无机及分析化学答案(第二版)第一章

第一章物质的聚集状态（部分）1-3．用作消毒剂的过氧化氢溶液中过氧化氢的质量分数为0.030，这种水溶液的密度为1.0gmL1，请计算这种水溶液中过氧化氢的质量摩尔浓度、物质的量浓度和摩尔分数。解：1L溶液中，m(H2O2)=1000mL1.0gmL10.030=30gm(H2O)=1000mL1.0gmL1(10.030)=9.7102gn(H2O2)=30g/34gmoL1=0.88moln(H2O)=970g/18g.mol1=54molb(H2O2)=0.88mol/0.97kg=0.91molkg1c(H2O2)=0.88mol/1L=0.88molL1x(H2O2)=0.88/(0.88.+54)=0.0161-4．计算5.0%的蔗糖(C12H22O11)水溶液与5.0%的葡萄糖(C6H12O6)水溶液的沸点。解：b(C12H22O11)=5.0g/(342g.mol10.095kg)=0.15molkg1b(C6H12O6)=5.0g/(180g.mol10.095kg)=0.29molkg1蔗糖溶液沸点上升Tb=Kbb(C12H22O11)=0.52Kkgmol10.15molkg1=0.078K蔗糖溶液沸点为：373.15K+0.078K=373.23K葡萄糖溶液沸点上升Tb=Kbb(C6H12O6)=0.52Kkgmol10.29molkg1=0.15K葡萄糖溶液沸点为：373.15K+0.15K=373.30K1-5．比较下列各水溶液的指定性质的高低(或大小)次序。(l)凝固点:0.1molkg1C12H22O11溶液，0.1molkg1CH3COOH溶液，0.1molkg1KCl溶液。(2)渗透压：0.1molL1C6H12O6溶液，0.1molL1CaCl2溶液，0.1molL1KCl溶液，1molL1CaCl2溶液。（提示：从溶液中的粒子数考虑。）解：凝固点从高到低：0.1molkg1C12H22O11溶液0.1molkg1CH3COOH溶液0.1molkg1KCl溶液渗透压从小到大：0.1molL1C6H12O6溶液0.1molL1KCl溶液0.1molL1CaCl2溶液1molL1CaCl2溶液1-6．在20℃时，将5.0g血红素溶于适量水中，然后稀释到500mL,测得渗透压为0.366kPa。试计算血红素的相对分子质量。解：=cRTc=/RT=[0.366/(8.314293.15)]molL1=1.50104molL1500103L1.50104molL1=5.0g/MM=6.7104gmol11-7．在严寒的季节里为了防止仪器中的水冰结，欲使其凝固点下降到3.00℃，试问在500g水中应加甘油(C3H8O3)多少克?解：ΔTf=Kf(H2O)b(C3H8O3)b(C3H8O3)=ΔTf/Kf(H2O)=[3.00/1.86]molkg1=1.61molkg1m(C3H8O3)=1.610.50092.09g=74.1g1-8．硫化砷溶胶是通过将硫化氢气体通到H3AsO3溶液中制备得到：2H3AsO3+3H2S=As2S3+6H2O试写出该溶胶的胶团结构式。解：[(As2S3)mnHS(nx)H+]xxH+1-9．将10.0mL0.01molL1的KCl溶液和100mL0.05mo1L1的AgNO3溶液混合以制备AgCl溶胶。试问该溶胶在电场中向哪极运动?并写出胶团结构。解：AgNO3是过量的，胶团结构为：[(AgCl)mnAg+(nx)NO3]x+xNO31-14．医学上用的葡萄糖(C6H12O6)注射液是血液的等渗溶液，测得其凝固点下降为0.543℃。(l)计算葡萄糖溶液的质量分数。(2)如果血液的温度为37℃,血液的渗透压是多少?解：(1)Tf=Kf(H2O)b(C6H12O6)b(C6H12O6)=Tf/Kf(H2O)=0.543K/1.86Kkgmol1=0.292molkg1w=0.292180/(0.292180+1000)=0.0499(2)=cRT=0.292molL18.314kPaLmol1K1(273.15+37)K=753kPa1-15．孕甾酮是一种雌性激素,它含有(质量分数)9.5%H、10.2%O和80.3%C，在5.00g苯中含有0.100g的孕甾酮的溶液在5.18℃时凝固，孕甾酮的相对分子质量是多少？写出其分子式。解：Tf=TfTf=[278.66(273.15+5.18)]K=0.33KTf=Kf(苯)b(孕甾酮)=Kf(苯)m(孕甾酮)/[M(孕甾酮)m(苯)]M(孕甾酮)=Kf(苯)m(孕甾酮)/[ΔTfm(苯)]=[5.120.100/(0.330.00500)]gmol1=3.1102gmol1CHO=310.3080.3%/12.011:310.309.5%/1.008:310.3010.2%/16.00=21:29:2所以孕甾酮的相对分子质量是3.1102gmol1，分子式是C21H29O2。1-16．海水中含有下列离子，它们的质量摩尔浓度如下：b(Cl)=0.57molkg1、b(SO42)=0.029molkg1、b(HCO3)=0.002molkg1、b(Na+)=0.49molkg1、b(Mg2+)=0.055molkg1、b(K+)=0.011molkg1和b(Ca2+)=0.011molkg1，请计算海水的近似凝固点和沸点。解：Tf=Kf(H2O)b=[1.86(0.57+0.029+0.002+0.49+0.055+0.011+0.011)]K=2.17KTf=273.15K–2.17K=270.98KΔTb=Kb(H2O)·b=[0.52(0.57+0.029+0.002+0.49+0.055+0.011+0.011)]K=0.61KTb=373.15K+0.61K=373.76K1-17．三支试管中均放入20.00mL同种溶胶。欲使该溶胶聚沉，至少在第一支试管加入0.53mL4.0mo1L1的KCl溶液，在第二支试管中加入1.25mL0.050mo1L1的Na2SO4溶液，在第三支试管中加入0.74mL0.0033mo1L1的Na3PO4溶液,试计算每种电解质溶液的聚沉值,并确定该溶胶的电性。解：第一支试管聚沉值：4.00.531000/(20.00+0.53)=1.0102(mmo1L1)第二支试管聚沉值：0.0501.251000/(20+1.25)=2.9(mmo1L1)第三支试管聚沉值：0.00330.741000/(20+0.74)=0.12(mmo1L1)溶胶带正电。1-18．Thesugarfructosecontains40.0%C,6.7%Hand53.3%Obymass.Asolutionof11.7goffructosein325gofethanolhasaboilingpointof78.59C.Theboilingpointofethanolis78.35C,andKbforethanolis1.20Kkg.mol1.Whatisthemolecularformulaoffructose?Solution:Tb=TbTb=[78.5978.35]K=0.24KTb=Kb(ethanol)b(fructose)=Kb(ethanol)m(fructose)/M(fructose)m(ethanol)M(fructose)=Kb(ethanol)m(fructose)/Tbm(ethanol)=[1.2011.7/(0.240.325)]gmol1=180gmol1C:H:O=18040%/12.011:1806.75%/1.008:18053.32%/16.00=6:12:6MolecularformulaoffructoseisC6H12O6.1-19．AsampleofHgCl2weighing9.41gisdissolvedin32.75gofethanol,C2H5OH.Theboiling-pointelevationofthesolutionis1.27C.IsHgCI2anelectrolyteinethanol?Showyourcalculations.(Kb=1.20Kkgmol1)Solution:IfHgCl2isnotanelectrolyteinethanolb(HgCl2)=[9.41/(271.50.03275)]molkg1=1.05molkg1now,Tb=Kb(ethanol)b(HgCl2)b(HgCl2)=Tb/Kb(ethanol)=[1.27/1.20]molkg1=1.05molkg1Therefore,HgCl2isnotanelectrolyteinethanol.1-20.CalculatethepercentbymassandthemolalityintermsofCuSO4forasolutionpreparedbydissolving11.5gofCuSO45H2Oin0.1000kgofwater.Remembertoconsiderthewaterreleasedfromthehydrate.Solution:m(CuSO4)=[11.5159.6/249.68]g=7.35gm(H2O)=[11.57.35]g=4.15gPercentbymass:7.35/(100.0+4.15)=0.071b=[7.35/(159.60.1042)]molkg1=0.442molkg11-21．Thecellwallsofredandwhitebloodcellsaresemipermeablemembranes.Theconcentrationofsoluteparticlesinthebloodisabout0.6mo1L1.Whathappenstobloodcellsthatareplaceinpurewater?Ina1mo1L1sodiumchloridesolution?Solution:Inpurewater:Waterwillentrythecell.Ina1mo1L1sodiumchloridesolution:Thecellwilllosewater.