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

Transcription

1 CHTER 17 ROERTIES OF SOUTIONS Slutin Cmpsitin 1. Mass percent: the percent by mass f the slute in the slutin. Mle fractin: the rati f the number f mles f a given cmpnent t the ttal number f mles f slutin. Mlarity: the number f mles f slute per liter f slutin. Mlality: the number f mles f slute per kilgram f slvent. lume is temperature-dependent, whereas mass and the number f mles are nt. Only mlarity has a vlume term, s nly mlarity is temperature-dependent. 1. a. HNO (l) H + (aq) + NO (aq) b. Na SO 4 (s) Na + (aq) + SO 4 (aq) c. l(no ) (s) l + (aq) + NO (aq) d. SrBr (s) Sr + (aq) + Br (aq) e. KClO 4 (s) K + (aq) + ClO 4 (aq) f. NH 4 Br(s) NH 4 + (aq) + Br (aq) g. NH 4 NO (s) NH 4 + (aq) + NO (aq) h. CuSO 4 (s) Cu + (aq) + SO 4 (aq) i. NaOH(s) Na + (aq) + OH (aq) 14. Ml Na CO ml Na CO 0.1 ml Na CO Na CO (s) Na + (aq) + CO (aq); ml Na + (0.1) 0.4 ml Ml NaHCO ml NaHCO 0.00 ml NaHCO NaHCO (s) Na + (aq) + HCO (aq); ml Na ml + ttal ml Na 0.4 ml ml 0.45 ml M M Na + Na ttal vlume

2 CHTER 17 ROERTIES OF SOUTIONS Mlality 40.0 g EG 60.0 g H O 1000 g 1 ml EG 10.7 ml/kg kg 6.07 g Mlarity 40.0 g EG g slutin 1.05 g 1000 cm 1 ml 6.77 ml/ cm 6.07 g 40.0 g EG 1 ml 6.07 g ml EG; 60.0 g H O 1 ml 18.0 g. ml H O EG mle fractin ethylene glycl Hydrchlric acid (HCl): mlarity 8 g HCl 100. g sln 1.19 g sln 1000 cm 1 ml HCl 1 ml/ cm sln 6.5 g mlality 8 g HCl 6 g slvent 1000 g 1 ml HCl 17 ml/kg kg 6.5 g 8 g HCl 1 ml 6.5 g 1.0 ml HCl; 6 g H O 1 ml 18.0 g.4 ml H O mle fractin f HCl 1.0 HCl Nitric acid (HNO ): 70. g HNO 1.4 g sln 1000 cm 1 ml HNO 16 ml/ 100. g sln cm sln 6.0 g 70. g HNO 1000 g 1 ml HNO 7 ml/kg 0. g slvent kg 6.0 g 70. g HNO 1 ml 6.0 g 1.1 ml HNO ; 0. g H O 1 ml 18.0 g 1.7 ml H O HNO

3 670 CHTER 17 ROERTIES OF SOUTIONS Sulfuric acid (H SO 4 ): 95 g HSO g sln 1.84 g sln 1000 cm 1 ml HSO4 18 ml/ cm sln 98.1g HSO4 95 g H SO 5 g H O 1000 g 1 ml 194 ml/kg 00 ml/kg kg 98.1g 4 95 g H SO 4 1 ml 98.1g 0.97 ml H SO 4 ; 5 g H O 1 ml 18.0 g 0. ml H O H SO cetic cid (CH CO H): 99 g CH COH 100. g sln 99 g CH CO 1g H O 1.05 g sln 1000 cm 1 ml 17 ml/ cm sln g H 1000 g 1 ml 1600 ml/kg 000 ml/kg kg g 99 g CH CO H CH COH 1 ml g mmnia (NH ): 1.6 ml CH CO H; 1 g H O 8 g NH 0.90 g 1000 cm 1 ml 15 ml/ 100. g sln cm 17.0 g 8 g NH 7 g H O 1000 g 1 ml ml/kg kg 17.0 g 1 ml 18.0 g 0.06 ml H O 8 g NH H 1 ml 17.0 g 1.6 ml NH ; 7 g H O H 1 ml 18.0 g 4.0 ml H O NH m C 5 H g m 16 g C 5 H 1 ; 5 m 0.6 g 1 ml 0. ml C 5 H 1 m 7.15 g

4 CHTER 17 ROERTIES OF SOUTIONS m C 6 H g m 0. g C 6 H 14 ; 45 m 0.66 g m 1 ml 0.4 ml C 6 H g Mass % pentane mass pentane ttal mass g 16 g + 0. g 100 5% pentane ml pentane ttal ml 0. ml 0. ml ml 0.9 Mlality ml pentane kg hexane 0. ml 0.00 kg 7. ml/kg Mlarity ml pentane slutin 0. ml 5 m + 45 m 1000 m 1.1 ml/ 18. If we have m f wine: 1.5 m C H 5 OH g m 9.86 g C H 5 OH and 87.5 m H O 1.00 g 87.5 g H O m Mass % ethanl 9.86 g % by mass 87.5 g g Mlality 9.86 g C H5OH 1 ml.45 ml/kg kg H O g 19. If we have 1.00 f slutin: 1.7 ml citric acid m slutin 19.1 g ml 1.10 g m 6 g Mass % f citric acid g citric acid g slutin 100.9% g In 1.00 f slutin, we have 6 g citric acid and ( ) 840 g f H O. 1.7 ml citric acid Mlality 0.84 kg H O 1.6 ml/kg 840 g H O 1 ml 18.0 g 47 ml H O; citric acid

5 67 CHTER 17 ROERTIES OF SOUTIONS ml acetne 1.00 kg ethanl 1.00 mlal; g C H 5 OH 1 ml g 1.7 ml C H 5 OH acetne ml CH COCH H g CH COCH 1 m 7.7 m CH COCH ml CH COCH g g ethanl 1 m g 170 m; ttal vlume m Mlarity 1.00 ml M 1. Because the density f water is 1.00 g/m, m f water has a mass f 100. g. Density mass 10.0 g H O g HO 1.06 g/m 1.06 g/cm vlume 104 m Ml H O g 1 ml g 0.10 ml H O 4 Ml H O 100. g 1 ml 18.0 g 5.55 ml H O Mle fractin f H O ml HO4 ( ) ml H O Mlarity Mlality ml H O ml/ sln ml H O 1.0 ml/kg kg slvent. a. If we use 100. m (100. g) f H O, we need: kg H O.0 ml KCl kg g 14.9 g 15 g KCl ml KCl

6 CHTER 17 ROERTIES OF SOUTIONS 67 Disslve 15 g KCl in 100. m H O t prepare a.0 m KCl slutin. This will give us slightly mre than 100 m, but this will be the easiest way t make the slutin. Because we dn t knw the density f the slutin, we can t calculate the mlarity and use a vlumetric flask t make exactly 100 m f slutin. b. If we tk 15 g NaOH and 85 g H O, the vlume wuld prbably be less than 100 m. T make sure we have enugh slutin, let s use 100. m H O (100. g H O). et x mass f NaCl. Mass % 15 x , x (100.)x, x 17.6 g 18 g x Disslve 18 g NaOH in 100. m H O t make a 15% NaOH slutin by mass. c. In a fashin similar t part b, let s use 100. m CH OH. et x mass f NaOH m CH OH 0.79 g m 79 g CH OH Mass % 5 x 79 + Disslve 6 g NaOH in 100. m CH OH. 100, 5(79) + 5x (100.)x, x 6. g 6 g x d. T make sure we have enugh slutin, let s use 100. m (100. g) f H O. et x ml C 6 H 1 O 6. 1 ml H O 100. g H O 5.55 ml H O 18.0 g 0.10 C6H1O6 0.6 ml C 6 H 1 O 6 x x g ml, (0.10)x x, x 0.6 ml C 6 H 1 O g C 6 H 1 O 6 Disslve 110 g C 6 H 1 O 6 in 100. m f H O t prepare a slutin with C6H1O6 Thermdynamics f Slutins and Slubility. ike disslves like refers t the nature f the intermlecular frces. lar slutes and inic slutes disslve in plar slvents because the types f intermlecular frces present in slute and slvent are similar. When they disslve, the strengths f the intermlecular frces in slutin are abut the same as in pure slute and pure slvent. The same is true fr nnplar slutes in nnplar slvents. The strengths f the intermlecular frces (ndn dispersin frces) are abut the same in slutin as in pure slute and pure slvent. In all cases f like disslves like, the magnitude f ΔH sln is either a small psitive number (endthermic) r a

7 674 CHTER 17 ROERTIES OF SOUTIONS small negative number (exthermic), with a value clse t zer. Fr plar slutes in nnplar slvents and vice versa, ΔH sln is a very large, unfavrable value (very endthermic). Because the energetics are s unfavrable, plar slutes d nt disslve in nnplar slvents, and vice versa. 4. The disslving f an inic slute in water can be thught f as taking place in tw steps. The first step, called the lattice energy term, refers t breaking apart the inic cmpund int gaseus ins. This step, as indicated in the prblem requires a lt f energy and is unfavrable. The secnd step, called the hydratin energy term, refers t the energy released when the separated gaseus ins are stabilized as water mlecules surrund the ins. Because the interactins between water mlecules and ins are strng, a lt f energy is released when ins are hydrated. Thus the disslutin prcess fr inic cmpunds can be thught f as cnsisting f an unfavrable and a favrable energy term. These tw prcesses basically cancel each ther ut, s when inic slids disslve in water, the heat released r gained is minimal, and the temperature change is minimal. 5. Using Hess s law: NaI(s) Na + (g) + I (g) Na + (g) + I (g) Na + (aq) + I (aq) ΔH ΔH E ( 686 kj/ml) ΔH ΔH hyd 694 kj/ml NaI(s) Na + (aq) + I (aq) ΔH sln 8 kj/ml ΔH sln refers t the heat released r gained when a slute disslves in a slvent. Here, an inic cmpund disslves in water. 6. a. CaCl (s) Ca + (g) + Cl (g) ΔH ΔH E ( 47 kj) Ca + (g) + Cl (g) Ca + (aq) + Cl (aq) ΔH ΔH hyd CaCl (s) Ca + (aq) + Cl (aq) 46 kj 47 kj + ΔH hyd, ΔH hyd 9 kj CaI (s) Ca + (g) + I (g) Ca + (g) + I (g) Ca + (aq) + I (aq) ΔH sln 46 kj ΔH ΔH E (-059 kj) ΔH ΔH hyd CaI (s) Ca + (aq) + I (aq) 104 kj 059 kj + ΔH hyd, ΔH hyd 16 kj ΔH sln 104 kj b. The enthalpy f hydratin fr CaCl is mre exthermic than fr CaI. ny differences must be due t differences in hydratins between Cl and I. Thus the chlride in is mre strngly hydrated than the idide in. 7. Bth l(oh) and NaOH are inic cmpunds. Because the lattice energy is prprtinal t the charge f the ins, the lattice energy f aluminum hydrxide is greater than that f sdium hydrxide. The attractin f water mlecules fr l + and OH - cannt vercme the larger lattice energy, and l(oh) is insluble. Fr NaOH, the favrable hydratin energy is large enugh t vercme the smaller lattice energy, and NaOH is sluble.

8 CHTER 17 ROERTIES OF SOUTIONS a. Mg + ; smaller size, higher charge b. Be + ; smaller size c. Fe + ; smaller size, higher charge d. F ; smaller size e. Cl ; smaller size f. SO 4 ; higher charge 9. Water is a plar mlecule capable f hydrgen bnding. lar mlecules, especially mlecules capable f hydrgen bnding, and ins can be hydrated. Fr cvalent cmpunds, as plarity increases, the attractin t water (hydratin) increases. Fr inic cmpunds, as the charge f the ins increase and/r the size f the ins decrease, the attractin t water (hydratin) increases. a. CH CH OH; CH CH OH is plar, whereas CH CH CH is nnplar. b. CHCl ; CHCl is plar, whereas CCl 4 is nnplar. c. CH CH OH; CH CH OH is much mre plar than CH (CH ) 14 CH OH. 0. Water is a plar slvent and disslves plar slutes and inic slutes. Carbn tetrachlride (CCl 4 ) is a nnplar slvent and disslves nnplar slutes (like disslves like). T predict the plarity f the fllwing mlecules, draw the crrect ewis structure and then determine if the individual bnd diples cancel r nt. If the bnd diples are arranged in such a manner that they cancel each ther ut, then the mlecule is nnplar. If the bnd diples d nt cancel each ther ut, then the mlecule is plar. a. KrF, 8 + (7) e b. SF, 6 + (7) 0 e F Kr F F S F nnplar; sluble in CCl 4 plar; sluble in H O c. SO, 6 + (6) 18 e d. CO, 4 + (6) 16 e O S O + 1 mre O C O plar; sluble in H O nnplar; sluble in CCl 4 e. MgF is an inic cmpund s it is sluble in water.

9 676 CHTER 17 ROERTIES OF SOUTIONS f. CH O, 4 + (1) e g. C H 4, (4) + 4(1) 1 e H O C H H H C C H H plar; sluble in H O nnplar (like all cmpunds made up f nly carbn and hydrgen); sluble in CCl 4 1. s the length f the hydrcarbn chain increases, the slubility decreases. The OH end f the alchls can hydrgen-bnd with water. The hydrcarbn chain, hwever, is basically nnplar and interacts prly with water. s the hydrcarbn chain gets lnger, a greater prtin f the mlecule cannt interact with the water mlecules, and the slubility decreases; i.e., the effect f the OH grup decreases as the alchls get larger.. The main intermlecular frces are: hexane (C 6 H 14 ): ndn dispersin; chlrfrm (CHCl ): diple-diple, ndn dispersin; methanl (CH OH): H-bnding; and H O: H-bnding (tw places) There is a gradual change in the nature f the intermlecular frces (weaker t strnger). Each preceding slvent is miscible in its predecessr because there is nt a great change in the strengths f the intermlecular frces frm ne slvent t the next.. Structure effects refer t slute and slvent having similar plarities in rder fr slutin frmatin t ccur. Hydrphbic slutes are mstly nnplar substances that are waterfearing. Hydrphilic slutes are mstly plar r inic substances that are water-lving. ressure has little effect n the slubilities f slids r liquids; it des significantly affect the slubility f a gas. Henry s law states that the amunt f a gas disslved in a slutin is directly prprtinal t the pressure f the gas abve the slutin (C k). The equatin fr Henry s law wrks best fr dilute slutins f gases that d nt dissciate in r react with the slvent. HCl(g) des nt fllw Henry s law because it dissciates int H + (aq) and Cl (aq) in slutin (HCl is a strng acid). Fr O and N, Henry s law wrks well since these gases d nt react with the water slvent. n increase in temperature can either increase r decrease the slubility f a slid slute in water. It is true that a slute disslves mre rapidly with an increase in temperature, but the amunt f slid slute that disslves t frm a saturated slutin can either decrease r increase with temperature. The temperature effect is difficult t predict fr slid slutes. Hwever, the temperature effect fr gas slutes is easier t predict because the slubility f a gas typically decreases with increasing temperature. 4. Henry s law is beyed mst accurately fr dilute slutins f gases that d nt dissciate in r react with the slvent. O will bind t hemglbin in the bld. Given this reactin in the slvent, O (g) in bld des nt fllw Henry s law.

10 CHTER 17 ROERTIES OF SOUTIONS gas kc, atm k ml, k 96 atm/ml gas kc, 1.10 atm 96 atm C, C ml/ ml m grape juice 1 m C H5OH 0.79 g CH5OH 1 ml CH5OH 100. m juice m g ml CO 1.54 ml CO (carry extra significant figure) ml C H OH ml CO ttal ml CO ml CO (g) + ml CO (aq) n g + n aq atm ng (98 K) n RT g ml K 6n g CO atm n aq kc (4.7)n aq ml CO CO 6n g (4.7)n aq and frm abve n aq 1.54 n g ; slving: 6n g 4.7(1.54 n g ), 69n g 65.8, n g 0.18 ml CO 6(0.18) 59 atm in gas phase; 59 atm atm C, C 1.8 ml CO / in wine ml 7. s the temperature increases the gas mlecules will have a greater average kinetic energy. greater fractin f the gas mlecules in slutin will have a kinetic energy greater than the attractive frces between the gas mlecules and the slvent mlecules. Mre gas mlecules are able t escape t the vapr phase and the slubility f the gas decreases. apr ressures f Slutin 8. Ml C H 8 O 164 g 1 ml 9.09 g 1.78 ml C H 8 O Ml H O 8 m 0.99 g 1 ml 18.6 ml H O m 18.0 g sln H O HO 18.6 ml ( ) ml trr trr 50.0 trr

11 678 CHTER 17 ROERTIES OF SOUTIONS 9. ml CH5OH in slutin sln C H OH C H OH ; C H OH ttal ml in slutin 5.6 g C H 8 O 1 ml C H8O 0.58 ml C H 8 O 9.09 g 1.7 g C H 5 OH 11 trr.90 ml.48 ml 1 5 ml C H OH.90 ml C H 5 OH; ttal ml ml g C H OH, 5 CH5OH 16 trr 40. Cmpared t H O, slutin d (methanl-water) will have the highest vapr pressure since methanl is mre vlatile than water ( H O.8 trr at 5 C). Bth slutin b (glucsewater) and slutin c (NaCl-water) will have a lwer vapr pressure than water by Rault's law. NaCl disslves t give Na + ins and Cl ins; glucse is a nnelectrlyte. Because there are mre slute particles in slutin c, the vapr pressure f slutin c will be the lwest. 41. Slutin d (methanl-water); methanl is mre vlatile than water, which will increase the ttal vapr pressure t a value greater than the vapr pressure f pure water at this temperature. 4. ttal ml CH Cl + CH Br ; ; CH Cl ml ttal CH Cl ttal 0.75(1 trr) + ( )(11.4 trr) trr In the vapr: CH 49.9 trr Cl 0.875; 57.0 trr CH Cl ttal CHBr Nte: In the Slutins Guide we added r t the mle fractin symbl t emphasize which value we are slving. If the r is mitted, then the liquid phase is assumed. 4., / atm/0.90 atm B BB B B B ml benzene 1 ml ; ml benzene g C 6 H ml ttal ml g ml et x ml slute; then: B 0.968, (0.968)x 1.000, x 0.0 ml x Mlar mass 10.0 g 0.0 ml 0 g/ml.0 10 g/ml 44. CS CS ttal 0.855(6 trr) 5 trr

12 CHTER 17 ROERTIES OF SOUTIONS 679 CS 5 trr CS CS, CS trr CS CS 45. a. 5 m C 5 H 1 45 m C 6 H g m 0.66 g m 1 ml 0. ml C 5 H ml 0.4 ml C 6 H 14 ; ttal ml ml ml pentane in slutin 0. ml pen 0.9, ttal ml in slutin 0.56 ml hex pen pen pen 0.9(511 trr).0 10 trr; hex 0.61(150. trr) 9 trr ttal pen + hex trr 90 trr b. Frm Chapter 5 n gases, the partial pressure f a gas is prprtinal t the number f mles f gas present. Fr the vapr phase: pen ml pentane in vapr pen.0 10 trr 0.69 ttal ml vapr 90 trr ttal Nte: In the Slutins Guide, we added r t the mle fractin symbl t emphasize which value we are slving. If the r is mitted, then the liquid phase is assumed. 46. tl tltl, pen benben; fr the vapr, / ttal. Because the mle fractins f benzene and tluene are equal in the vapr phase,. tl ben tl tl ben ben (1.00 ), (8 trr) tl ben tl (1.00 tl )95 trr 1 95, 0.77; tl tl ben 47. ttal meth + prp, 174 trr meth (0 trr) prp prp + (44.6 trr); meth meth + (1.000 meth )44.6 trr, meth prp n ideal liquid-liquid slutin fllws Rault s law: + ttal BB

13 680 CHTER 17 ROERTIES OF SOUTIONS nnideal liquid-liquid slutin des nt fllw Rault s law, giving a ttal pressure either greater than predicted by Rault s law (psitive deviatin) r less than predicted (negative deviatin). In an ideal slutin, the strengths f the intermlecular frces in slutin are equal t the strengths f the intermlecular frces in pure slute and pure slvent. When this is true, ΔH sln 0 and ΔT sln 0. Fr psitive deviatins frm Rault s law, the slutin has weaker intermlecular frces in slutin than in pure slute and pure slvent. sitive deviatins have ΔH sln > 0 (are endthermic) and ΔT sln < 0. Fr negative deviatins, the slutin has strnger intermlecular frces in slutin than in pure slute r pure slvent. Negative deviatins have ΔH sln < 0 (are exthermic) and ΔT sln > 0. Examples f each type f slutin are: ideal: psitive deviatins: negative deviatins: benzene-tluene ethanl-hexane acetne-water g CH COCH 1 ml g ml acetne 50.0 g CH OH 1 ml.04 g 1.56 ml methanl acetne 0.56; methanl acetne ttal methanl + acetne 0.644(14 trr) (71 trr) 9.1 trr trr trr Because partial pressures are prprtinal t the mles f gas present, in the vapr phase: acetne 96.5 trr acetne 0.51; methanl trr ttal The actual vapr pressure f the slutin (161 trr) is less than the calculated pressure assuming ideal behavir (188.6 trr). Therefre, the slutin exhibits negative deviatins frm Rault s law. This ccurs when the slute-slvent interactins are strnger than in pure slute and pure slvent. 50. a. n ideal slutin wuld have a vapr pressure at any mle fractin f H O between that f pure prpanl and pure water (between 74.0 trr and 71.9 trr). The vapr pressures f the varius slutins are nt between these limits, s water and prpanl d nt frm ideal slutins. b. Frm the data, the vapr pressures f the varius slutins are greater than if the slutins behaved ideally (psitive deviatin frm Rault s law). This ccurs when the intermlecular frces in slutin are weaker than the intermlecular frces in pure slvent and pure slute. This gives rise t endthermic (psitive) ΔH sln values.

14 CHTER 17 ROERTIES OF SOUTIONS 681 c. The interactins between prpanl and water mlecules are weaker than between the pure substances because the slutins exhibit a psitive deviatin frm Rault s law. d. t H O 0.54, the vapr pressure is highest as cmpared t the ther slutins. Because a slutin bils when the vapr pressure f the slutin equals the external pressure, the H O 0.54 slutin shuld have the lwest nrmal biling pint; this slutin will have a vapr pressure equal t 1 atm at a lwer temperature as cmpared t the ther slutins. 51. N, the slutin is nt ideal. Fr an ideal slutin, the strengths f intermlecular frces in the slutin are the same as in pure slute and pure slvent. This results in ΔH sln 0 fr an ideal slutin. ΔH sln fr methanl-water is nt zer. Because ΔH sln < 0, this slutin exhibits negative deviatin frm Rault s law. 5. Because the slute is vlatile, bth the water and slute will transfer back and frth between the tw beakers. The vlume in each beaker will becme cnstant when the cncentratins f slute in the beakers are equal t each ther. Because the slute is less vlatile than water, ne wuld expect there t be a larger net transfer f water mlecules int the right beaker than the net transfer f slute mlecules int the left beaker. This results in a larger slutin vlume in the right beaker when equilibrium is reached, i.e., when the slute cncentratin is identical in each beaker. 5. Slutins f and B have vapr pressures less than ideal (see Figure f the text), s this plt shws negative deviatins frm Rault s law. Negative deviatins ccur when the intermlecular frces are strnger in slutin than in pure slvent and slute. This results in an exthermic enthalpy f slutin. The nly statement that is false is e. substance bils when the vapr pressure equals the external pressure. Because B 0.6 has a lwer vapr pressure at the temperature f the plt than either pure r pure B, ne wuld expect this slutin t require the highest temperature in rder fr the vapr pressure t reach the external pressure. Therefre, the slutin with B 0.6 will have a higher biling pint than either pure r pure B. (Nte that because B >, B is mre vlatile than.) Clligative rperties 54. Clligative prperties are prperties f a slutin that depend nly n the number, nt the identity, f the slute particles. slutin f sme cncentratin f glucse (C 6 H 1 O 6 ) has the same clligative prperties as a slutin f sucrse (C 1 H O 11 ) having the same cncentratin. substance freezes when the vapr pressure f the liquid and slid are identical t each ther. dding a slute t a substance lwers the vapr pressure f the liquid. lwer temperature is needed t reach the pint where the vapr pressures f the slutin and slid are identical. Hence the freezing pint is depressed when a slutin frms. substance bils when the vapr pressure f the liquid equals the external pressure. Because a slute lwers the vapr pressure f the liquid, a higher temperature is needed t reach the

15 68 CHTER 17 ROERTIES OF SOUTIONS pint where the vapr pressure f the liquid equals the external pressure. Hence the biling pint is elevated when a slutin frms. 55. Osmtic pressure: the pressure that must be applied t a slutin t stp smsis; smsis is the flw f slvent int the slutin thrugh a semipermeable membrane. The equatin t calculate smtic pressure π is: π MRT where M is the mlarity f the slutin, R is the gas cnstant, and T is the Kelvin temperature. The mlarity f a slutin apprximately equals the mlality f the slutin when 1 kg slvent 1 slutin. This ccurs fr dilute slutins f water because d H O 1.00 g/cm. 56. This is true if the slute will disslve in camphr. Camphr has the largest K b and K f cnstants. This means that camphr shws the largest change in biling pint and melting pint as a slute is added. The larger the change in ΔT, the mre precise is the measurement, and the mre precise is the calculated mlar mass. Hwever, if the slute wn t disslve in camphr, then camphr is n gd, and anther slvent must be chsen that will disslve the slute. 57. Mlality m ml slute 7.0 g N H4CO 1000 g 1 ml NH4CO kg slvent g H O kg.00 mlal g N H CO 4 ΔT b K b m 0.51 C.00 mlal 1.5 C mlal The biling pint is raised frm C t C (assuming 1 atm). 58. ΔT f K f m, ΔT f 1.50 C 1.86 C m, m ml/kg mlal 0.00 kg H O ml CH8O kg H O 9.09 g C H O g C H 8 O ml CH8O 59. Mlality m 50.0 g C H6O 50.0 g H O 1000 g 1 ml 16.1 ml/kg kg 6.07 g ΔT f K f m 1.86 C/mlal 16.1 mlal 9.9 C; T f 0.0 C 9.9 C 9.9 C ΔT b K b m 0.51 C/mlal 16.1 mlal 8. C; T b C + 8. C 108. C 60. ΔT 5.50 C 4.59 C 0.91 C K f m, m 0.91 C 0.10 ml/kg 9.1 C / mlal

16 CHTER 17 ROERTIES OF SOUTIONS m 0.10 ml H O 18.0 g HO Mass H O kg t-butanl g H O kg t - butanl ml HO 4.0 g kg 1 ml 58.0 g T b C C C ml/kg; ΔT b K b m 0.51 C kg/ml ml/kg 0.5 C 6. π MRT, M π RT 8.00 atm atm K ml 98 K 0.7 ml/ 6. ΔT b C C 1.5 C; m ΔT b 1.5 C 0.68 ml/kg K 5.0 C kg / ml b Ml bimlecule kg slvent 0.68 ml hydrcarbn kg slvent ml Frm the prblem,.00 g bimlecule was used, which must cntain ml bimlecule. The mlar mass f the bimlecule is:.00 g ml 498 g/ml 64. M π RT 1atm trr 760 trr ml/ atm 00. K K ml ml ml catalase Mlar mass g ml g/ml 65. ΔT f K f m, m ΔTf K f 0.00 C 5.1 C kg / ml ml thyrxine kg benzene The mles f thyrxine present are: kg benzene ml thyrxine kg benzene ml thyrxine

17 684 CHTER 17 ROERTIES OF SOUTIONS 4 Frm the prblem, g thyrxine was used; this must cntain ml thyrxine. The mlar mass f the thyrxine is: mlar mass g ml 776 g/ml 66. m ΔTf 0.0 C 16.1 ml C H 6 O /kg K 1.86 C kg / ml f Because the density f water is 1.00 g/cm, the mles f C H 6 O needed are: 15.0 H O 1.00 kg H O 16.1 ml CH6O 4 ml C H 6 O H O kg H O 6.07 g lume C H 6 O 4 ml C H 6 O ml C H O 6 1cm 1,500 cm g ΔT b K b m 0.51 C 16.1 mlal 8. C; T b C + 8. C 108. C mlal 67. M 1.0 g 1 ml g ml/; π MRT t 98 K: π ml atm K ml 98 K 760 trr atm, π 0.0 trr Because d 1.0 g/cm, 1.0 f slutin has a mass f 1.0 kg. Because nly 1.0 g f prtein is present per liter slutin, 1.0 kg f H O is present, and mlality equals mlarity t the crrect number f significant figures. ΔT f K f m 1.86 C mlal mlal C 68. Osmtic pressure is better fr determining the mlar mass f large mlecules. temperature change f 10-5 C is very difficult t measure. change in height f a clumn f mercury by 0. mm (0. trr) is nt as hard t measure precisely. 0.1 ml atm 69. π MRT 98 K.45 atm atm K ml π atm 760 mm Hg atm 000 mm m The smtic pressure wuld supprt a mercury clumn f apprximately m. The height f a fluid clumn in a tree will be higher because Hg is mre dense than the fluid in a tree. If we

18 CHTER 17 ROERTIES OF SOUTIONS 685 assume the fluid in a tree is mstly H O, then the fluid has a density f 1.0 g/cm. The density f Hg is 1.6 g/cm. Height f fluid m m 70. ΔT f C; m ΔT K f f.70 C 0.57 mlal 5.1 C/mlal et x mass f naphthalene (mlar mass 18. g/ml). Then 1.60 x mass f anthracene (mlar mass 178. g/ml). x 18. mles naphthalene and 1.60 x 178. mles anthracene x 1.60 x ml slute (178.) x (18.) (18.) x, kg slvent kg slvent 18.(178.) (50.0)x , (50.0)x , (50.0)x 5, x 0.70 g naphthalene S the mixture is: 0.70 g % naphthalene by mass and 56% anthracene by mass 1.60 g 71. With additin f salt r sugar, the smtic pressure inside the fruit cells (and bacteria) is less than utside the cell. Water will leave the cells, which will dehydrate any bacteria present, causing them t die. rperties f Electrlyte Slutins 7. strng electrlyte dissciates cmpletely int ins in slutin. weak electrlyte dissciates nly partially int ins in slutin. Clligative prperties depend n the ttal number f particles in slutin. By measuring a prperty such as freezing-pint depressin, biling-pint elevatin, r smtic pressure, we can calculate the van t Hff factr (i) t see if an electrlyte is strng r weak trr O (.8 trr) H H, O 0.84; slute is the mle fractin f all the slute particles present. Because NaCl disslves t prduce tw ins in slutin (Na + and Cl ), is the mle fractin f Na + and Cl ins present. The mle fractin f NaCl is 1/ (0.176) NaCl. t 45 C, sln 0.84(71.9 trr) 59. trr.

19 686 CHTER 17 ROERTIES OF SOUTIONS 74. If ideal, NaCl dissciates cmpletely, and i.00. ΔT f ik f m; assuming water freezes at 0.00 C: 1.8 C 1.86 C kg/ml m, m 0.44 ml NaCl/kg H O ssume an amunt f slutin that cntains 1.00 kg f water (slvent) g 0.44 ml NaCl 0.1 g NaCl ml 0.1g Mass % NaCl % g + 0.1g 75. Na O 4 (s) Na + (aq) + O 4 (aq), i 4.0; CaBr (s) Ca + (aq) + Br (aq), i.0 KCl(s) K + (aq) + Cl (aq), i.0 The effective particle cncentratins f the slutins are (assuming cmplete dissciatin): 4.0(0.010 mlal) mlal fr the Na O 4 slutin;.0(0.00 mlal) mlal fr the CaBr slutin;.0(0.00 mlal) mlal fr the KCl slutin; slightly greater than 0.00 mlal fr the HF slutin because HF nly partially dissciates in water (it is a weak acid). a. The m Na O 4 slutin and the 0.00 m KCl slutin bth have effective particle cncentratins f m (assuming cmplete dissciatin), s bth f these slutins shuld have the same biling pint as the m C 6 H 1 O 6 slutin (a nnelectrlyte). b. ; as the slute cncentratin decreases, the slvent s vapr pressure increases because increases. Therefre, the 0.00 m HF slutin will have the highest vapr pressure because it has the smallest effective particle cncentratin. c. ΔT K f m; the 0.00 m CaBr slutin has the largest effective particle cncentratin, s it will have the largest freezing pint depressin (largest ΔT). 76. The slutins f glucse, NaCl and CaCl will all have lwer freezing pints, higher biling pints, and higher smtic pressures than pure water. The slutin with the largest particle cncentratin will have the lwest freezing pint, the highest biling pint, and the highest smtic pressure. The CaCl slutin will have the largest effective particle cncentratin because it prduces three ins per mle f cmpund. a. pure water b. CaCl slutin c. CaCl slutin d. pure water e. CaCl slutin 77. The van t Hff factr i is the number f mles f particles (ins) prduced fr every mle f slute disslved. Fr NaCl, i since Na + and Cl are prduced in water; fr l(no ), i 4 since l + and NO ins are prduced when l(no ) disslves in water. In real life, the van t Hff factr is rarely the value predicted by the number f ins a salt disslves int; i is

20 CHTER 17 ROERTIES OF SOUTIONS 687 generally smething less than the predicted number f ins. This is due t a phenmenn called in pairing, where at any instant a small percentage f ppsitely charged ins pair up and act like a single slute particle. In pairing ccurs mst when the cncentratin f ins is large. Therefre, dilute slutins behave mst ideally; here, i is clse t that determined by the number f ins in a salt. 78. a. s discussed in Figure f the text, the water wuld migrate frm right t left. Initially, the level f liquid in the right arm wuld g dwn, and the level in the left arm wuld g up. t sme pint the rate f slvent transfer will be the same in bth directins, and the levels f the liquids in the tw arms will stabilize. The height difference between the tw arms will be a measure f the smtic pressure f the NaCl slutin. b. Initially, H O mlecules will have a net migratin int the NaCl side. Hwever, NaCl mlecules can nw migrate int the H O side. Because slute and slvent transfer are bth pssible, the levels f the liquids will be equal nce the rate f slute and slvent transfer is equal in bth directins. t this pint the cncentratin f NaCl will be equal in bth chambers, and the levels f liquid will be equal. 79. NaCl(s) Na + (aq) + Cl (aq), i.0 π imrt ml atm 9 K 4.8 atm K ml pressure greater than 4.8 atm shuld be applied t ensure purificatin by reverse smsis. 80. a. MgCl (s) Mg + (aq) + Cl (aq), i.0 ml ins/ml slute ΔT f ik f m C/mlal mlal 0.8 C; T f -0.8 C (ssuming water freezes at 0.00 C.) ΔT b ik b m C/mlal mlal C; T b C (ssuming water bils at C.) b. FeCl (s) Fe + (aq) + Cl (aq), i 4.0 ml ins/ml slute ΔT f ik f m C/mlal mlal 0.7 C; T f 0.7 C ΔT b ik b m C/mlal mlal 0.10 C; T b C 81. a. MgCl, i (bserved).7 ΔT f ik f m C/mlal mlal 0.5 C; T f 0.5 C ΔT b ik b m C/mlal mlal C; T b C b. FeCl, i (bserved).4 ΔT f ik f m C/mlal mlal 0. C; T f -0. C ΔT b ik b m C/mlal mlal C; T b C

21 688 CHTER 17 ROERTIES OF SOUTIONS 8. ΔT f ik f m, i ΔTf K m f C 1.86 C / mlal 0.05 mlal.6 fr 0.05 m CaCl i fr m CaCl ; i fr 0.78 m CaCl Nte that i is less than the ideal value f.0 fr CaCl. This is due t in pairing in slutin. π 8. π imrt, M irt atm atm K ml 98 K ml/ Mlar mass f cmpund g ml 97.8 g/ml 84. a. T C 5(T F )/9 5( 9 )/9 4 C ssuming the slubility f CaCl is temperature independent, the mlality f a saturated CaCl slutin is: 74.5 g CaCl 1000 g g H O kg 1 ml CaCl g CaCl 6.71 ml CaCl kg H O ΔT f ik f m C kg/ml 6.71 ml/kg 7.4 C ssuming i.00, a saturated slutin f CaCl can lwer the freezing pint f water t 7.4 C. ssuming these cnditins, a saturated CaCl slutin shuld melt ice at 4 C ( 9 F). b. Frm Exercise 17.8, i.6; ΔT f ik f m C; T f C. ssuming i.6, a saturated CaCl slutin will nt melt ice at 4 C ( 9 F). dditinal Exercises 85. Bth slutins and cllids have suspended particles in sme medium. The majr difference between the tw is the size f the particles. cllid is a suspensin f relatively large particles cmpared t a slutin. Because f this, cllids will scatter light, whereas slutins will nt. The scattering f light by a cllidal suspensin is called the Tyndall effect The micelles frm s that the inic ends f the detergent mlecules, the SO 4 ends, are expsed t the plar water mlecules n the utside, whereas the nnplar hydrcarbn chains frm the detergent mlecules are hidden frm the water by pinting tward the inside f the micelle. Dirt, which is basically nnplar, is stabilized in the nnplar interir f the micelle and is washed away.

22 CHTER 17 ROERTIES OF SOUTIONS 689 detergent mlecule SO 4 - nnplar hydrcarbn dirt 87. Cagulatin is the destructin f a cllid by the aggregatin f many suspended particles t frm a large particle that settles ut f slutin. 88. The main factr fr stabilizatin seems t be electrstatic repulsin. The center f a cllid particle is surrunded by a layer f same charged ins, with ppsitely charged ins frming anther charged layer n the utside. Overall, there are equal numbers f charged and ppsitely charged ins, s the cllidal particles are electrically neutral. Hwever, since the uter layers are the same charge, the particles repel each ther and d nt easily aggregate fr precipitatin t ccur. Heating increases the velcities f the cllidal particles. This causes the particles t cllide with enugh energy t break the in barriers, allwing the cllids t aggregate and eventually precipitate ut. dding an electrlyte neutralizes the adsrbed in layers, which allws cllidal particles t aggregate and then precipitate ut prf ethanl slutin is 46% C H 5 OH by vlume. ssuming m f slutin: ml ethanl 46 m C H 5 OH mlarity 0.79 ml g m 7.9 M ethanl 1 ml C H OH 0.79 ml C H 5 OH g g 90. Mass f H O 160. m 159 g kg m 0.78 ml Ml NaDTZ kg ml kg Mlar mass f NaDTZ 8.4 g ml 69 g/ml

23 690 CHTER 17 ROERTIES OF SOUTIONS sln OH O H ; ml H O 159 g 1 ml 8.8 ml 18.0 g Sdium diatrizate is a salt because there is a metal (sdium) in the cmpund. Frm the shrt-hand ntatin fr sdium diatrizate, NaDTZ, we can assume this salt breaks up int Na + and DTZ ins. S the mles f slute particles are (0.0601) 0.10 ml slute particles. H O 8.8 ml 0.10 ml ml 0.987; sln trr.7 trr 91. ΔT K f m, m ΔT K f.79 C 1.50 mlal 1.86 C/mlal a. ΔT K b m, ΔT (0.51EC/mlal)(1.50 mlal) 0.77EC, T b EC b. sln waterwater, water ml H ml H O O + ml slute ssuming 1.00 kg f water, we have 1.50 ml slute, and: ml H O ml H O g H O 55.5 ml H O 18.0 g H O 55.5 ml water 0.974; sln (0.974)(.76 mm Hg).1 mm Hg c. We assumed ideal behavir in slutin frmatin, we assumed the slute was nnvlatile, and we assumed i 1 (n ins frmed). 9. ΔT imk f, i ΔT mk f 0.50 ml kg.79 C 1.86 C kg ml.00 We have three ins in slutins, and we have twice as many anins as catins. Therefre, the frmula f Q is MCl. ssuming g f cmpund: 1 ml Cl 8.68 g Cl ml Cl 5.45 g ml M ml Cl 1 ml M ml M ml Cl Mlar mass f M 61. g M ml M 11.4 g/ml; M is Cd, s Q CdCl.

24 CHTER 17 ROERTIES OF SOUTIONS O C O H Benzic acid is capable f hydrgen-bnding, but a significant part f benzic acid is the nnplar benzene ring. In benzene, a hydrgen-bnded dimer frms. C O H O O H O C The dimer is relatively nnplar and thus mre sluble in benzene than in water. Because benzic acid frms dimers in benzene, the effective slute particle cncentratin will be less than 1.0 mlal. Therefre, the freezing-pint depressin wuld be less than 5.1 C (ΔT f K f m). 94. Benzic acid (see Exercise 9) wuld be mre sluble in a basic slutin because f the reactin: C 6 H 5 CO H + OH C 6 H 5 CO + H O By remving the prtn frm benzic acid, an anin frms, and like all anins, the species becmes mre sluble in water. 95. a. NH 4 NO (s) NH 4 + (aq) + NO (aq) ΔH sln? Heat gain by disslutin prcess heat lss by slutin; We will keep all quantities psitive in rder t avid sign errrs. Because the temperature f the water decreased, the disslutin f NH 4 NO is endthermic (ΔH is psitive). Mass f slutin g Heat lss by slutin 4.18 J 76.6 g (5.00 C.4 C) 5 J C g 5 J g NH4NO ΔH sln J/ml 6.6 kj/ml 1.60 g NH4NO ml NH4NO b. We will use Hess s law t slve fr the lattice energy. The lattice energy equatin is: NH 4 + (g) + NO (g) NH 4 NO (s) ΔH lattice energy NH + 4 (g) + NO (g) NH + 4 (aq) + NO (aq) ΔH ΔH hyd -60. kj/ml NH + 4 (aq) + NO (aq) NH 4 NO (s) ΔH ΔH sln -6.6 kj/ml NH + 4 (g) + NO (g) NH 4 NO (s) ΔH ΔH hyd ΔH sln -657 kj/ml

25 69 CHTER 17 ROERTIES OF SOUTIONS 96. a. The average values fr each in are: 00. mg Na +, 15.7 mg K +, 5.45 mg Ca +, 88 mg Cl, and 46 mg lactate (C H 5 O ) Nte: Because we can precisely weigh t ±0.1 mg n an analytical balance, we'll carry extra significant figures and calculate results t ±0.1 mg. The nly surce f lactate is NaC H 5 O. 46 mg C H 5 O mg NaCH5O mg C H O mg sdium lactate The nly surce f Ca + is CaCl CH O mg Ca mg CaCl HO mg Ca r 0.0 mg CaCl CH O The nly surce f K + is KCl mg K mg KCl mg K 9.9 mg KCl Frm what we have used already, let's calculate the mass f Na + added mg sdium lactate 46.0 mg lactate 6.5 mg Na + Thus we need t add an additinal 6.5 mg Na + t get the desired 00. mg. 6.5 mg Na mg NaCl +.99 mg Na 601. mg NaCl Nw let's check the mass f Cl added: 0.0 mg CaCl CH O mg Cl mg CaCl H O 9.6 mg Cl 0.0 mg CaCl CH O 9.6 mg Cl 9.9 mg KCl! 15.7 mg K mg Cl 601. mg NaCl! 6.5 mg Na mg Cl Ttal Cl 88.5 mg Cl This is the quantity f Cl we want (the average amunt f Cl ). n analytical balance can weigh t the nearest 0.1 mg. We wuld use 09.5 mg sdium lactate, 0.0 mg CaCl CH O, 9.9 mg KCl, and 601. mg NaCl.

26 CHTER 17 ROERTIES OF SOUTIONS 69 b. T get the range f smtic pressure, we need t calculate the mlar cncentratin f each in at its minimum and maximum values. t minimum cncentratins, we have: 85 mg Na 100. m + 1 mml.99 mg 0.14 M; 14.1 mg K 100. m + 1 mml 9.10 mg M 4.9 mg Ca 100. m + 1 mml mg M; 68 mg Cl 100. m 1 mml 5.45 mg M 1 mg C H5O 100. m 1 mml mg M (Nte: mlarity ml/ mml/m.) Ttal M π MRT 0.59 ml atm 10. K 6.59 atm K ml Similarly, at maximum cncentratins, the cncentratin fr each in is: Na + : 0.17 M; K + : M; Ca + : M; Cl : M; C H 5 O : 0.09 M The ttal cncentratin f all ins is 0.87 M. π 0.87 ml atm 10. K 7.0 atm K ml Osmtic pressure ranges frm 6.59 atm t 7.0 atm. pen 97. pen 0.15 ; pen penpen; ttal pen + hex pen (511) + hex (150.) ttal Because : (511) + (1.000 )(150.) hex pen ttal pen pen pen pen pen (511) pen, 0.15, 0.15( pen ) ttal pen 511 pen + 54 pen 511 pen, pen a. Water bils when the vapr pressure equals the pressure abve the water. In an pen pan, atm 1.0 atm. In a pressure cker, inside > 1.0 atm, and water bils at a higher temperature. The higher the cking temperature, the faster is the cking time. b. Salt disslves in water, frming a slutin with a melting pint lwer than that f pure water (ΔT f K f m). This happens in water n the surface f ice. If it is nt t cld, the ice melts. This wn't wrk if the ambient temperature is lwer than the depressed freezing pint f the salt slutin.

27 694 CHTER 17 ROERTIES OF SOUTIONS c. When water freezes frm a slutin, it freezes as pure water, leaving behind a mre cncentrated salt slutin. d. On the CO phase diagram, the triple pint is abve 1 atm, and CO (g) is the stable phase at 1 atm and rm temperature. CO (l) can't exist at nrmal atmspheric pressures. Therefre, dry ice sublimes instead f bils. In a fire extinguisher, > 1 atm and CO (l) can exist. When CO is released frm the fire extinguisher, CO (g) frms as predicted frm the phase diagram mg CO 1.01 mg C mg CO.88 mg C; % C.88 mg 4.80 mg % C 1.65 mg H O.016 mg H 18.0 mg H O mg H; % H mg 4.80 mg % H Mass % O ( ) 15.4% O Out f g: 80.8 g C 1 ml 1.01 g 6.7 ml C; g H 1 ml g.8 ml H; g O 1 ml g 0.96 ml O; Therefre, the empirical frmula is C 7 H 4 O. ΔT f K f m, m ΔTf. C 0.56 mlal K 40. C / mlal f Ml anthraquinne kg camphr 0.56 ml anthraquinne kg camphr ml Mlar mass 1. g ml 10 g/ml The empirical mass f C 7 H 4 O is 7(1) + 4(1) g/ml. Because the mlar mass is twice the empirical mass, the mlecular frmula is C 14 H 8 O Out f g, there are: 1.57 g C 1 ml C g.68 ml C;

28 CHTER 17 ROERTIES OF SOUTIONS g H 1 ml H g 5.6 ml H; g O 1 ml O g.946 ml O; Empirical frmula: C H 4 O ; use the freezing-pint data t determine the mlar mass. m ΔTf 5.0 C.80 mlal K 1.86 C/mlal f Ml slute kg.80 ml slute kg ml slute Mlar mass g ml 151 g/ml The empirical frmula mass f C H 4 O g/ml. Because the mlar mass is abut twice the empirical mass, the mlecular frmula is C 4 H 8 O 6, which has a mlar mass f g/ml. Nte: We use the experimental mlar mass t determine the mlecular frmula. Knwing this, we calculate the mlar mass precisely frm the mlecular frmula using atmic masses a. m ΔTf 1. C 0.58 ml/kg K 5.1 C kg / ml f Ml unknwn kg Mlar mass f unknwn Uncertainty in temperature 0.58 ml unknwn kg 1. g ml % 1. % uncertainty in 0 g/ml 9 g/ml. S mlar mass 0 ±9 g/ml. 0 g/ml ml b. N, cdeine culd nt be eliminated since its mlar mass is in the pssible range including the uncertainty. c. We wuld like the uncertainty t be ±1 g/ml. We need the freezing-pint depressin t be abut 10 times what it was in this prblem. Tw pssibilities are:

29 696 CHTER 17 ROERTIES OF SOUTIONS 1. make the slutin 10 times mre cncentrated (may be slubility prblem). use a slvent with a larger K f value, e.g., camphr 10. MX M + + X K sp [M + ][X ]; ΔT K f m, m ml kg ΔT K f ml/kg kg 50 g ml ttal slute particles (carrying extra sig. fig.) 1000 g ssume a slutin density f 1.0 g/m s that vlume f slutin 50 m. [M + ] (0.0075/) M, [X (0.0075/) ] M K sp [M + ][X ] ( )( ) M X (s) M + (aq) + X (aq) K sp [M + ] [X ] Initial s slubility (ml/) 0 0 Equil. s s K sp (s) (s) 108s 5 ; ttal in cncentratin s + s 5s. π imrt, im ttal in cncentratin π RT atm K 1 ml atm 1 98 K ml/ 5s ml/, s ml/ K sp 108s 5 108( ) m ΔTf 0.46 C 0.9 mlal K 1.86 C/ mlal f ssuming a slutin density 1.00 g/m, then 1.00 cntains 0.9 ml slute. NaCl Na + + Cl i ; s: (ml NaCl) + ml C 1 H O ml Mass NaCl + mass C 1 H O g n NaCl + Slving: n 0.9 and 58.44(n NaCl ) + 4.( n ) 0.0 n C C 1HO11 1HO11 C 1HO ml 14.5 g and n NaCl 0.09 ml 5.45 g

30 CHTER 17 ROERTIES OF SOUTIONS 697 Mass % C 1 H O g 0.0 g % and 7.5% NaCl by mass C 1HO ml ml ml m g kg 1 ml g ml/kg ml/ (dilute slutin) ΔT f ik f m, C i(1.86 C/mlal)( mlal), i 1.5 If i 1.0, percent dissciatin 0%, and if i.0, percent dissciatin 100%. Because i 1.5, the weak acid is 50.% dissciated. H H + + K a + [H ][ ] [H] Because the weak acid is 50.% dissciated: [H + ] [ ] [H] M M [H] [H] 0 amunt H reacted M M M K a + [H ][ ] [H] ( )( ) ΔT f K f m, m ΔTf 5.40 C.90 mlal K 1.86 C / mlal f.90 ml slute n, n ml f ins in slutin kg slvent kg Because NaNO and Mg(NO ) are strng electrlytes: n (x ml f NaNO ) + [y ml Mg(NO ) ] ml ins g In additin: 6.50 g x ml NaNO ml 148. g + y ml Mg(NO ) ml We have tw equatins: x + y and (85.00)x + (148.)y 6.50 Slving by simultaneus equatins:

31 698 CHTER 17 ROERTIES OF SOUTIONS (85.00)x (17.5)y 6.16 (85.00)x + (148.)y 6.50 (0.8)y 0.4, y ml Mg(NO ) Mass f Mg(NO ) ml 148. g/ml.4 g Mg(NO ), r 7% Mg(NO ) by mass Mass f NaNO 6.50 g -.4 g 4.1 g NaNO, r 6% NaNO by mass 107. im π atm ml/ ttal in cncentratin 1 1 RT atm K ml (98. K) Mg Na Cl Cl Mg Na ml/ M + + M + + M ; M M + + M + (charge balance) Cmbining: M M Mg Na et x mass MgCl and y mass NaCl; then x + y g. x y M and Mg + M + (Because ) Na Ttal in cncentratin x y ml/ Rearranging: x + (.585)y 1.57 Slving by simultaneus equatins: x + (.585)y 1.57 (x + y) (0.5000) (0.585)y 0.07, y 0.14 g NaCl Mass MgCl g 0.14 g 0.6 g; mass % MgCl 0.6 g g 100 7% 108. Use the thermdynamic data t calculate the biling pint f the slvent. ΔH J/ml t biling pint, ΔG 0 ΔH TΔS, ΔH TΔS, T 5. K 1 1 ΔS J K ml ΔT K b m, (55.4 K 5. K) (.5 K kg/ml)(m), m Mass slvent 150. m g 1 kg 0.1 kg m 1000 g ml/kg.5

32 CHTER 17 ROERTIES OF SOUTIONS 699 Mass slute 0.1 kg slvent 0.84 ml slute kg slvent 14 g 15.7 g 16 g slute ml Challenge rblems 109. Frm the prblem, C H 6 6 CCl We need the pure vapr pressures ( ) in rder t 4 calculate the vapr pressure f the slutin. C 6 H 6 (l) C 6 H 6 (g) K at 5 C C H C 6H6 6 6 ΔG ΔG ΔG rxn f, C6H6 (g) f, C6H6 (l) kj/ml kj/ml 5.16 kj/ml ΔG J/ml ΔG RT ln K, ln K 1 1 RT (8.145 J K ml )(98 K).08 K H C 6 6 e atm Fr CCl 4 : ΔG kj/ml ( 65.1 kj/ml) 4.6 kj/ml rxn ΔG f, CCl (g) ΔG 4 f, CCl4 (l) K CCl 4 ΔG 460 J/ml exp exp 1 1 RT J K ml 98 K atm 0.500(0.15 atm) atm; C H C6H6 C6H6 6 6 CCl (0.155 atm) atm C H C H ttal atm atm atm ; CCl et mle fractin in slutin, s B. Frm the prblem,. ttal (50.0 trr) (50.0 trr) + (1.000 )(100.0 trr) (50.0) (50.0) , (50.0) 75.0, 0.00 The mle fractin f in slutin is 0.00.

33 700 CHTER 17 ROERTIES OF SOUTIONS 111. Fr the secnd vapr cllected, B, and T, et B, mle fractin f benzene in the secnd slutin and mle fractin f tluene in the secnd slutin. T, B, T, B, B ttal + B B T B, B, (750.0 trr) (750.0 trr) + (1.000 B, )(00.0 trr) Slving: B, T, This secnd slutin came frm the vapr cllected frm the first (initial) slutin, s, T, et B,1 mle fractin benzene in the first slutin and f tluene in first slutin B,1 T,1 B,1 T,1 mle fractin B, B ttal + B B T B,1 B,1 (750.0 trr) (750.0 trr) + (1.000 B,1 )(00.0 trr) Slving: B, The riginal slutin had B 0.86 and T a. Freezing-pint depressin is determined using mlality fr the cncentratin units, whereas mlarity units are used t determine smtic pressure. We need t assume that the mlality f the slutin equals the mlarity f the slutin. b. Mlarity mles slvent ; mlality liters slutin mles slvent kg slvent When the liters f slutin equal the kilgrams f slvent present fr a slutin, then mlarity equals mlality. This ccurs fr an aqueus slutin when the density f the slutin is equal t the density f water, 1.00 g/cm. The density f a slutin is clse t 1.00 g/cm when nt a lt f slute is disslved in slutin. Therefre, mlarity and mlality values are clse t each ther nly fr dilute slutins. c. ΔT K f m, m Δ T 0.61 C 1.86 C kg/ml K f 0.4 ml/kg ssuming 0.4 ml/kg 0.4 ml/: π MRT 0.4 ml atm 98 K 8.17 atm K ml

34 CHTER 17 ROERTIES OF SOUTIONS 701 d. m Δ T.0 C 0.51 C kg/ml K b.9 ml/kg This slutin is much mre cncentrated than the istnic slutin in part c. Here, water will leave the plant cells in rder t try t equilibrate the in cncentratin bth inside and utside the cell. Because there is such a large cncentratin discrepancy, all the water will leave the plant cells, causing them t shrivel and die. 11. a. ssuming MgCO (s) des nt dissciate, the slute cncentratin in water is: 560 μg MgCO (s) 560 mg g 1 ml MgCO m 84. g ml MgCO / n applied pressure f 8.0 atm will purify water up t a slute cncentratin f: M π RT atm K 8.0 atm 1 1 ml 00. K 0. ml 114. m When the cncentratin f MgCO (s) reaches 0. ml/, the reverse smsis unit can n lnger purify the water. et vlume () f water remaining after purifying 45 f H O. When + 45 f water has been prcessed, the mles f slute particles will equal: ml/ (45 + ) 0. ml/ Slving: 0.0 ( ), 0.96 The minimum ttal vlume f water that must be prcessed is Nte: If MgCO des dissciate int Mg + and CO ins, then the slute cncentratin increases t M, and at least 47 f water must be prcessed. b. N; a reverse smsis system that applies 8.0 atm can nly purify water with a slute cncentratin f less than 0. ml/. Salt water has a slute cncentratin f (0.60 M) 1. ml/ ins. The slute cncentratin f salt water is much t high fr this reverse smsis unit t wrk. ΔT K f C 0.18 ml/kg 1.86 C/mlal π MRT, where M ml/; we must assume that mlarity mlality s that we can calculate the smtic pressure. This is a reasnable assumptin fr dilute slutins when 1.00 kg f water 1.00 f slutin. ssuming cmplete dissciatin f NaCl, a 0.18 m slutin crrespnds t 6.7 g NaCl disslved in 1.00 kg f water. The vlume f slutin may be a

35 70 CHTER 17 ROERTIES OF SOUTIONS little larger than 1.00 but nt by much (t three sig. figs.). The assumptin that mlarity mlality will be gd here. π (0.18 M)( atm K 1 ml 1 )(98 K) 5. atm 115. a. ssuming n in assciatin between SO 4 (aq) and Fe + (aq), then i 5 fr Fe (SO 4 ). π imrt 5( ml/)( atm K 1 ml 1 )(98 K) 6.11 atm b. Fe (SO 4 ) (aq) Fe + (aq) + SO 4 (aq) Under ideal circumstances, /5 f π calculated abve results frm Fe + and /5 results frm SO 4. The cntributin t π frm SO 4 is / atm.67 atm. Because SO 4 is assumed unchanged in slutin, the SO 4 cntributin in the actual slutin will als be.67 atm. The cntributin t the actual π frm the Fe(H O) 6 + dissciatin reactin is atm. The initial cncentratin f Fe(H O) + 6 is (0.0500) M. The setup fr the weak acid prblem is: Fe(H O) 6 H + + [H ][Fe(OH)(HO) 5 ] + Fe(OH)(H O) 5 K a + [Fe(H O) ] Initial M ~0 0 x ml/ f Fe(H O) 6 + reacts t reach equilibrium Equil x x x π.06 atm Ttal in cncentratin im 0.15 M 1 1 RT atm K ml (98 K) 0.15 M x + x + x x, x 0.05 M 6 K a [H + ][Fe(OH)(H [Fe(H O) + 6 O) ] + 5 ] x x (0.05) ( ) (0.05) K a Initial mles Cl g Cl 4 1 ml Cl 4 /19.74 g Cl ml Cl 4 ΔT 5.97 C Ttal mlality f slute particles im, 0.00 ml/kg Kf 9.8 C kg / ml Because we have kg CCl 4, the ttal mles f slute particles present is: 0.00 ml/kg ( kg) ml