Chemistry 11 ANSWERE KEY REVIEW QUESIONS Chapter 1 1. Cmmercial nitric acid has a density 1.2 g/ml and is 16. M. Calculate the mass percent and mlality HNO in this slutin. Assume 1 L slutin: 6.2 g 16. ml HNO x = 18. g HNO 1 ml 1.2 g 1. L sl'n x x = 12 g sl'n 1 L 1. ml 18. g HNO mass percent = x1 = 71.% 12 g sl'n mass slvent = 12 g - 18. g = 12 g =.12 kg 16. ml HNO mlality = = 8.8 m.12 kg slvent 2. he density a 1.8 M slutin LiBr in acetnitrile (CHCN) is.826 g/ml. Calculate the cncentratin this slutin in (a) mlality, (b) mle ractin LiBr, and (c) mass percent CHCN. a) Mlality Assume 1 L slutin (cntains 1.8 ml slute) Mass slutin = 1 ml Mass slute = 1.8 ml.826 g x 1 ml = 826 g 86.85 g x =156 g Mass slvent = 826 g - 156 g = 67. g 1.8 ml LiBr Mlality = = 2.69 m.67 kg b) mle ractin ml slvent = 67. g x 1.5 g 1.8 LiBr = =.99 1.8 + 16. = 16. ml 67. g c) Mass % CHCN= x1 = 81.1% 826 g 1
. Bth methanl (CHOH) and ethylene glycl (C2H6O2) are used as antireeze. Which is mre eective that is, which prduces a lwer reezing pint i equal amunts each are added t the same amunt water? Methanl wuld be mre eective as antireeze because it has a lwer mlar mass (2. g/ml) cmpared with ethylene glycl (62.1 g/ml)). herere the same mass methanl prduces mre mles per gram slute, a greater mlality and a greater eect n biling and reezing pints the slvent.. Arrange the llwing slutins in the rder increasing biling pint. (Assume cmplete dissciatin r strng electrlytes). m glycerl (CH8O) i = 1.25 m KBr i = 2.1 m CaCl2 i = Slutin Cnc. particles. m glycerl. m.25 m KBr.5 m.1 m CaCl2. m he greater the cncentratin the particles in slutin, the greater the biling pint. herere,.1 m CaCl2 <. m glycerl <.25 m KCl 5. Fr each pair slutins listed belw, determine which will have the higher biling pint. (Assume cmplete dissciatin r strng electrlytes) a) 1.5 M NaCl and.5 M Al(NO) NaCl has greater particle cncentratin (. M vs. 2. M) b) 2. M NaOH and 2. M C6H12O6 NaOH has greater particle cncentratin (. M vs. 2. M) c). M Na2CO and.7 M KCl KCl has greater particle cncentratin (.1 M vs..12 M) 2
6. What are the reezing and biling pints an aqueus slutin cntaining 2.2 g urea, (NH2)2CO, in 15 ml slutin? (Density slutin = 1. g/ml) ml slute = 2.2 g x 6.7 g =.6 ml mass slutin = 15 ml 1. g x 1 ml = 19.5 g mass slvent = 19.5 g - 2.2 g = 17. g =.17 kg.6 ml mlality = =.228 m.17 kg = mk = (.228 m)(.512 C/m) =.117 C b b b = mk = (.228 m)(1.86 C/m) =.2 C = 1.117 C = -.2 C 7. A slutin an unknwn nnvlatile, nn-electrlyte cmpund was prepared by disslving.25 g the unknwn in. g CCl. he biling pint the resultant slutin was measured t be.57 C higher than the pure slvent. Calculate the mlar mass the unknwn slute. (Kb= 5.2 C/m).57 C K 5.2 b m = = =.711 b ml slute = m x kg slvent = (.711)(. kg) =.28 ml.25 g Mlar mass = = 87.9 g/ml.28 ml 8. Stearic acid (C18H6O2) and palmitic acid (C16H2O2) are cmmn atty acids. Cmmercial grades stearic acid usually cntain palmitic acid as well. A 1.115-g sample cmmercial grade stearic acid is disslved in 5. ml benzene (d=.879 g/ml). he reezing pint the slutin is und t be 5.72 C. he reezing pint pure benzene is 5.5 C and K r benzene is 5.12 C/m. What is the mass percent palmitic acid in the stearic acid sample? = 5.5-5.72 =.61 C.61 C m = = =.9 m K 5.12 C/m ml slute = m x kg slvent = (.9)(5. ml.879 g x 1 ml 256.8 g mass slute =.96 ml x =.116 g slute.116 g % palmitic acid = x1 = 9.11% 1.115 g 1 kg x ) =.96 ml 1 g
9. What is the minimum mass ethylene glycl (C2H6O2) that must be disslved in 1.5 kg water t prevent the slutin rm reezing at 1. F? = -1. F = -2. C 2. C m = = = 12.5 m K 1.86 C/m ml slute = m x kg slvent = (12.5)(1.5 kg) = 181.7 ml 62.8 g 1 kg mass slute = 181.7 ml x x = 11. kg slute 1 g 1. Calculate the vapr pressure a slutin prepared by adding 2.5 g glycerin (CH8O) t 1. g water at 7 C. ( =2 trr) ml CH8O = 2.5 g 7.778.5 + 7.778 x =.5 ml ml H2O = 1 g 92.1 g x 18. g = 7.778 ml H2O = =.957 sl'n = H2O = (.957)(2 trr) = 22 trr 11. A mixture (C8H8, 8%) and (C8H1, 62%) is separated by ractinal distillatin at 9 C. What is the cmpsitin the vapr in equilibrium with this mixture at 9 C, given the llwing vapr pressure the tw cmpnents at this temperature: = 1 mmhg and ethyl benzene = 182 mmhg. ml = 8. g x =.65 ml 1.1 g ml = 62. g x 16.2 g =.58 ml.65 = =.85 = 1-.85 =.615.65 +.58 In the vapr abve the mixture, = = (.85)(1 mmhg) = = = (.615)(182 mmhg) = 112 mmhg = 112 + 51.6 = 16.6 mmhg ttal Cmpsitin the vapr 51.6 mmhg = =.15 16.6 mmhg 51.6 mmhg = 1-.15 =.685
12. Calculate the vapr pressure a slutin prepared by adding a) 2.5 g glycerin (CH8O) t 1. g water at 7 C. ( =2 trr) ml CH8O = 2.5 g 7.778.5 + 7.778 x =.5 ml ml H2O = 1 g 92.1 g x 18. g = 7.778 ml H2O = =.957 sl'n = H2O = (.957)(2 trr) = 22 trr b) 5. g Na2SO t 92. g water at 55 C. ( =118. trr) ml Na2SO = 5. g x =.52 ml ml H2O = 92. g 12. g 5.111 HO = =.99 2 (.52) + 5.111 sl'n = H2O = (.99)(118. trr) = 117.2 trr x 18. g = 5.11 1. he vapr pressure CCl is.5 atm and the vapr pressure is.526 atm at C. A slutin is prepared rm equal masses these cmpunds at this temperature. a) Calculate the mle ractin in the vapr abve this slutin. ml CCl = 5. g x =.25 ml ml = 5. g x 15.81 g 119.7 g =.19 ml.25 CCl = =.7 = 1-.7 =.56.25 +.19 In the vapr abve the mixture, CCl = CCl = (.7)(.5 atm) =.155 atm = = (.56)(.526 atm) =.296 atm =.155 +.296 =.51 ttal.296 atm = =.656.51 atm b) I the vapr abve the riginal slutin is cndensed and islated in a separate lask, what wuld be the vapr pressure abve this new slutin? In the cndensed liquid, =.656 = = (.65)(.526 atm) =.5 atm 5