1. Mass percent: the percent by mass of the solute in the solution. Molarity: the number of moles of solute per liter of solution.

Transcription

1 CHATER EEEN ROERTIES OF SOUTIONS Fr Review 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.. KF(s) K + (aq) + F (aq) H = H sln ; K + (g) + Cl (g) K + (aq) + F (aq) H = H hyd KF(s) K + (g) + F (g) H 1 = H E K + (g) + F (g) K + (aq) + F (aq) H = H hyd KF(s) K + (aq) + F (aq) H = H sln = H E + H hyd It is true that H 1 and H have large magnitudes fr their values; hwever, the signs are ppsite (H 1 is large and psitive because it is the reverse f the lattice energy and H, the hydratin energy, is large and negative). These tw H values basically cancel ut each ther giving a H sln value clse t zer.. 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 strength 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 strength 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 H sln is either a small psitive number (endthermic) r a small negative number (exthermic). Fr plar slutes in nnplar slvents and vice versa, H 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. 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. 7

2 CHATER 11 ROERTIES OF SOUTIONS 7 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. An 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 as the slubility f a gas typically decreases with increasing temperature. 5. Rault s law: sln = slvent slvent ; When a slute is added t a slvent, the vapr pressure f a slutin is lwered frm that f the pure slvent. The quantity slvent, the mle fractin f slvent, is the fractin that the slutin vapr pressure is lwered. Fr the experiment illustrated in Fig. 11.9, the beaker f water will stp having a net transfer f water mlecules ut f the beaker when the equilibrium vapr pressure, H O, is reached. This can never happen. The beaker with the slutin wants an equilibrium vapr pressure f H O, which is less than H O. When the vapr pressure ver the slutin is abve H O, a net transfer f water mlecules int the slutin will ccur in rder t try t reduce the vapr pressure t H O. The tw beakers can never btain the equilibrium vapr pressure they want. The net transfer f water mlecules frm the beaker f water t the beaker f slutin stps after all f the water has evaprated. If the slute is vlatile, then we can get a transfer f bth the slute and slvent back and frth between the beakers. A state can be reached in this experiment where bth beakers have the same slute cncentratin and hence the same vapr pressure. When this state is reached, n net transfer f slute r water mlecules ccurs between the beakers s the levels f slutin remain cnstant. When bth substances in a slutin are vlatile, then Rault s law applies t bth. The ttal vapr pressure abve the slutin is the equilibrium vapr pressure f the slvent plus the equilibrium vapr pressure f the slute. Mathematically: TOT A B AA BB where A is either the slute r slvent and B is the ther ne. 6. An ideal liquid-liquid slutin fllws Rault s law: TOT AA BB A nnideal liquid-liquid slutin des nt fllw Rault s law, either giving a ttal pressure greater than predicted by Rault s law (psitive deviatin) r less than predicted (negative deviatin).

3 74 CHATER 11 ROERTIES OF SOUTIONS In an ideal slutin, the strength f the intermlecular frces in slutin are equal t the strength f the intermlecular frces in pure slute and pure slvent. When this is true, H sln = 0 and T 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 H sln > 0 (are endthermic) and T sln < 0. Fr negative deviatins, the slutin has strnger intermlecular frces in slutin than in pure slute r pure slvent. Negative deviatins have H sln < 0 (are exthermic) and T sln > 0. Examples f each type f slutin are: ideal: psitive deviatins: negative deviatins: benzene-tluene ethanl-hexane acetne-water 7. Clligative prperties are prperties f a slutin that depend nly n the number, nt the identity, f the slute particles. A 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. A substance freezes when the vapr pressure f the liquid and slid are identical t each ther. Adding a slute t a substance lwers the vapr pressure f the liquid. A 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. A 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 pint where the vapr pressure f the liquid equals the external pressure. Hence, the biling pint is elevated when a slutin frms. The equatin t calculate the freezing pint depressin r biling pint elevatin is: T = Km where K is the freezing pint r biling pint cnstant fr the slvent and m is the mlality f the slute. Table 11.5 lists the K values fr several slvents. The slvent which shws the largest change in freezing pint fr a certain cncentratin f slute is camphr; it has the largest K f value. Water, with the smallest K b value, will shw the smallest increase in biling pint fr a certain cncentratin f slute. T calculate mlar mass, yu need t knw the mass f the unknwn slute, the mass and identity f slvent used, and the change in temperature (T) f the freezing r biling pint f the slutin. Since the mass f unknwn slute is knwn, ne manipulates the freezing pint data t determine the number f mles f slute present. Once the mass and mles f slute are knwn, ne can determine the mlar mass f the slute. T determine the mles f slute present, ne determines the mlality f the slutin frm the freezing pint data and multiplies this by the kilgrams f slvent present; this equals the mles f slute present. 8. 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,, is: = MRT

4 CHATER 11 ROERTIES OF SOUTIONS 75 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 since d H O = 1.00 g/cm. 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 bacteria present, causing them t die. Dialysis allws the transfer f slvent and small slute mlecules thrugh a membrane. T purify bld, the bld is passed thrugh a cellphane tube (the semipermeable membrane); this cellphane tube is immersed in a dialyzing slutin which cntains the same cncentratins f ins and small mlecules as in bld, but has nne f the waste prducts nrmally remved by the kidney. As bld is passed thrugh the dialysis machine, the unwanted waste prducts pass thrugh the cellphane membrane, cleansing the bld. Desalinatin is the remval f disslved salts frm an aqueus slutin. Here, a slutin is subjected t a pressure greater than the smtic pressure and reverse smsis ccurs, i.e., water passes frm the slutin thrugh the semipermeable membrane back int pure water. Desalinatin plants can turn sea-water with its high salt cntent int drinkable water. 9. A strng electrlyte cmpletely dissciates int ins in slutin, a weak electrlyte nly partially dissciates int ins in slutin, and a nnelectrlyte des nt dissciate int ins when disslved in slutin. Clligative prperties depend n the ttal number f slute particles in slutin. By measuring a prperty such as freezing pint depressin, biling pint elevatin, r smtic pressure, we can determine the number f slute particles present frm the slute and thus characterize the slute as a strng, weak, r nnelectrlyte. The van t Hff factr, i, is the number f mles f particles (ins) prduced fr every ml f slute disslved. Fr NaCl, i = since Na + and Cl are prduced in water; fr Al(NO ), i = 4 since Al + and NO ins are prduced when Al(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 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. 10. A cllidal dispersin is a suspensin f particles in a dispersing medium. See Table 11.7 fr sme examples f different types f cllids. Bth slutins and cllids have suspended particles in sme medium. The majr difference between the tw is the size f the particles. A cllid is a suspensin f relatively large particles as cmpared t a slutin. Because f this, cllids will scatter light while slutins will nt. The scattering f light by a cllidal suspensin is called the Tyndall effect. Cagulatin is the destructin f a cllid by the aggregatin f many suspended particles t frm a large particle that settles ut f slutin.

5 76 CHATER 11 ROERTIES OF SOUTIONS Slutin Review 9. 1 mlch7oh 585g CH7OH 60.09g CH7OH 1.00 = 9.74 M g CaCl 1mlCaCl g CaCl m = 19.9 m 0.580mlCaCl 11. 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 + = 0.00 ml ttalml Na 0.4ml 0.00ml 0.45ml M = 4.5 M Na + Na ttalvlume a. HNO (l) H + (aq) + NO (aq) b. Na SO 4 (s) Na + (aq) + SO 4 (aq) Questins c. Al(NO ) (s) Al + (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) 1. As the temperature increases, the gas mlecules will have a greater average kinetic energy. A greater fractin f the gas mlecules in slutin will have kinetic energy greater than the attractive frces between the gas mlecules and the slvent mlecules. Mre gas mlecules will escape t the vapr phase and the slubility f the gas will decrease. 14. Henry s law is beyed mst accurately fr dilute slutins f gases that d nt dissciate in r react with the slvent. NH is a weak base and reacts with water by the fllwing reactin: NH (aq) + H O(l) NH 4 + (aq) + OH (aq) O will bind t hemglbin in the bld. Due t these reactins in the slvent, NH (g) in water and O (g) in bld d nt fllw Henry s law.

6 CHATER 11 ROERTIES OF SOUTIONS 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. 16. Slutins f A and B have vapr pressures less than ideal (see Figure 11.1 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. A substance bils when the vapr pressure equals the external pressure. Since B = 0.6 has a lwer vapr pressure at the temperature f the plt than either pure A r pure B, then 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 A r pure B. (Nte that because B > A, B is mre vlatile than A, and B will have a lwer biling pint temperature than A). 17. N, the slutin is nt ideal. Fr an ideal slutin, the strength f intermlecular frces in slutin is 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 (heat is released), this slutin shws a negative deviatin frm Rault s law. 18. The micelles frm s the inic ends f the detergent mlecules, the SO 4 ends, are expsed t the plar water mlecules n the utside, while 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. = detergent mlecule = SO 4 - = nnplar hydrcarbn = dirt 19. Nrmality is the number f equivalents per liter f slutin. Fr an acid r a base, an equivalent is the mass f acid r base that can furnish 1 ml f prtns (if an acid) r accept 1

7 78 CHATER 11 ROERTIES OF SOUTIONS ml f prtns (if a base). A prtn is an H + in. Mlarity is defined as the mles f slute per liter f slutin. When the number f equivalents equals the number f mles f slute, then nrmality = mlarity. This is true fr acids which nly have ne acidic prtn in them and fr bases that accept nly ne prtn per frmula unit. Examples f acids where equivalents = mles slute are HCl, HNO, HF, and HC H O. Examples f bases where equivalents = mles slute are NaOH, KOH, and NH. When equivalents mles slute, then nrmality mlarity. This is true fr acids that dnate mre than ne prtn (H SO 4, H O 4, H CO, etc.) and fr bases that react with mre than ne prtn per frmula unit [Ca(OH), Ba(OH), Sr(OH), etc.]. 0. It is true that the sdium chlride lattice must be brken in rder t disslve in water, but a lt f energy is released when the water mlecules hydrate the Na + and Cl ins. These tw prcesses have relatively large values fr the amunt f energy assciated with them, but they are ppsite in sign. The end result is they basically cancel each ther ut resulting in a H sln 0. S energy is nt the reasn why inic slids like NaCl are s sluble in water. The answer lies in nature s tendency tward the higher prbability f the mixed state. rcesses, in general, are favred that result in an increase in disrder because the disrdered state is the easiest (mst prbable) state t achieve. The tendency f prcesses t increase disrder will be discussed in Chapter 16 when entrpy, S, is intrduced. 1. Only statement b is true. A substance freezes when the vapr pressure f the liquid and slid are the same. When a slute is added t water, the vapr pressure f the slutin at 0C is less than the vapr pressure f the slid and the net result is fr any ice present t cnvert t liquid in rder t try t equalize the vapr pressures (which never can ccur at 0C). A lwer temperature is needed t equalize the vapr pressure f water and ice, hence the freezing pint is depressed. Fr statement a, the vapr pressure f a slutin is directly related t the mle fractin f slvent (nt slute) by Rault s law. Fr statement c, clligative prperties depend n the number f slute particles present and nt n the identity f the slute. Fr statement d, the biling pint f water is increased because the sugar slute decreases the vapr pressure f the water; a higher temperature is required fr the vapr pressure f the slutin t equal the external pressure s biling can ccur.. 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 T, the mre precise the measurement and the mre precise the calculated mlar mass. Hwever, if the slute wn t disslve in camphr, then camphr is n gd and anther slvent must be chsen which will disslve the slute.. Istnic slutins are thse which have identical smtic pressures. Crenatin and hemlysis refer t phenmena that ccur when red bld cells are bathed in slutins having a mismatch in smtic pressures inside and utside the cell. When red bld cells are in a slutin having a higher smtic pressure than that f the cells, the cells shrivel as there is a net transfer f

8 CHATER 11 ROERTIES OF SOUTIONS 79 water ut f the cells. This is called crenatin. Hemlysis ccurs when the red bld cells are bathed in a slutin having lwer smtic pressure than that inside the cell. Here, the cells rupture as there is a net transfer f water t int the red bld cells. 4. In pairing is a phenmenn that ccurs in slutin when ppsitely charged ins aggregate and behave as a single particle. Fr example, when NaCl is disslved in water, ne wuld expect sdium chlride t exist as separate hydrated Na + ins and Cl ins. A few ins, hwever, stay tgether as NaCl and behave as just ne particle. In pairing increases in a slutin as the in cncentratin increases (as the mlality increases). Exercises Cncentratin f Slutins 5. Because the density f water is 1.00 g/m, m f water has a mass f 100. g. density = mass 10.0 g HO g HO = 1.06 g/m = 1.06 g/cm vlume 104m ml H O 4 = 10.0 g 1 ml 97.99g = 0.10 ml H O 4 ml H O = 100. g 1 ml 18.0g = 5.55 ml H O mle fractin f H O 4 = 0.10mlHO4 ( ) ml = H O = = mlarity = mlality = 6. mlality = mlarity = mlH O = ml/ mlH O = 1.0 ml/kg 0.100kg 40.0 g EG 1000g 1mlEG = 10.7 ml/kg 60.0 g H O kg 6.07g where EG = ethylene glycl (C H 6 O ) 40.0 g EG g slutin 1.05g 1000cm cm 1ml 6.07g = 6.77 ml/

9 80 CHATER 11 ROERTIES OF SOUTIONS 40.0 g EG 1 ml 6.07g = ml EG; 60.0 g H O 1 ml 18.0g =. ml H O EG = 0.16 = mle fractin ethylene glycl Hydrchlric acid: mlarity = 8g HCl 100. g s ln 1.19g s ln 1000cm cm s ln 1mlHCl = 1 ml/ 6.46g mlality = 8g HCl 6g slvent 1000g kg 1mlHCl 6.46g = 17 ml/kg 8 g HCl 1 ml 6.46g = 1.0 ml HCl; 6 g H O 1 ml 18.0g =.4 ml H O mle fractin f HCl = HCl 1.0 = Nitric acid: 70. g HNO 1.4g s ln 1000cm 1mlHNO 100. g s ln cm s ln = 16 ml/ 6.0g 70. g HNO 1000g 1mlHNO 0. g slvent kg 6.0g = 7 ml/kg 70. g HNO 1 ml 6.0g = 1.1 ml HNO ; 0. g H O 1 ml 18.0g = 1.7 ml H O HNO Sulfuric acid: 95g HSO 100. g s ln 95g H SO 5 g H O 1.1 = g s ln 1000cm 1mlHSO 4 = 18 ml/ cm s ln 98.09g HSO g kg 4 1ml 98.09g = 194 ml/kg 00 ml/kg 95 g H SO 4 1 ml 98.09g = 0.97 ml H SO 4 ; 5 g H O 1 ml 18.0g = 0. ml H O H SO =

10 CHATER 11 ROERTIES OF SOUTIONS 81 Acetic Acid: 99g HC HO 100. g s ln 99g HC HO 1g H O 1.05g s ln 1000cm cm s ln 1000g kg 1ml 60.05g 1ml 60.05g H = 17 ml/ = 1600 ml/kg 000 ml/kg 99 g HC H O HC HO 1 ml 60.05g 1.6 = = 1.6 ml HC H O ; 1 g H O 1 ml 18.0g = 0.06 ml H O Ammnia: 8g NH 100. g s ln 8g NH 7g H O 0.90g 1000cm cm 1000g kg 1ml 17.0g 1ml 17.0g = ml/kg = 15 ml/ 8 g NH 1 ml 17.0g = 1.6 ml NH ; 7 g H O 1 ml 18.0g = 4.0 ml H O NH 1.6 = a. If we use 100. m (100. g) f H O, we need: kg H O.0 mlkcl kg 74.55g ml KCl = 14.9 g = 15 g KCl 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. Since 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). et x = mass f NaCl. mass % NaOH = 15 = x 100, x = 100. x, x = 17.6 g 18 g 100. x Disslve 18 g NaOH in 100. m H O t make a 15% NaOH slutin by mass.

11 8 CHATER 11 ROERTIES OF SOUTIONS c. In a fashin similar t part b, let s use 100. m CH OH. et x = mass f NaOH m CH OH mass % NaOH = 5 = 0.79g = 79 g CH OH m Disslve 6 g NaOH in 100. m CH OH. x 100, 5(79) + 5 x = 100. x, x = 6. g 6 g 79 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 mlh O 100. g H O = 5.55 ml H O 18.0g C 6H1O6 x 0.10 ; 0.10 x = x, x = 0.6 ml C 6 H 1 O 6 x ml C 6 H 1 O g = 110 g C 6 H 1 O 6 ml Disslve 110 g C 6 H 1 O 6 in 100. m f H O t prepare a slutin with = C 6H1O m C 5 H 1 45 m C 6 H 14 mass % pentane = 0.6g = 16 g C 5 H 1 ; 5 m m 0.66g = 0. g C 6 H 14 ; 45 m m mass pentane 100 = ttalmass 16g 16g 0.6g m 0.66g m = 5% 0. g 1ml = 0. ml C 5 H g 1ml = 0.4 ml C 6 H g pentane = ml pentane = ttalml 0.ml = ml 0.4ml mlality = ml pentane = kg hexane 0.ml = = 7. ml/kg 0.00kg mlarity = ml pentane = slutin 0. If there are m f wine: 1.5 m C H 5 OH 0.ml 1000m =.1 ml/ 5m 45m g = 9.86 g C H 5 OH and 87.5 m H O m 1.00g = 87.5 g H O m

12 CHATER 11 ROERTIES OF SOUTIONS 8 mass % ethanl = = 10.1% by mass mlality = 9.86g CH5OH 1ml =.45 ml/kg kg H O 46.07g 1. If we have 1.00 f slutin: 1.7 ml citric acid m slutin 19.1g = 6 g citric acid (H C 6 H 5 O 7 ) ml 1.10g = g slutin m mass % f citric acid = 6g 100 =.9% g In 1.00 f slutin, we have 6 g citric acid and ( ) = 840 g f H O. mlality = 1.7 mlcitricacid 0.84kg H O = 1.6 ml/kg 840 g H O 1 ml 18.0g = 47 ml H O; citric acid 1.7 = Since citric acid is a triprtic acid, the number f prtns citric acid can prvide is three times the mlarity. Therefre, nrmality = mlarity: nrmality = 1.7 M = 4.11 N. 1.00mlacetne = 1.00 mlal; g C H 5 OH 1.00kg ethanl 1 ml 46.07g = 1.7 ml C H 5 OH acetne = 1.00 = ml CH COCH 58.08g CH COCH 1m = 7.7 m CH COCH mlch COCH 0.788g g ethanl 1 m 0.789g = 170 m; Ttal vlume = = 140 m mlarity = 1.00ml = M 1.4

13 84 CHATER 11 ROERTIES OF SOUTIONS Energetics f Slutins and Slubility. Using Hess s law: NaI(s) Na + (g) + I (g) ΔH = ΔH E = (686 kj/ml) Na + (g) + I (g) Na + (aq) + I (aq) 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. 4. 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) ΔH sln = 46 kj 46 kj = 47 kj + ΔH hyd, ΔH hyd = 9 kj CaI (s) Ca + (g) + I (g) ΔH = ΔH E = (059 kj) Ca + (g) + I (g) Ca + (aq) + I (aq) ΔH = ΔH hyd CaI (s) Ca + (aq) + I (aq) ΔH sln = 104 kj 104 kj = 059 kj + ΔH hyd, ΔH hyd = 16 kj b. The enthalpy f hydratin fr CaCl is mre exthermic than fr CaI. Any differences must be due t differences in hydratin between Cl and I. Thus, the chlride in is mre strngly hydrated as cmpared t the idide in. 5. Bth Al(OH) and NaOH are inic cmpunds. Since 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 Al + and OH cannt vercme the larger lattice energy and Al(OH) is insluble. Fr NaOH, the favrable hydratin energy is large enugh t vercme the smaller lattice energy and NaOH is sluble. 6. 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. Since 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.

14 CHATER 11 ROERTIES OF SOUTIONS 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. f. CH O, 4 + (1) + 6 = 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 8. a. water b. water c. hexane d. water 9. Water is a plar mlecule capable f hydrgen bnding. lar mlecules, especially mlecules capable f hydrgen bnding, and ins are all attracted t water. Fr cvalent cmpunds, as plarity increases, the attractin t water increases. Fr inic cmpunds, as the charge f the ins increases and/r the size f the ins decreases, the attractin t water increases. a. CH CH OH; CH CH OH is plar while CH CH CH is nnplar. b. CHCl ; CHCl is plar while CCl 4 is nnplar. c. CH CH OH; CH CH OH is much mre plar than CH (CH ) 14 CH OH.

15 86 CHATER 11 ROERTIES OF SOUTIONS 40. Fr inic cmpunds, as the charge f the ins increases and/r the size f the ins decreases, the attractin t water (hydratin) increases. a. Mg + ; smaller size, higher charge b. Be + ; smaller c. Fe + ; smaller size, higher charge d. F ; smaller e. Cl ; smaller f. SO 4 - ; higher charge 41. As 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. As 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. 4. Benzic acid is capable f hydrgen bnding, but a significant part f benzic acid is the nnplar benzene ring which is cmpsed f nly carbn and hydrgen. In benzene, a hydrgen bnded dimer frms: C O H O O H O C The dimer is relatively nnplar since the plar part f benzic acid is hidden in the dimer frmatin. Thus, benzic acid is mre sluble in benzene than in water due t the dimer frmatin. Benzic acid wuld be mre sluble in 0.1 M NaOH 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 ml 4. C = k, = k atm, k = ml C = k, C = 1.10 atm = 1.14 atm 10 ml/atm 10 ml/ m grape juice 1mCH5OH 0.79g CH5OH 1mlCH5OH 100. m juice m 46.07g mlco mlc H OH 5 = 1.54 ml CO (carry extra significant figure) 1.54 ml CO = ttal ml CO = ml CO (g) + ml CO (aq) = n g + n aq

16 CHATER 11 ROERTIES OF SOUTIONS 87 CO = n g RT = C CO = = k atm ng 98K mlk 7510 n aq = 4.0 n aq = 4.0 n aq.110 ml atm = 6 n g CO = 6 n g = 4.0 n aq and frm abve n aq = 1.54 n g ; Slving: CO 6 n g = 4.0(1.54 n g ), 69 n g = 66., n g = 0.18 ml = 6(0.18) = 59 atm in gas phase C = k CO =.110 ml 59 atm, C = 1.8 ml CO / in wine atm apr ressures f Slutins 45. ml C H 8 O = 164 g 1 ml 9.09g = 1.78 ml C H 8 O ml H O = 8 m 0.99g 1ml = 18.6 ml H O m 18.0g H O H O HO = 18.6 ml ( ) ml trr = trr = 50.0 trr 46. H OH = H OHC H OH; C H OH = C 5 C mlch5oh slutin ttalmlin slutin 5.6 g C H 8 O 1 mlc H8O = 0.58 ml C H 8 O 9.09g 1.7 g C H 5 OH 11 trr =.90ml.48ml 1 5 mlc H OH =.90 ml C H 5 OH; ttal ml = g =.48 ml H OH, C H OH C 5 5 = 16 trr

17 88 CHATER 11 ROERTIES OF SOUTIONS 47. B, / = atm/0.90 atm = BB B B B = ml benzene ; ml benzene = g C 6 H 6 ttalml 1 ml 78.11g = ml 1.000ml et x = ml slute, then: B = =, x = 1.000, x = 0.0 ml x mlar mass = 10.0 g 0.0ml = 0 g/ml.0 10 g/ml trr = Η Ο (.8 trr), Η Ο = 0.84; slute = = is the ml fractin f all the slute particles present. Since NaCl disslves t prduce tw ins in slutin (Na + and Cl ), is the mle fractin f Na + and Cl ins present (assuming cmplete dissciatin f NaCl). At 45 C, 49. a. 5 m C 5 H 1 H O = 0.84 (71.9 trr) = 59. trr 45 m C 6 H 14 pen 0.6g 1ml = 0. ml C 5 H 1 m 7.15g 0.66g m 1ml 86.17g ml pentanein slutin 0.ml = 0.9, ttalmlin slutin 0.56ml = 0.4 ml C 6 H 14 ; ttal ml = = 0.56 ml hex = = 0.61 pen penpen = 0.9(511 trr) =.0 10 trr; hex = 0.61(150. trr) = 9 trr ttal pen hex = = 9 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 pentanein vapr pen.010 trr = 0.69 ttalml vapr 90trr ttal Nte: In the Slutins Guide, we have added r t the mle fractin symbl t emphasize the value fr which we are slving. If the r is mitted, then the liquid phase is assumed. 50. = ttal CH Cl CH Br ; ; CHCl 0.000mlCH Cl mlttal = 0.75 ttal = 0.75 (1 trr) + ( ) (11.4 trr) = = 57.0 trr

18 CHATER 11 ROERTIES OF SOUTIONS 89 In the vapr: CH 49.9 trr Cl = 0.875; 57.0 trr CH Cl ttal CH Br = = , 174 trr = ttal meth 174 = 0 meth prp + (1.000 meth (0 trr) + meth ) 44.6, prp (44.6 trr); prp = meth 19 = 0.500; meth prp = = ; ; Fr the vapr, /. Since the mle fractins f tl tl tl ben ben ben A benzene and tluene are equal in the vapr phase, then. A ttal tl ben tl tl ben ben = (1.00 tl ) ben, tl (8 trr) = (1.00 tl ) 95 trr 1 tl = 95, tl = 0.77; ben = = Cmpared t H O, slutin d (methanl/water) will have the highest vapr pressure because methanl is mre vlatile than water. Bth slutin b (glucse/water) 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. Since there are mre slute particles in slutin c, the vapr pressure f slutin c will be the lwest. 54. 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 g CH COCH 1 ml 58.08g = ml acetne 50.0 g CH OH 1 ml.04g = 1.56 ml methanl acetne = = 0.56; methanl = acetne = ttal = methanl + acetne = 0.644(14 trr) (71 trr) = 9.1 trr trr Because partial pressures are prprtinal t the mles f gas present, then in the vapr phase: acetne acetne 96.5 trr = 0.51; trr ttal methanl = = = trr 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

19 90 CHATER 11 ROERTIES OF SOUTIONS frm Rault s law. This ccurs when the slute-slvent interactins are strnger than in pure slute and pure slvent. 56. a. An 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 slutins are nt between these limits, s water and prpanl d nt make ideal slutins. b. Frm the data, the vapr pressures f the varius slutins are greater than in the ideal slutin (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. c. The interactins between prpanl and water mlecules are weaker than between the pure substances since this slutin exhibits a psitive deviatin frm Rault s law. d. At H O = 0.54, the vapr pressure is highest as cmpared t the ther slutins. Since 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 than the ther slutins. Clligative rperties 57. mlality = m = ΔT b = K b m = mlslute 7.0g N H4CO 1000g kgslvent 150.0g H O kg 0.51C.00 mlal = 1.5 C mlal 1ml NH4CO =.00 mlal 60.06g N H CO The biling pint is raised frm C t C (assuming = 1 atm) ΔT b = C C = 1.5 C; m = ΔT K b b 1.5C = 0.68 ml/kg 5.0C kg/ ml ml bimlecule = kg slvent 0.68ml hydrcarbn = ml kgslvent Frm the prblem,.00 g bimlecule was used that must cntain ml bimlecule. The mlar mass f the bimlecule is:.00g = 498 g/ml ml 59. ΔT f = K f m, ΔT f = 1.50 C = 1.86C m, m = ml/kg mlal 0.00 kg H O 0.806mlCH8O kg H O 9.09g C H O 8 = 14.8 g C H 8 O mlch8o

20 CHATER 11 ROERTIES OF SOUTIONS ΔT f = 5.50 C 4.59 C = 0.91 C = K f m, m = 0.91C = 0.10 ml/kg 9.1C/ mlal mass H O = kg t-butanl 0.10ml H O 18.0g HO = g H O kg t butanl mlho 61. mlality = m = 50.0g C H6O 50.0g H O 1000g kg 1ml 6.07g = 16.1 ml/kg 6. m = 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 ΔTf K f 5.0C = 1.4 ml C H 6 O /kg 1.86C kg/ ml Since the density f water is 1.00 g/cm, the mles f C H 6 O needed are: 1.00kg HO 1.4mlCH6O 15.0 H O = 01 ml C H 6 O H O kg H O lume C H 6 O = 01 ml C H 6 O 6.07g mlc H O 6 1cm = 11,00 cm = g ΔT b = K b m = 0.51C mlal 1.4 mlal = 6.8 C, T b = C C = C 6. ΔT f = K f m, m = ΔTf K f 0.00C ml thyrxine 5.1C kg/ ml kg benzene The ml f thyrxine present is: kg benzene ml thyrxine 4 = ml thyrxine kg benzene 4 Frm the prblem, g thyrxine were used; this must cntain ml thyrxine. The mlar mass f the thyrxine is: mlar mass = 0.455g ml = 776 g/ml 64. empirical frmula mass 7(1) + 4(1) + 16 = 104 g/ml ΔT f = K f m, m = ΔTf K f.c = 0.56 mlal 40. C/ mlal

21 9 CHATER 11 ROERTIES OF SOUTIONS ml anthraquinne = kg slvent mlar mass = 65. a. M = 1. g = 10 g/ml ml 0.56mlanthraquinne = ml kgslvent mlar mass 10 =.0; mlecular frmula = C 14 H 8 O empiricalfrmula mass g prtein At 98 K: π = 1ml g = ml/; π = MRT ml atm 98 K mlk 760trr, π = 0.0 trr atm Because d = 1.0 g/cm, 1.0 slutin has a mass f 1.0 kg. Because nly 1.0 g f prtein is present per liter f slutin, 1.0 kg f H O is present and mlality equals mlarity. ΔT f = K f m = 1.86C mlal mlal =.0 10 C b. Osmtic pressure is better fr determining the mlar mass f large mlecules. A 5 temperature change f 10 C is very difficult t measure. A change in height f a clumn f mercury by 0. mm (0. trr) is nt as hard t measure precisely. 66. M = π RT 1atm 0.74 trr 760trr = atm 00. K mlk 5 10 ml/ 1.00 mlar mass = π = MRT, M = 68. M = π RT ml 5 = ml catalase 10.00g π RT 5 ml = g/ml 8.00atm = 0.7 ml/ atm 98K mlk 15atm = 0.6 M slute particles atm 95K mlk This represents the ttal mlarity f the slute particles. NaCl is a sluble inic cmpund that breaks up int tw ins, Na + and Cl. Therefre, the cncentratin f NaCl needed is 0.6/ = 0.1 M; this NaCl cncentratin will prduce a 0.6 M slute particle slutin assuming cmplete dissciatin.

22 CHATER 11 ROERTIES OF SOUTIONS 9 0.1ml NaCl 58.44g NaCl 1.0 = g NaCl ml NaCl Disslve 18 g f NaCl in sme water and dilute t 1.0 in a vlumetric flask. T get 0.1 ± 0.01 ml/, we need 18.1 g ± 0.6 g NaCl in 1.00 slutin. rperties f Electrlyte Slutins O Na O 4 (s) Na + (aq) + (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: 4.0(0.010 mlal) = mlal fr Na O 4 slutin;.0(0.00 mlal) = mlal fr CaBr slutin;.0(0.00 mlal) = mlal fr KCl slutin; slightly greater than 0.00 mlal fr HF slutin since 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 since increases. Therefre, the 0.00 m HF slutin will have the highest vapr pressure since 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). 70. The slutins f C 1 H O 11, 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 ml f cmpund. a. pure water b. CaCl slutin c. CaCl slutin d. pure water e. CaCl slutin 71. 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 (Assuming water freezes at 0.00 C.) ΔT b = ik b m = C/mlal mlal = C; T b = C (Assuming water bils at C.)

23 94 CHATER 11 ROERTIES OF SOUTIONS 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 7. NaCl(s) Na + (aq) + Cl (aq), i =.0 π = imrt = ml atm 9 K = 4.8 atm mlk A pressure greater than 4.8 atm shuld be applied t insure purificatin by reverse smsis. 7. ΔT f = ik f m, i = ΔTf K m f C =.6 fr 0.05 m CaCl 1.86 C/ mlal 0.05mlal i = =.60 fr m CaCl ; i = i ave = ( )/ = =.57 fr 0.78 m CaCl 0.78 Nte that i is less than the ideal value f.0 fr CaCl. This is due t in pairing in slutin. Als nte that as mlality increases, i decreases. Mre in pairing ccurs as the slute cncentratin increases. 74. 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 75. a. T C = 5(T F )/9 = 5(9 )/9 = 4 C; Assuming the slubility f CaCl is temperature independent, the mlality f a saturated CaCl slutin is: 74.5 g CaCl g 1000g 1mlCaCl kg g CaCl 6.71mlCaCl kg H O ΔT f = ik f m = C kg/ml 6.71 ml/kg = 7.4 C Assuming i =.00, a saturated slutin f CaCl can lwer the freezing pint f water t 7.4 C. Assuming these cnditins, a saturated CaCl slutin shuld melt ice at 4 C (9 F).

24 CHATER 11 ROERTIES OF SOUTIONS 95 b. Frm Exercise 11.7, i ave =.60; ΔT f = ik f m = =.4 C T f =.4 C Assuming i =.60, a saturated CaCl slutin will nt melt ice at 4 C(9 F). π 76. = imrt, M = irt.50atm = = ml/ atm.00 98K K ml mlar mass f cmpund = Additinal Exercises 0.500g ml = 97.8 g/ml 77 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. Since the temperature f the water decreased, the disslutin f NH 4 NO is endthermic (ΔH is psitive). Mass f slutin = = 76.6 g. heat lss by slutin = 4.18J 76.6 g (5.00 C.4 C) = 5 J g C ΔH sln = 5J 1.60g NH 80.05g NH NO 4 4NO ml NH 4NO = J/ml = 6.6 kj/ml 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) NH 4 + (g) + NO (g) NH 4 + (aq) + NO (aq) ΔH = lattice energy 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 78. The main intermlecular frces are: hexane (C 6 H 14 ): ndn dispersin; chlrfrm (CHCl ): diple-diple; ndn dispersin; methanl (CH OH): H bnding; 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. 79. 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 the cking time.

25 96 CHATER 11 ROERTIES OF SOUTIONS 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. c. When water freezes frm a slutin, it freezes as pure water, leaving behind a mre cncentrated salt slutin. Therefre, the melt f frzen sea ice is pure water. d. On the CO phase diagram in chapter 10, the triple pint is abve 1 atm, s 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. e. Adding a slute t a slvent increases the biling pint and decreases the freezing pint f the slvent. Thus, the slvent is a liquid ver a wider range f temperatures when a slute is disslved. 80. A 9 prf ethanl slutin is 46% C H 5 OH by vlume. Assuming m f slutin: ml ethanl = 46 m C H 5 OH mlarity = 0.79ml g m = 7.9 M ethanl 1mlCH OH = 0.79 ml C H 5 OH 46.07g 0 5 CS CS 81. Because partial pressures are prprtinal t the mles f gas present, then /. CS = (6 trr) = 5 trr CS tt CS 5trr CS CS, CS = trr CS CS 0.1ml atm 8. π = MRT = 98 K =.45 atm atm mlk 760mm Hg π = atm 000 mm m atm The smtic pressure wuld supprt a mercury clumn f 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 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 8. Out f g, there are: tt 1.57 g C 1 mlc 1.01g =.69 ml C;.69 =

26 CHATER 11 ROERTIES OF SOUTIONS g H 1 mlh = 5.6 ml H; 1.008g 5.6 = g O 1 mlo =.946 ml O; 16.00g.946 = empirical frmula: C H 4 O ; Use the freezing pint data t determine the mlar mass. m = ΔTf K f 5.0C 1.86C/ mlal =.80 mlal ml slute = kg.80mlslute = ml slute kg mlar mass = 10.56g = 151 g/ml ml The empirical frmula mass f C H 4 O = g/ml. Since 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 the atmic masses in the peridic table. 84. a. As 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. At sme pint, the rate f slvent transfer wuld be the same in bth directins and the levels f the liquids in the tw arms wuld stabilize. The height difference between the tw arms is 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, Na + and Cl ins 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. At this pint, the cncentratin f Na + and Cl ins will be equal in bth chambers and the levels f liquid will be equal. 85. 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 Assume an amunt f slutin which cntains 1.00 kg f water (slvent) ml NaCl 58.44g = 0.1 g NaCl; mass % NaCl = ml 0.1g g 0.1g = 1.97%

27 98 CHATER 11 ROERTIES OF SOUTIONS 86. 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 ppsite 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. Adding an electrlyte neutralizes the adsrbed in layers which allws cllidal particles t aggregate and then precipitate ut. 87. T = K f m, m = ΔT K f.79 C = 1.50 mlal 1.86 C / mlal a. T = K b m, T = (0.51C/mlal)(1.50 mlal) = 0.77C, T b = C b. water water water, water = mlh O mlh O mlslute Assuming 1.00 kg f water, we have 1.50 ml slute and: ml H O = g H O 1 mlh 18.0g H O O = 55.5 ml H O water = 55.5 ml = 0.974; water = (0.974)(.76 mm Hg) =.1 mm Hg c. We assumed ideal behavir in slutin frmatin and assumed i = 1 (n ins frm). Challenge rblems B, 88. Fr the secnd vapr cllected, = and = 0.86, where = mle fractin f benzene in the secnd slutin and B, T, = B, = = B TOT B = B T T, T, B, = mle fractin f tluene in the secnd slutin. 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,1 = et B,1 T,1 = mle fractin f tluene in first slutin. = mle fractin benzene in the first slutin and B,1 T,1 = B,1

28 CHATER 11 ROERTIES OF SOUTIONS 99 B,1 = = Slving: B TOT B,1 = 0.86 B = B T B,1 B,1 (750.0 trr) (750.0 trr) (1.000 B,1 )(00.0 trr) The riginal slutin had B = 0.86 and T = Fr 0.% A by mles in the vapr, 0. = 0.0 = A Ax x B, y 0.0 A A A Ax x (1.00 B A )y 100: A x = 0.0( A x) y 0.0 A y, A x 0.0 A x A y = 0.0 y 0.0 y A (x 0.0 x y) = 0.0 y, A = ; 0.70x 0.0y B = 1.00 A Similarly, if vapr abve is 50.% A: A y x 0.80 y If vapr abve is 80%A: A = ; 0.0x 0.80y ; y B 1.00 B = 1.00 A y x y If the liquid slutin is 0.%A by mles, A = 0.0. Thus, A A A If slutin is 50.%A: If slutin is 80.%A: B =.0 x and 0.0x 0.70y A A 0 B x x y and B 0.0 x x 0.70y x and 0.80x 0.0y B A 1.00 A 90. m = 91. m = ΔT K f C = 0.18 ml/kg 1.86 C / mlal = MRT where M = ml/; We must assume that mlarity = mlality s we can calculate the smtic pressure. This is a reasnable assumptin fr dilute slutins when 1.00 kg f water 1.00 f slutin. Assuming 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 little larger than 1.00, but nt by much (t three sig figs). The assumptin that mlarity = mlality will be gd here. = (0.18 M)( atm/kml)(98 K) = 5. atm ΔT 0.46 C = 0.9 mlal K 1.86 C/ mlal f

29 400 CHATER 11 ROERTIES OF SOUTIONS et x = ml NaCl in mixture and y = ml C 1 H O 11. NaCl(aq) Na + (aq) + Cl (aq); In slutin, NaCl exists as separate Na + and Cl ins. 0.9 ml = ml Na + + ml Cl + ml C 1 H O 11 = x + x + y = x + y The mlar mass f NaCl is g/ml and the mlar mass f C 1 H O 11 = 4. g/ml. Setting up anther equatin fr the mass f the mixture: 0.0 g = x(58.44) + y(4.) Substituting: 0.0 = x + (0.9 x) 4. Slving: x = ml NaCl = 0.09 ml and y = ml C 1 H O 11 = 0.9 (0.09) = 0.04 ml mass NaCl = 0.09 ml g/ml = 5.45 g NaCl mass% NaCl = 5.45g 0.0 g 100 = 7.%, mass % C 1 H O 11 = = 7.7% sucrse = 0.04ml ( ) ml = a. π = imrt, im = π RT Assuming f slutin: 7.8atm atm 98K mlk = 0.0 ml/ ttal ml slute particles = ml Na + + ml Cl + ml NaCl = 0.0 ml mass slutin = m 1.071g = 1071 g slutin m mass NaCl in slutin = g = 10.7 g NaCl ml NaCl added t slutin = 10.7 g 1 ml 58.44g = 0.18 ml NaCl Sme f this NaCl dissciates int Na + and Cl (tw ml ins per ml NaCl) and sme remains undissciated. et x = ml undissciated NaCl = ml in pairs. ml slute particles = 0.0 ml = (0.18 x) + x 0.0 = 0.66 x, x = ml in pairs

30 CHATER 11 ROERTIES OF SOUTIONS 401 fractin f in pairs = = 0.5, r 5% 0.18 b. ΔT = K f m where K f = 1.86 C kg/ml; Frm part a, f slutin cntains 0.0 ml f slute particles. T calculate the mlality f the slutin, we need the kg f slvent present in slutin. mass f slutin = 1071 g; mass f NaCl = 10.7 g mass f slvent in slutin = 1071 g 10.7 g = g ΔT = 1.86 C kg/ml 0.0ml 1.060kg = 0.56 C Assuming water freezes at C, then T f = 0.56 C. 9. pen = 0.15 = pen ttal ; pen = penpen = pen (511 trr); ttal = pen + hex = pen (511) + hex (150.) Since pen = hex = pen ttal + 54 pen = 511 pen, pen, then: ttal = pen(511), 0.15 = pen pen pen = = (511 trr) + (1.000, 0.15 ( pen ) = 511 pen pen )(150.) = pen 94. a. m = ΔTf K f 1.C 5.1C kg/ ml = 0.58 ml/kg ml unknwn = kg mlar mass f unknwn = Uncertainty in temperature = S, mlar mass = 0 ± 9 g/ml. 0.58ml unknwn kg 1.g = 0 g/ml ml = ml = %; A % uncertainty in 0 g/ml 1. = 9 g/ml. b. N, cdeine culd nt be eliminated since its mlar mass is in the pssible range including the uncertainty.