oework #4 hapter 17 roperties o Solutios 1. a) NO(s) + (aq) + NO - (aq) ) NaSO4(s) Na + (aq) + SO4 - (aq) c) Al(NO)(s) Al + (aq) + NO - (aq) d) SrBr(s) Sr + (aq) + Br - (aq) e) KlO4(s) K + (aq) + lo4 - (aq) ) N4Br(s) N4 + (aq) + Br - (aq) ) N4NO(s) N4 + (aq) + NO - (aq) h) uso4(s) u + (aq) + SO4 - (aq) i) NaO(s) Na + (aq) + O - (aq) 15. Need to id arity (), ality (), ad e ractio (χ) o O Kow ass % O = 40.0% d 1.05 solutio 18.0 c.08 All o the quatities that eed to e calculated are itrisic properties (do ot deped o saple size), thereore, assue saple size o 0.. alculate arity O solutio alculate the es o O O O alculate the ass o O O O 40.0% 0.% 0.% 0. O O Note: It was assued that the total ass was 0 40.0 O 40.0 0.44 O O.08 O alculate the volue o solutio solutio 0. solutio 95.c 0.095L d 1.05 solutio c 1
0.44 O.7 solutio 0.095L alculate ality O alculate the ass o solvet (O) 0.0 40.0 0.0 0.000k O O solutio O 0.44 O.7 0.000k alculate the e ractio O O alculate es o O O 0.0 O. 18.0 O O O 0.44 O 0.1 O 0.44. 1. Need to id arity (), ality (), ad e ractio (χ) o O I Geeral Kow ass % solvet d solvet o the quatities that eed to e calculated are itrisic properties (do ot deped o saple size), thereore, assue saple size o 0.. olarity solutio olality solvet ole Fractio solvet
l alculate es o (l) usi ass % 8% 0.% 0.% solvet 0. 8 8 1.0 1 l.4 l alculate the olue o solutio usi desity solutio 0. solutio 84.0c 0.0840L d 1.19 solutio c alculate the es o solvet () 0. 8 solvet solutio solvet 1 O 18.0.4 O olarity 1.0 1 solutio 0.0840L olality 1.0 1 solvet 0.0k ole Fractio 1.0 0. 1.0.4 solvet NO alculate es o (NO) usi ass % 70.% 0.% 0.% solvet 0. 70. 1 NO 70..0 NO 1.1 alculate the olue o solutio usi desity solutio 0. solutio 70.4c 0.0704L d 1.4 solutio c alculate the es o solvet () 0. 70. 0. solvet solutio solvet 1 O 0. 18.0 1.7 O olarity 1.1 1 0.0704L solutio
olality 1.1 7 solvet 0.00k ole Fractio 1.1 0.9 1.1 1.7 solvet SO4 alculate es o (SO4) usi ass % 95% 0.% 0.% solvet 0. 95. 1 SO4 95. 98.09 0.97 SO 4 alculate the olue o solutio usi desity solutio 0. solutio 54.c 0.054L d 1.84 solutio c alculate the es o solvet () 0. 95 5 solvet solutio solvet 1 O 5 18.0 0. O olarity 0.97 18 solutio 0.054L olality 0.97 00 solvet 0.005k ole Fractio 0.97 0.7 0.97 0. solvet O alculate es o (O) usi ass % 99% 0.% 0.% solvet 0. 99 1 O 99 1.07 1. O alculate the olue o solutio usi desity solutio 0. solutio 95.c 0.095L d 1.05 solutio c 4
alculate the es o solvet () 0. 99 1 solvet solutio solvet 1 O 1 18.0 0.0 O olarity 1. 17 solutio 0.095L olality 1. 000 solvet 0.001k ole Fractio 1. 0.9 1. 0.0 solvet N alculate es o (N) usi ass % 8% 0.% 0.% solvet 0. 8 1 N 17.04 N 8 1. alculate the olue o solutio usi desity solutio 0. solutio 1c 0.1L d 0.90 solutio c alculate the es o solvet () 0. 8 7 solvet solutio solvet 1 O 7 18.0 4.0 O olarity 1. 15 solutio 0.1L olality 1. solvet 0.07k ole Fractio 1. 0.9 1. 4.0 solvet 5
17. Need to id arity (), ality (), e ractio (χ), ad ass % o 5 Kow 50L d d alculate ass% ass ass 5 5 alculate e ractio 5 0.87 15L 0.874 c c 5 % 5 0% total alculate the ass o toluee d 50c 0.87 4.4 5 5 c alculate the ass o ezee d 15c 0.874 9. c alculate the total ass 4.4 9. 15 total 5 4.4 5 % 5 0% 0% 8.% total 15 5 5 alculate the es o toluee 1 5 4.4 0.471 9.15 alculate the es o ezee 1 9 1.40 78.1 0.471 5 0.5 5 0.471 1.40 5 alculate ality 0.471 4. 0.9k solvet alculate arity 0.471.9 0.175L solutio 5 5
0. The extet o hydratio icreases with icreasi chare desity. I eeral, the saller the size, the larer the chare desity. Also, the larer the chare, the hiher the chare desity. a) + ) Be + c) Fe + d) F - e) l - ) SO4-1. a) N N ca -od with water while caot ) N N is polar ad is ot c) OO OO ca -od with water ad OO caot. a) KrF (liear) opolar dissolves i l4 ) SF (et) polar dissolves i O c) SO (et) polar dissolves i O d) O (liear) opolar dissolves i l4 e) F ioic dissolves i O ) O polar dissolves i O ) opolar dissolves i l4 7. k 0.790 8.1 k k 4 at k 9 Lat 1.at 9 0.00114 Lat 40. solutio solvet solvet alculate the e ractio o O O O 8 alculate the es o O d 8c 0.99 5 O O c 5 18. 1 O 18.0 alculate the es o 8O 14 1.78 18. O 0.91 18. 1.78 1 8O 8 9.11 8 7
alculate the vapor pressure o O/8O 0.91 54.74torr 50. torr solutio solvet solvet 0 4. 45. solutio solvet solvet Sallest apor ressure Larest apor ressure c<<a<d The vapor pressure o ethaol is larer tha that o water, thereore, solutio d will have the larest vapor pressure. ure water will have the ext larest vapor pressure. Glucose is a o-electrolyte, thereore, it will ot dissociate i water 0. 99 O. Whe Nal dissociates i water it will or ios Na + ad l -, thereore, 0. 99 O causi solutio c to have the sallest vapor pressure. solutio solvet 0.900at solvet 0.98 solvet solvet? 0.90at Deterie es o 78.11 1.00 0.0 1 78.1 1.00 0.98 1.00? Deterie ar ass o sustace.0? 00 0.0?? 47. a) solutio 51 51 14 14 51 14 solutio 5 1 14 5 1 14 14 51 alculate es o 51 d 5c 0. 1 51 O c 1 0. alculate es o 14 1 51 5 1 7.1 5 1 8
d 45c 0. 0. 14 14 c 0. 0.5 1 14 14 8.0 14 51 14 solutio 5 1 14 5 1 14 14 51 0. 0.5 solutio 511torr 150. torr 90torr 0.5 0. 0.5 0. ) We eed to calculate the e ractio o petae i the vapor phase. You caot use the e ractio that was calculated i part (a) ecause that is the e ractio i the liquid phase. I the as phase RT RT 5 1.0 torr 90torr 5 1 5 1 5 1 5 1 14 RT total RT total 0.9 49. We eed to id χprop ad χeth What we kow solutio eth eth prop prop 174torr 0torr 44. torr (equ. a) eth 1 (equ. ) prop eth Solve or χeth i equatio 1 eth prop 1 eth prop prop lu ito equatio (a) ad solve or χprop 174torr 1 0torr prop Solve or χeth 1 0.499 1 0.499 eth prop 0.501 44.torr eth eth prop 50. lot a shows a positive deviatio ro Raoult s law. This happes whe there are uavorale relatioships etwee the solvet ad the. This is the situatio or situatio d i which the solvet/ have diereces i polarity is opolar ad O is polar. Thereore, each o the sustaces prop 9
would e ore stale y theselves ad ot i solutio. It is also the case or. Both o these species are capale o -odi idividually, thereore, addi the toether will ot ake a ore stale situatio. lot shows a eative deviatio ro Raoult s law. This happes whe there are avorale relatioships etwee the solvet ad the. This is the situatio or situatio a, ecause O would ot e ale to -od o its ow. owever, whe water is preset it ca participate i -odi aki the iterecular orces i the solutio stroer tha the idividual iterecular orces. lot c shows the ideal Raoult s law. This happes whe the iterecular orces are the sae etwee the idividual parts o the solutios as with the solutio. This is the case or c. Both ad have early idetical iterecular orces resulti i a ideal situatio whe they are ixed. 51. Need to id phase) solutio solutio solutio, O O O O O O, ad l14 O l14 O ( stads or the e ractio i the vapor O O O alculate es o acetoe 1 O 50.0 58.09 0.81 alculate es o ethaol 50.0 1.5 1 O.05 0.81 1.5 solutio 71torr 14torr 188.5torr 0.81 1.5 1.5 0.81 alculate the e ractio o acetoe i the vapor phase O O RT RT RT total RT O alculate the e ractio o ethaol i the vapor phase 9.1 O O torr 0.488 O 188.torr total total 9.5torr 188.torr 0.51
apor ressure (torr) Sice the actual vapor pressure is less tha the total vapor pressure (eative deviatio to Raoult s Law) there ust e avorale -solvet iteractio. Solute-solvet iteractios are stroer tha - ad solvet-solvet iteractios. 5. 84 8 80 78 7 74 7 70 0 0. 0.4 0. 0.8 1 ole Fractio O a) I order to e a ideal solutio whe you plot vapor pressure vs. e ractio a straiht lie should e oserved. Thereore, the solutio is ot ideal. ) Whe there is a positive deviatio ro Raoult s law there are uavorale iteractios etwee ad solvet, thereore, Δ is positive. c) The positive deviatio ro Raoult s law shows that the vapor pressure is hiher tha the ideal case; this iplies weaker iteractio etwee propaol ad the water tha the pure sustace. d) The hiher the vapor pressure, the lower the oili poit. Thereore, the lowest oili poit is whe 0. 54 O 5. I the cotaier eels war it eas that the solutio released heat whe it was ored. At costat pressure q, thereore, Δ is eative. This iplies that there will e a eative deviatio to Raoult s law ad the solutio is ot ideal. Note: This prole is icorrect ecause ethaol ad water would have a positive deviatio which would ake the solutio cool to the touch. 55. a) True ) True c) True d) True e) False The saller the vapor pressure, the hiher the oili poit. 11
59. They wat you to id T The oili poit o pure water is 0 T 0. 0 T alculate the chae i oili poit T ik alculate the ality T ik T T alculate the es o urea 1 N O 0.07 7.0 N 0.449 N 0.449. 00.1500k I= 1 ecause o electrolyte k 0. 51 K ik k 1.000.51 1.5 0.0 T 0.0 1.5 1. 5 0. Need to deterie 8 O Fid Fid T T 1.50 0.00 T T 0.00 T 1.50 T ik 8O is a o-electrolyte i=1 k 1.50 1 1.8 0.80 Fid 8O 8O 8O O 8O 0.80 0.00k 0.11 8 0.11 14.8 9.11 8 1 8O 8 1
1. Need to id T ad T T 0. 00 T alculate ΔT T ik Ethylee lycol is a o-electrolyte, i=1 T T T alculate T Assue 1.00 k o solutio 0.500 k O 0.500 k O 500. 8.05 ik 1 O.08 8.04 1.1 0.500k k 1 1.1 1.8 0.0 0.00 T 0.00 0.0 0. 0 0. 0 T alculate ΔT T ik T ik k 1 1.1 0.51 8.1 0.0 T 0.0 8.1 8.. Need to id T T 99. 75 T T alculate ΔT T ik No-electrolyte, i=1 alculate 1 58.0 4.0 0.414 0.414 0.90 0.00k O k T ik 1 0.90 0.51 0.5 99.75 T 99.75 0.5 0. 08 1
8. Need to calculate T T O 0.00 T 0.0 0.00 T 0.0 alculate O T ik Ethylee lycol is a o-electrolyte, i=1 k 0.0 1 1.8 1.1 alculate alculate O 00L 1c 1.00 15.0L 15000 15.0k O O 1L 1L 1c O O 1.1 15.0k alculate ass o O 4 4 1.50 alculate O.08 4 1 O c L 1.50 1500L 1.5L 4 1 1 1.11 1c alculate the oili poit o the solutio T 0.00 T alculate ΔT k T ik 1 1.1 0.51 8. T 0.00 8. 8. 9. Need to calculate ΔT ad π T ik alculate protei 14
T d solutio 1 c.0 Thereore, i you have 1.0 L o solutio it will weiht 00 I 1 L o solutio there is 1.0 o protei Thereore, i a 1.0 L solutio there is 1.0 o protei ad 999 o O alculate protei Assue 1 L solutio 1 5 1.0 1.1 4 9.0 5 protei 1.1 5 1.1.999k ik irt irt alculate 5 k 5 1 1.1 1.8.1 Solutio protei 1 5 5.1 1.1 1.0L 5 L at 4 1 1.1 0.080 98K.7 at 0. 0 torr K 7. Need to calculate ass% 8, ad 14 8 8 ass% 8 0% 0% 1.0 total 14 14 ass% 14 0% 0% 1.0 total Fid ass o 8 Kow.0 1 8 14 T.81 alculate ΔT T 5.51 T.81 5.51 T T.70 Solve or ass o 8 15
T ik 8 14.70 1 5.1 0.000k 8 14 k 178.4 18.18 0.05 18.18 178.4 84 8 14 8 14 8 14 40. 178.4 18.18 1.0 14 8 8 8 8 40. 178.4 18.18 1.0 5 50.0 ass% 0.70 Fid ass o 8 Kow 1.0 ass% 8 0.70 8 8 0% 0% 44% total 1.0 1.0 0.70 14 8 0.90 14 14 0% 0% 5% total 1.0 14 14 0.90 75. Need to calculate χnal ad vapor pressure o solutio Kow solutio 19. torr at 5 8 9. torr at 5 71. torr at 45 14 8 1
solutio solvet solvet 19.torr.8torr O O O 0.84 1 O Na l 1 0.84 Na l Na 0.17 O Na l l Nal Na + + l - Thereore, the e ractios o Na + ad l - ust e equal ad it ust equal the e ractio o Nal 0.17 Nal 0.0880 Na l 0.84 71.9torr 59.torr solutio solvet solvet 77. 0.0 NaO4 i= 4 0.040 (or all ios/ecules) 0.00 abr i= 0.00 (or all ios/ecules) 0.00 Kl i= 0.040 (or all ios/ecules) 0.00 F i~1 0.00 (or all ios/ecules) Sice F is a weak acid ost o the F will e F ad ot + ad F - a) 1O is a o-electrolyte, i=1 0.040 (or all ios/ecules) Thereore the NaO4 ad the Kl solutio will have the sae oili poits the 1O solutio. ) The solutio with the hihest vapor pressure will e the solutio that has the sallest ality or all ios/ecules. Thereore, the F solutio will have the hihest vapor pressure. c) The solutio with the larest reezi poit depressio will e the solutio that has the larest ality or all ios/ecules. Thereore, the abr solutio will have the larest reezi poit depressio. 78. ure Water χ=0.00 (all ios/ecules) 1O (χ=0.01) i=1 χ=0.01 (all ios/ecules) Nal (χ=0.01) i=1 χ=0.0 (all ios/ecules) al(χ=0.01) i= χ=0.0 (all ios/ecules) a) lowest χ pure water ) hihest χ al solutio c) hihest χ al solutio d) lowest χ pure water c) hihest χ al solutio 17
111. Need to calculate ass% l l l ass% l 0% 0% 0.500 total Nal Nal ass% Nal 0% 0% total 0.500 Fid ass o l Kow.500 0 l 0.950at irt i l l i Nal Nal Nal RT The upper case is the arity l Nal l Nal Lat 0.950at 0.080 K 98K solutio solutio The upper case is the ar ass l Nal l Nal Lat 0.950at 0.080 K 98K 1.0000L 1.0000L 175. 190.44 0.0115 95.18 58.44 554.8 89.87 175. l Nal l Nal l 190.44 Nal 0.500 lu Nal l or ass o l 89.87 175. 190.44 0.500 5.5 l 15.11 0.54 l l l 0.54 ass% l 0% 0% 70.8% total 0.500 l 18