the stic pressure f the slutin, we can calculate the larity f the slutin. Then, fr the larity, we can deterine the nuber f les in 0.8330 g f the plyer. What units shuld we use fr π and teperature? Slutin: First, we calculate the larity using Equatin (1.8) f the text. π MRT M 1 at 5.0 Hg π 760 Hg l K.80 10 RT 98 K 0.081 L at 4 M Multiplying the larity by the vlue f slutin (in L) gives les f slute (plyer).? l f plyer (.80 10 4 l/l)(0.170 L) 4.76 10 5 l plyer
318 CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS Lastly, dividing the ass f plyer (in g) by the les f plyer, gives the lar ass f the plyer. 0.8330 g plyer 4 lar ass f plyer 1.75 10 g/l 5 4.76 10 l plyer 1.67 Methd 1: First, find the cncentratin f the slutin, then wrk ut the lar ass. The cncentratin is: Mlarity π 1.43 at RT (0.081 L at/k l)(300 K) 0.0581 l/l The slutin vlue is 0.3000 L s the nuber f les f slute is: 0.0581 l 0.3000 L 0.0174 l 1L The lar ass is then: 7.480 g 0.0174 l 430 g/l The epirical frula can be fund st easily by assuing a 100.0 g saple f the substance. 1l Mles C 41.8 g 3.48 l C 1.01 g 1l Mles H 4.7 g 4.7 l H 1.008 g 1l Mles O 37.3 g.33 l O 16.00 g 1l Mles N 16.3 g 1.16 l N 14.01 g The gives the frula: C 3.48 H 4.7 O.33 N 1.16. Dividing thrugh by the sallest subscript (1.16) gives the epirical frula, C 3 H 4 O N, which has a ass f 86.0 g per frula unit. The lar ass is five ties this aunt (430 86.0 5.0), s the lecular frula is (C 3 H 4 O N) 5 r C 15 H 0 O 10 N 5. METHOD : Use the larity data as abve t deterine the lar ass. lar ass 430 g/l Multiply the ass % (cnverted t a decial) f each eleent by the lar ass t cnvert t gras f each eleent. Then, use the lar ass t cnvert t les f each eleent. 1lC n C (0.418) (430 g) 15.0 l C 1.01 g C 1lH n H (0.047) (430 g) 0 l H 1.008 g H 1lO n O (0.373) (430 g) 10.0 l O 16.00 g O 1lN n N (0.163) (430 g) 14.01 g N 5.00 l N
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 319 Since we used the lar ass t calculate the les f each eleent present in the cpund, this ethd directly gives the lecular frula. The frula is C 15 H 0 O 10 N 5. 1.68 We use the stic pressure data t deterine the larity. M π 4.61 at l K RT (0 + 73) K 0.081 L at 0.19 l/l Next we use the density and the slutin ass t find the vlue f the slutin. ass f sln 6.85 g + 100.0 g 106.9 g sln 1L vlue f sln 106.9 g sln 104.4 L 0.1044 L 1.04 g Multiplying the larity by the vlue (in L) gives les f slute (carbhydrate). l f slute M L (0.19 l/l)(0.1044 L) 0.000 l slute Finally, dividing ass f carbhydrate by les f carbhydrate gives the lar ass f the carbhydrate. lar ass 6.85 g carbhydrate 0.000 l carbhydrate 343 g/l 1.73 CaCl is an inic cpund (why?) and is therefre an electrlyte in water. Assuing that CaCl is a strng electrlyte and cpletely dissciates (n in pairs, van't Hff factr i 3), the ttal in cncentratin will be 3 0.35 1.05, which is larger than the urea (nnelectrlyte) cncentratin f 0.90. (a) (c) The CaCl slutin will shw a larger biling pint elevatin. The CaCl slutin will shw a larger freezing pint depressin. The freezing pint f the urea slutin will be higher. The CaCl slutin will have a larger vapr pressure lwering. 1.74 Biling pint, vapr pressure, and stic pressure all depend n particle cncentratin. Therefre, these slutins als have the sae biling pint, stic pressure, and vapr pressure. 1.75 Assue that all the salts are cpletely dissciated. Calculate the lality f the ins in the slutins. (a) 0.10 Na 3 PO 4 : 0.10 4 ins/unit 0.40 0.35 NaCl: 0.35 ins/unit 0.70 (c) 0.0 MgCl : 0.0 3 ins/unit 0.60 (d) 0.15 C 6 H 1 O 6 : nnelectrlyte, 0.15 (e) 0.15 CH 3 COOH: weak electrlyte, slightly greater than 0.15 The slutin with the lwest lality will have the highest freezing pint (sallest freezing pint depressin): (d) > (e) > (a) > (c) >.
30 CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 1.76 The freezing pint will be depressed st by the slutin that cntains the st slute particles. Yu shuld try t classify each slute as a strng electrlyte, a weak electrlyte, r a nnelectrlyte. All three slutins have the sae cncentratin, s cparing the slutins is straightfrward. HCl is a strng electrlyte, s under ideal cnditins it will cpletely dissciate int tw particles per lecule. The cncentratin f particles will be 1.00. Acetic acid is a weak electrlyte, s it will nly dissciate t a sall extent. The cncentratin f particles will be greater than 0.50, but less than 1.00. Glucse is a nnelectrlyte, s glucse lecules reain as glucse lecules in slutin. The cncentratin f particles will be 0.50. Fr these slutins, the rder in which the freezing pints bece lwer is: 0.50 glucse > 0.50 acetic acid > 0.50 HCl In ther wrds, the HCl slutin will have the lwest freezing pint (greatest freezing pint depressin). 1.77 (a) NaCl is a strng electrlyte. The cncentratin f particles (ins) is duble the cncentratin f NaCl. Nte that 135 L f water has a ass f 135 g (why?). The nuber f les f NaCl is: 1l 1. g NaCl 0.363 l NaCl 58.44 g Next, we can find the changes in biling and freezing pints (i ) 0.363 l 0.135 kg.70 ΔT b ik b (0.5 C/)(.70 ).8 C ΔT f ik f (1.86 C/)(.70 ) 10.0 C The biling pint is 10.8 C; the freezing pint is 10.0 C. Urea is a nnelectrlyte. The particle cncentratin is just equal t the urea cncentratin. The lality f the urea slutin is: 1 l urea les urea 15.4 g urea 0.56 l urea 60.06 g urea 0.56 l urea 0.0667 kg H O 3.84 ΔT b ik b 1(0.5 C/)(3.84 ).0 C ΔT f ik f 1(1.86 C/)(3.84 ) 7.14 C The biling pint is 10.0 C; the freezing pint is 7.14 C. 1.78 Using Equatin (1.5) f the text, we can find the le fractin f the NaCl. We use subscript 1 fr H O and subscript fr NaCl. Δ P ΔP P1 P1
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 31 3.76 Hg.98 Hg 0.0383 3.76 Hg Let s assue that we have 1000 g (1 kg) f water as the slvent, because the definitin f lality is les f slute per kg f slvent. We can find the nuber f les f particles disslved in the water using the definitin f le fractin. n n + n 1 1lH O 1000 g H O 55.49 l H O 1 18.0 g HO n n 55.49 + n 0.0383 n 1.884 l Since NaCl dissciates t fr tw particles (ins), the nuber f les f NaCl is half f the abve result. 1lNaCl Mles NaCl 1.884 l particles 0.940 l l particles The lality f the slutin is: 0.940 l 1.000 kg 0.940 1.79 Bth NaCl and CaCl are strng electrlytes. Urea and sucrse are nnelectrlytes. The NaCl r CaCl will yield re particles per le f the slid disslved, resulting in greater freezing pint depressin. Als, sucrse and urea wuld ake a ess when the ice elts. 1.80 Strategy: We want t calculate the stic pressure f a NaCl slutin. Since NaCl is a strng electrlyte, i in the van't Hff equatin is. π imrt Since, R is a cnstant and T is given, we need t first slve fr the larity f the slutin in rder t calculate the stic pressure (π). If we assue a given vlue f slutin, we can then use the density f the slutin t deterine the ass f the slutin. The slutin is 0.86% by ass NaCl, s we can find gras f NaCl in the slutin. Slutin: T calculate larity, let s assue that we have 1.000 L f slutin (1.000 10 3 L). We can use the slutin density as a cnversin factr t calculate the ass f 1.000 10 3 L f slutin. 3 1.005 g sln (1.000 10 L sln) 1005 g f sln 1Lsln Since the slutin is 0.86% by ass NaCl, the ass f NaCl in the slutin is: 0.86% 1005 g 8.6 g NaCl 100%
3 CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS The larity f the slutin is: 8.6 g NaCl 1 l NaCl 0.15 M 1.000 L 58.44 g NaCl Since NaCl is a strng electrlyte, we assue that the van't Hff factr is. Substituting i, M, R, and T int the equatin fr stic pressure gives: π 0.15 l 0.081 L at imrt () (310 K) L l K 7.6 at 1.81 The teperature and larity f the tw slutins are the sae. If we divide Equatin (1.1) f the text fr ne slutin by the sae equatin fr the ther, we can find the rati f the van't Hff factrs in ters f the stic pressures (i 1 fr urea). π CaCl π urea imrt 0.605 at i MRT 0.45 at.47 1.8 Fr Table 1.3 f the text, i 1.3 π imrt 0.0500 l 0.081 L at π (1.3) (98 K) L l K π 1.6 at 1.85 Fr this prble we ust find the slutin le fractins, the lality, and the larity. Fr larity, we can assue the slutin t be s dilute that its density is 1.00 g/l. We first find the nuber f les f lyszye and f water. n lyszye 1l 6 0.100 g 7.18 10 l 13930 g n water 1l 150 g 8.3 l 18.0 g Vapr pressure lwering: lyszyepwater Δ P n n lyszye lyszye + n water (3.76 Hg) 6 7.18 10 l 5 Δ P (3.76 Hg).05 10 Hg 6 [(7.18 10 ) + 8.3] l Freezing pint depressin: Biling pint elevatin: 6 7.18 10 l 5 Δ Tf K f (1.86 C/ ) 8.90 10 C 0.150 kg 6 7.18 10 l 5 Δ Tb Kb (0.5 C/ ).5 10 C 0.150 kg
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 33 Ostic pressure: As stated abve, we assue the density f the slutin is 1.00 g/l. The vlue f the slutin will be 150 L. 6 7.18 10 l 3 π MRT (0.081 L at/l K)(98 K) 1.17 10 at 0.889 Hg 0.150 L Nte that nly the stic pressure is large enugh t easure. 1.86 At cnstant teperature, the stic pressure f a slutin is prprtinal t the larity. When equal vlues f the tw slutins are ixed, the larity will just be the ean f the larities f the tw slutins (assuing additive vlues). Since the stic pressure is prprtinal t the larity, the stic pressure f the slutin will be the ean f the stic pressure f the tw slutins. π.4 at + 4.6 at 3.5 at 1.87 Water igrates thrugh the seiperiable cell walls f the cucuber int the cncentrated salt slutin. When we g swiing in the cean, why dn't we shrivel up like a cucuber? When we swi in fresh water pl, why dn't we swell up and burst? 1.88 (a) We use Equatin (1.4) f the text t calculate the vapr pressure f each cpnent. P1 1P1 First, yu ust calculate the le fractin f each cpnent. Siilarly, A na 1.00 l na + n B 1.00l+1.00l 0.500 B 0.500 Substitute the le fractin calculated abve and the vapr pressure f the pure slvent int Equatin (1.4) t calculate the vapr pressure f each cpnent f the slutin. PA APA (0.500)(76 Hg) 38 Hg PB BPB (0.500)(13 Hg) 66 Hg The ttal vapr pressure is the su f the vapr pressures f the tw cpnents. P Ttal P A + P B 38 Hg + 66 Hg 104 Hg This prble is slved siilarly t part (a). Siilarly, A na.00 l na + n B.00l+5.00l 0.86 B 0.714 PA APA (0.86)(76 Hg) Hg
34 CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS PB BPB (0.714)(13 Hg) 94 Hg P Ttal P A + P B Hg + 94 Hg 116 Hg 1.89 ΔT f ik f i ΔTf.6 3.5 K f (1.86)(0.40) 1.90 Fr the stic pressure, yu can calculate the larity f the slutin. M 1at 30.3 Hg π 760 Hg l K 3 1.58 10 l/l RT 308 K 0.081 L at Multiplying larity by the vlue f slutin in liters gives the les f slute. (1.58 10 3 l slute/l sln) (0.6 L sln) 4.14 10 4 l slute Divide the gras f slute by the les f slute t calculate the lar ass. 1. g 3 lar ass f slute.95 10 g/l 4 4.14 10 l 1.91 One aneter has pure water ver the ercury, ne aneter has a 1.0 M slutin f NaCl and the ther aneter has a 1.0 M slutin f urea. The pure water will have the highest vapr pressure and will thus frce the ercury clun dwn the st; clun X. Bth the salt and the urea will lwer the verall pressure f the water. Hwever, the salt dissciates int sdiu and chlride ins (van't Hff factr i ), whereas urea is a lecular cpund with a van't Hff factr f 1. Therefre the urea slutin will lwer the pressure nly half as uch as the salt slutin. Y is the NaCl slutin and Z is the urea slutin. Assuing that yu knew the teperature, culd yu actually calculate the distance fr the tp f the slutin t the tp f the aneter? 1.9 Slve Equatin (1.7) f the text algebraically fr lality (), then substitute ΔT f and K f int the equatin t calculate the lality. Yu can find the nral freezing pint fr benzene and K f fr benzene in Table 1. f the text. ΔT f 5.5 C 3.9 C 1.6 C Δ Tf 1.6 C 0.31 Kf 5.1 C/ Multiplying the lality by the ass f slvent (in kg) gives les f unknwn slute. Then, dividing the ass f slute (in g) by the les f slute, gives the lar ass f the unknwn slute. 0.31 l slute 3? l f unknwn slute (8.0 10 kg benzene) 1 kg benzene.5 10 3 l slute
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 35 0.50 g lar ass f unknwn.0 10 g/l 3.5 10 l The lar ass f ccaine C 17 H 1 NO 4 303 g/l, s the cpund is nt ccaine. We assue in ur analysis that the cpund is a pure, neric, nnelectrlyte. 1.93 The pill is in a hyptnic slutin. Cnsequently, by ssis, water ves acrss the seipereable ebrane int the pill. The increase in pressure pushes the elastic ebrane t the right, causing the drug t exit thrugh the sall hles at a cnstant rate. 1.94 The lality f the slutin assuing AlCl 3 t be a nnelectrlyte is: 1lAlCl3 l AlCl3 1.00 g AlCl3 0.00750 l AlCl3 133.3 g AlCl3 0.00750 l 0.0500 kg 0.150 The lality calculated with Equatin (1.7) f the text is: Δ Tf 1.11 C Kf 1.86 C/ 0.597 The rati 0.597 0.150 is 4. Thus each AlCl 3 dissciates as fllws: AlCl 3 (s) Al 3+ (aq) + 3Cl (aq) 1.95 Reverse ssis uses high pressure t frce water fr a re cncentrated slutin t a less cncentrated ne thrugh a seipereable ebrane. Desalinatin by reverse ssis is cnsiderably cheaper than by distillatin and avids the technical difficulties assciated with freezing. T reverse the stic igratin f water acrss a seipereable ebrane, an external pressure exceeding the stic pressure ust be applied. T find the stic pressure f 0.70 M NaCl slutin, we ust use the van t Hff factr because NaCl is a strng electrlyte and the ttal in cncentratin beces (0.70 M) 1.4 M. The stic pressure f sea water is: π imrt (0.70 l/l)(0.081 L at/l K)(98 K) 34 at T cause reverse ssis a pressure in excess f 34 at ust be applied. 1.96 First, we tabulate the cncentratin f all f the ins. Ntice that the chlride cncentratin ces fr re than ne surce. MgCl : If [MgCl ] 0.054 M, [Mg + ] 0.054 M [Cl ] 0.054 M Na SO 4 : if [Na SO 4 ] 0.051 M, [Na + ] 0.051 M [SO 4 ] 0.051 M CaCl : if [CaCl ] 0.010 M, [Ca + ] 0.010 M [Cl ] 0.010 M NaHCO 3 : if [NaHCO 3 ] 0.000 M [Na + ] 0.000 M [HCO 3 ] 0.000 M KCl: if [KCl] 0.0090 M [K + ] 0.0090 M [Cl ] 0.0090 M
36 CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS The subttal f chlride in cncentratin is: [Cl ] ( 0.0540) + ( 0.010) + (0.0090) 0.137 M Since the required [Cl ] is.60 M, the difference (.6 0.137.46 M) ust ce fr NaCl. The subttal f sdiu in cncentratin is: [Na + ] ( 0.051) + (0.000) 0.104 M Since the required [Na + ] is.56 M, the difference (.56 0.104.46 M) ust ce fr NaCl. Nw, calculating the ass f the cpunds required: NaCl: 58.44 g NaCl.46 l 143.8 g 1lNaCl 95.1g MgCl MgCl : 0.054 l 5.14 g 1lMgCl 14.1 g Na Na SO 4 : SO 0.051 l 4 7.5 g 1lNa SO 4 111.0 g CaCl CaCl : 0.010 l 1.11 g 1 l CaCl KCl: NaHCO 3 : 74.55 g KCl 0.0090 l 0.67 g 1lKCl 84.01 g NaHCO3 0.000 l 0.17 g 1lNaHCO3 1.97 (a) Using Equatin (1.8) f the text, we find the larity f the slutin. M π 0.57 at RT (0.081 L at/l K)(98 K) 0.0105 l/l This is the cbined cncentratins f all the ins. The aunt disslved in 10.0 L (0.01000 L) is 0.0105 l 4? les 0.0100 L 1.05 10 l 1L Since the ass f this aunt f prtein is 0.5 g, the apparent lar ass is 0.5 g 4 1.05 10 l 3.14 10 g/l We need t use a van t Hff factr t take int accunt the fact that the prtein is a strng electrlyte. The van t Hff factr will be i 1 (why?). M π 0.57 at 4 5.00 10 l/l irt (1)(0.081 L at/l K)(98 K)
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 37 This is the actual cncentratin f the prtein. The aunt in 10.0 L (0.0100 L) is 4 5.00 10 l 6 0.0100 L 5.00 10 l 1L Therefre the actual lar ass is: 0.5 g 6 5.00 10 l 4.50 10 g/l 4 1.98 Slutin A: Let lar ass be M. Δ P APA n ass lar ass (760 754.5) A (760) A 7.37 10 3 na 5.00 / M 3 A 7.37 10 na + nwater 5.00 / M + 100 /18.0 M 14 g/l Slutin B: Let lar ass be M Δ P BPB B 7.37 10 3 n ass lar ass nb.31/ M 3 B 7.37 10 nb + nbenzene.31/ M + 100 / 78.11 M 48 g/l The lar ass in benzene is abut twice that in water. This suggests se srt f dierizatin is ccurring in a nnplar slvent such as benzene. 1.99 H O H O + O 3.0 g HO 1 l HO 1 l O 3 10 L 4.4 10 l O 100 L 34.0 g HO l HO (a) Using the ideal gas law: V 3 nrt (4.4 10 l O )(0.081 L at/l K)(73 K) 99 L P 1.0 at
38 CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS The rati f the vlues: 99 L 10 L 9.9 Culd we have ade the calculatin in part (a) sipler if we used the fact that 1 le f all ideal gases at STP ccupies a vlue f.4 L? 1.100 As the chain beces lnger, the alchls bece re like hydrcarbns (nnplar) in their prperties. The alchl with five carbns (n-pentanl) wuld be the best slvent fr idine (a) and n-pentane (c) (why?). Methanl (CH 3 OH) is the st water like and is the best slvent fr an inic slid like KBr. 1.101 (a) Biling under reduced pressure. CO bils ff, expands and cls, cndensing water vapr t fr fg. 1.10 I H O: Diple - induced diple. I 3 H O: In - diple. Strnger interactin causes re I t be cnverted t I 3. 1.103 At equilibriu, the cncentratin in the beakers is equal. Let x L be the change in vlue. (50 x)(1.0 M) (50 + x)(.0 M) 3x 50 x 16.7 L The final vlues are: (50 16.7) L 33.3 L (50 + 16.7) L 66.7 L 1.104 (a) If the ebrane is pereable t all the ins and t the water, the result will be the sae as just reving the ebrane. Yu will have tw slutins f equal NaCl cncentratin. (c) This part is tricky. The veent f ne in but nt the ther wuld result in ne side f the apparatus acquiring a psitive electric charge and the ther side becing equally negative. This has never been knwn t happen, s we ust cnclude that igrating ins always drag ther ins f the ppsite charge with the. In this hypthetical situatin nly water wuld ve thrugh the ebrane fr the dilute t the re cncentrated side. This is the classic ssis situatin. Water wuld ve thrugh the ebrane fr the dilute t the cncentrated side. 1.105 T prtect the red bld cells and ther cells fr shrinking (in a hypertnic slutin) r expanding (in a hyptnic slutin). 1.106 First, we calculate the nuber f les f HCl in 100 g f slutin. n HCl 37.7 g HCl 1 l HCl 100 g sln 1.03 l HCl 100 g sln 36.46 g HCl
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 39 Next, we calculate the vlue f 100 g f slutin. V 1L 1L 100 g 0.0840 L 1.19 g 1000 L Finally, the larity f the slutin is: 1.03 l 0.0840 L 1.3 M 1.107 (a) Seawater has a larger nuber f inic cpunds disslved in it; thus the biling pint is elevated. (c) (d) (e) Carbn dixide escapes fr an pened sft drink bttle because gases are less sluble in liquids at lwer pressure (Henry s law). As yu prved in Prble 1.0, at dilute cncentratins lality and larity are alst the sae because the density f the slutin is alst equal t that f the pure slvent. Fr clligative prperties we are cncerned with the nuber f slute particles in slutin relative t the nuber f slvent particles. Since in clligative particle easureents we frequently are dealing with changes in teperature (and since density varies with teperature), we need a cncentratin unit that is teperature invariant. We use units f les per kilgra f ass (lality) rather than les per liter f slutin (larity). Methanl is very water sluble (why?) and effectively lwers the freezing pint f water. Hwever in the suer, the teperatures are sufficiently high s that st f the ethanl wuld be lst t vaprizatin. 1.108 Let the ass f NaCl be x g. Then, the ass f sucrse is (10. x)g. We knw that the equatin representing the stic pressure is: π MRT π, R, and T are given. Using this equatin and the definitin f larity, we can calculate the percentage f NaCl in the ixture. larity l slute Lsln Reeber that NaCl dissciates int tw ins in slutin; therefre, we ultiply the les f NaCl by tw. 1 l NaCl 1 l sucrse l slute x g NaCl + (10. x )g sucrse 58.44 g NaCl 34.3 g sucrse l slute 0.034x + 0.0980 0.0091x l slute 0.03130x + 0.0980 Mlarity f slutin l slute (0.03130 x + 0.0980) l Lsln 0.50L Substitute larity int the equatin fr stic pressure t slve fr x. π MRT
330 CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS (0. 03130 x + 0.0980) l L at 7.3 at 0.081 (96 K) 0.50 L l K 0.0753 0.03130x + 0.0980 x 1.45 g ass f NaCl 1.45 g Mass % NaCl 100% 14.% 10. g 1.109 ΔT f 5.5. 3.3 C C 10 H 8 : 18. g/l ΔTf 3.3 Kf 5.1 0.645 C 6 H 1 : 84.16 g/l Let x ass f C 6 H 1 (in gras). Using, l slute and l kg slvent ass lar ass 0.645 x 1.3 x + 84.16 18. 0.0189 kg 0.01 18.x + 111.1 84.16x (84.16)(18.) x 0.47 g %C6H1 0.47 100% 1.3 36% %C10H8 0.86 100% 1.3 65% The percentages dn t add up t 100% because f runding prcedures. 1.110 (a) Slubility decreases with increasing lattice energy. (c) Inic cpunds are re sluble in a plar slvent. Slubility increases with enthalpy f hydratin f the catin and anin. 1.111 The cpleted table is shwn belw: Attractive Frces Deviatin fr Rault s ΔH slutin A A, B B > A B Psitive Psitive (endtheric) A A, B B < A B Negative Negative (extheric) A A, B B A B Zer Zer
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 331 The first rw represents a Case 1 situatin in which A s attract A s and B s attract B s re strngly than A s attract B s. As described in Sectin 1.6 f the text, this results in psitive deviatin fr Rault s law (higher vapr pressure than calculated) and psitive heat f slutin (endtheric). In the secnd rw a negative deviatin fr Rault s law (lwer than calculated vapr pressure) eans A s attract B s better than A s attract A s and B s attract B s. This causes a negative (extheric) heat f slutin. In the third rw a zer heat f slutin eans that A A, B B, and A B interparticle attractins are all the sae. This crrespnds t an ideal slutin which beys Rault s law exactly. What srts f substances fr ideal slutins with each ther? 1.11 lality 1lHSO 98.0 g H 4 SO4 98.09 g HSO4 5.0 10 1kgHO.0 g HO 1000 g H O We can calculate the density f sulfuric acid fr the larity. larity 18 M The 18 l f H SO 4 has a ass f: 1 L 1000 L 18 l HSO4 1Lsln 98.0 g HSO4 3 18 l HSO4 1.8 10 g HSO4 1lHSO4 density 3 ass HSO4 1.8 10 g vlue 1000 L 1.80 g/l 1.113 Let's assue we have 100 g f slutin. The 100 g f slutin will cntain 70.0 g f HNO 3 and 30.0 g f H O. 1lHNO3 l slute (HNO 3) 70.0 g HNO3 1.11 l HNO3 63.0 g HNO kg slvent (H 1kg O) 30.0 g HO 0.0300 kg HO 1000 g 3 lality 1.11 l HNO3 0.0300 kg HO 37.0 T calculate the density, let's again assue we have 100 g f slutin. Since, d ass vlue we knw the ass (100 g) and therefre need t calculate the vlue f the slutin. We knw fr the larity that 15.9 l f HNO 3 are disslved in a slutin vlue f 1000 L. In 100 g f slutin, there are 1.11 les HNO 3 (calculated abve). What vlue will 1.11 les f HNO 3 ccupy?
33 CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 1000 L sln 1.11 l HNO3 69.8 L sln 15.9 l HNO3 Dividing the ass by the vlue gives the density. d 100 g 1.43 g/l 69.8 L 1.114 P P A A A P ethanl (0.6)(108 Hg) 67.0 Hg P 1-prpanl (0.38)(40.0 Hg) 15. Hg In the vapr phase: 67.0 ethanl 0.815 67.0 + 15. 1.115 Since the ttal vlue is less than the su f the tw vlues, the ethanl and water ust have an interlecular attractin that results in an verall saller vlue. 1.116 NH 3 can fr hydrgen bnds with water; NCl 3 cannt. (Like disslves like.) 1.117 In slutin, the Al(H O) 6 3+ ins neutralize the charge n the hydrphbic cllidal sil particles, leading t their precipitatin fr water. 1.118 We can calculate the lality f the slutin fr the freezing pint depressin. ΔT f K f 0.03 1.86 0.03 1.86 0.109 The lality f the riginal slutin was 0.106. Se f the slutin has inized t H + and CH 3 COO. CH 3 COOH CH 3 COO + H + Initial 0.106 0 0 Change x +x +x Equil. 0.106 x x x At equilibriu, the ttal cncentratin f species in slutin is 0.109. (0.106 x) + x 0.109 x 0.003 The percentage f acid that has undergne inizatin is: 0.003 0.106 100% 3%
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 333 1.119 Egg ylk cntains lecithins which slubilize il in water (See Figure 1.0 f the text). The nnplar il beces sluble in water because the nnplar tails f lecithin disslve in the il, and the plar heads f the lecithin lecules disslve in plar water (like disslves like). 1.10 First, we can calculate the lality f the slutin fr the freezing pint depressin. ΔT f (5.1) (5.5 3.5) (5.1) 0.39 Next, fr the definitin f lality, we can calculate the les f slute. 0.39 l slute kg slvent l slute 3 80 10 kg benzene l slute 0.031 l The lar ass (M) f the slute is: 3.8 g 0.031 l 1. 10 g/l The lar ass f CH 3 COOH is 60.05 g/l. Since the lar ass f the slute calculated fr the freezing pint depressin is twice this value, the structure f the slute st likely is a dier that is held tgether by hydrgen bnds. O H O H 3 C C C CH 3 A dier O H O 1.11 19 μg 19 10 6 g r 1.9 10 4 g 4 1.9 10 g 5 ass f lead/l 7.4 10 g/l.6 L Safety liit: 0.050 pp iplies a ass f 0.050 g Pb per 1 10 6 g f water. 1 liter f water has a ass f 1000 g. 0.050 g Pb 5 ass f lead 1000 g H O 5.0 10 g/l 6 1 10 g HO The cncentratin f lead calculated abve (7.4 10 5 g/l) exceeds the safety liit f 5.0 10 5 g/l. Dn t drink the water! 1.1 (a) ΔT f K f (1.86)() lality 1.1 This cncentratin is t high and is nt a reasnable physilgical cncentratin.
334 CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS Althugh the prtein is present in lw cncentratins, it can prevent the fratin f ice crystals. 1.13 If the can is tapped with a etal bject, the vibratin releases the bubbles and they ve t the tp f the can where they jin up t fr bigger bubbles r ix with the gas at the tp f the can. When the can is pened, the gas escapes withut dragging the liquid ut f the can with it. If the can is nt tapped, the bubbles expand when the pressure is released and push the liquid ut ahead f the. 1.14 As the water freezes, disslved inerals in the water precipitate fr slutin. The inerals refract light and create an paque appearance. 1.15 At equilibriu, the vapr pressure f benzene ver each beaker ust be the sae. Assuing ideal slutins, this eans that the le fractin f benzene in each beaker ust be identical at equilibriu. Cnsequently, the le fractin f slute is als the sae in each beaker, even thugh the slutes are different in the tw slutins. Assuing the slute t be nn-vlatile, equilibriu is reached by the transfer f benzene, via the vapr phase, fr beaker A t beaker B. The le fractin f naphthalene in beaker A at equilibriu can be deterined fr the data given. The nuber f les f naphthalene is given, and the les f benzene can be calculated using its lar ass and knwing that 100 g 7.0 g 93.0 g f benzene reain in the beaker. C10H8 0.15 l 1 l benzene 0.15 l + 93.0 g benzene 78.11 g benzene 0.11 Nw, let the nuber f les f unknwn cpund be n. Assuing all the benzene lst fr beaker A is transferred t beaker B, there are 100 g + 7.0 g 107 g f benzene in the beaker. Als, recall that the le fractin f slute in beaker B is equal t that in beaker A at equilibriu (0.11). The le fractin f the unknwn cpund is: unknwn n 1 l benzene n + 107 g benzene 78.11 g benzene 0.11 n n + 1.370 n 0.173 l There are 31 gras f the unknwn cpund disslved in benzene. The lar ass f the unknwn is: 31 g 0.173 l 1.8 10 g/l Teperature is assued cnstant and ideal behavir is als assued. 1.16 T slve fr the lality f the slutin, we need the les f slute (urea) and the kilgras f slvent (water). If we assue that we have 1 le f water, we knw the ass f water. Using the change in vapr pressure, we can slve fr the le fractin f urea and then the les f urea. Using Equatin (1.5) f the text, we slve fr the le fractin f urea. ΔP 3.76 Hg.98 Hg 0.78 Hg
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 335 Δ P P1 ureapwater urea P 0.78 Hg Δ P 3.76 Hg water 0.033 Assuing that we have 1 le f water, we can nw slve fr les f urea. urea 0.033 l urea l urea + l water n n urea urea + 1 0.033n urea + 0.033 n urea 0.033 0.967n urea n urea 0.034 l 1 le f water has a ass f 18.0 g r 0.0180 kg. We nw knw the les f slute (urea) and the kilgras f slvent (water), s we can slve fr the lality f the slutin. l slute 0.034 l kg slvent 0.0180 kg 1.9 H O H 1.17 (a) H C C C H S C S H H Acetne is a plar lecule and carbn disulfide is a nnplar lecule. The interlecular attractins between acetne and CS will be weaker than thse between acetne lecules and thse between CS lecules. Because f the weak attractins between acetne and CS, there is a greater tendency fr these lecules t leave the slutin cpared t an ideal slutin. Cnsequently, the vapr pressure f the slutin is greater than the su f the vapr pressures as predicted by Rault's law fr the sae cncentratin. Let acetne be cpnent A f the slutin and carbn disulfide cpnent B. Fr an ideal slutin, P A, PB B B, and P T P A + P B. APA P Pacetne APA (0.60)(349 Hg) 09.4 Hg PCS BPB (0.40)(501 Hg) 00.4 Hg P T (09.4 + 00.4) Hg 410 Hg Nte that the ideal vapr pressure is less than the actual vapr pressure f 615 Hg. (c) The behavir f the slutin described in part (a) gives rise t a psitive deviatin fr Rault's law [See Figure 1.8(a) f the text]. In this case, the heat f slutin is psitive (that is, ixing is an endtheric prcess).
336 CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 1.18 (a) The slutin is prepared by ixing equal asses f A and B. Let's assue that we have 100 gras f each cpnent. We can cnvert t les f each substance and then slve fr the le fractin f each cpnent. Since the lar ass f A is 100 g/l, we have 1.00 le f A. The les f B are: The le fractin f A is: 1lB 100 g B 0.909 l B 110 g B n A A 0.54 na + nb 1+ 0.909 1 Since this is a tw cpnent slutin, the le fractin f B is: B 1 0.54 0.476 We can use Equatin (1.4) f the text and the le fractins calculated in part (a) t calculate the partial pressures f A and B ver the slutin. P A P B APA (0.54)(95 Hg) BPB (0.476)(4 Hg) 50 Hg 0 Hg (c) Recall that pressure f a gas is directly prprtinal t les f gas (P n). The rati f the partial pressures calculated in part is 50 : 0, and therefre the rati f les will als be 50 : 0. Let's assue that we have 50 les f A and 0 les f B. We can slve fr the le fractin f each cpnent and then slve fr the vapr pressures using Equatin (1.4) f the text. The le fractin f A is: n 50 A A 0.71 na + nb 50 + 0 Since this is a tw cpnent slutin, the le fractin f B is: B 1 0.71 0.9 The vapr pressures f each cpnent abve the slutin are: P A P B APA (0.71)(95 Hg) BPB (0.9)(4 Hg) 67 Hg 1 Hg 1.19 The desired prcess is fr (fresh) water t ve fr a re cncentrated slutin (seawater) t pure slvent. This is an exaple f reverse ssis, and external pressure ust be prvided t verce the stic pressure f the seawater. The surce f the pressure here is the water pressure, which increases with increasing depth. The stic pressure f the seawater is: π MRT π (0.70 M)(0.081 L at/l K)(93 K) π 16.8 at The water pressure at the ebrane depends n the height f the sea abve it, i.e. the depth. P ρgh, and fresh water will begin t pass thrugh the ebrane when P π. Substituting π P int the equatin gives: and π ρgh
CHAPTER 1: PHYSICAL PROPERTIES OF SOLUTIONS 337 h π gρ Befre substituting int the equatin t slve fr h, we need t cnvert at t pascals, and the density t units f kg/ 3. These cnversins will give a height in units f eters. 5 1.0135 10 Pa 16.8 at 1.70 10 Pa 1at 1 Pa 1 N/ and 1 N 1 kg /s. Therefre, we can write 1.70 10 6 Pa as 1.70 10 6 kg/ s 3 3 1000 g 1 1.03 g 1 kg 100 c 1c 6 3 3 1.03 10 kg/ h kg 6 1.70 10 π s gρ 3 kg 9.81 1.03 10 3 s 168 1.130 T calculate the le fractin f urea in the slutins, we need the les f urea and the les f water. The nuber f les f urea in each beaker is: 0.10 l les urea (1) 0.050 L 0.0050 l 1L 0.0 l les urea () 0.050 L 0.010 l 1L The nuber f les f water in each beaker initially is: 1g 1l les water 50 L.8 l 1 L 18.0 g The le fractin f urea in each beaker initially is: 0.0050 l 0.0050 l.8 l 1.8 10 1 + 3 0.010 l 3.6 10 0.010 l +.8 l 3 Equilibriu is attained by the transfer f water (via water vapr) fr the less cncentrated slutin t the re cncentrated ne until the le fractins f urea are equal. At this pint, the le fractins f water in each beaker are als equal, and Rault s law iplies that the vapr pressures f the water ver each beaker are the sae. Thus, there is n re net transfer f slvent between beakers. Let y be the nuber f les f water transferred t reach equilibriu. 1 (equil.) (equil.) 0.0050 l 0.010 l 0.0050 l +.8 l y 0.010 l +.8 l + y