1、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1.0gmL-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) = 1000mL1.0gmL-10.030 = 30g m( H2O) = 1000mL1.0gmL-1(1-0.030) = 9.7102g n( H2O2) = 30g/34gmoL-1=0.88moln( H2O) = 970g/18g.mol-1=54molb( H2O2)= 0.88mol /0.97kg = 0.91molkg-1c( H2O2)= 0.
2、88mol/1L = 0.88molL-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-10.095kg)=0.15molkg-1b(C6H12O6)=5.0g/(180g.mol-10.095kg)=0.29molkg-1蔗糖溶液沸点上升DTb=Kbb(C12H22O11)= 0.52Kkgmol-10.15molkg-1=0.078K蔗糖溶液沸点为：373.15K+0.078K=373.23K葡萄糖溶液沸点
3、上升DTb=Kbb(C6H12O6)= 0.52Kkgmol-10.29molkg-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: 0.1molkg-1 C12H22O11溶液，0.1molkg-1 CH3COOH溶液，0.1molkg-1 KCl溶液。(2)渗透压：0.1molL-1 C6H12O6溶液，0.1molL-1CaCl2溶液，0.1molL-1 KCl溶液，1molL-1 CaCl2溶液。（提示：从溶液中的粒子数考虑。）解：凝固点从高到低：0.1molkg-1 C12H22O1
4、1溶液0.1molkg-1 CH3COOH溶液0.1molkg-1 KCl溶液渗透压从小到大：0.1molL-1 C6H12O6溶液0.1molL-1 KCl溶液0.1molL-1 CaCl2 溶液1molL-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： P = cRTc =P/RT = 0.366/(8.314293.15) molL-1 = 1.5010-4 molL-150010-3L1.5010-4molL-1 = 5.0g/MM = 6.7104gmol-11-7在严寒的季节里为了
5、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： Tf = Kf(H2O) b(C3H8O3)b(C3H8O3) =Tf / Kf(H2O) =3.00/1.86 molkg-1=1.61 molkg-1m(C3H8O3)=1.610.50092.09g=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)mnHS-(n-x)H+x-xH+1-9将10.0mL0.01molL-1的KCl溶液和100mL0.05
6、mo1L-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)mnAg+(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) DTf= Kf(H2O)b(C6H12O6)b(C6H12O6) = DTf/Kf(H2O) =0.543K/1.86 Kkgmol-1=0.292 molkg-1w=0.292180/(0.292180+1000)=
7、 0.0499(2) P = cRT= 0.292molL-18.314kPaLmol-1K-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： DTf =Tf -Tf=278.66-(273.15+5.18)K=0.33KDTf = 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)/Tfm(苯)=5.120.100/(0.330.0050
8、0)gmol-1=3.1102gmol-1C:H:O =310.3080.3%/12.011 : 310.309.5%/1.008 : 310.3010.2%/16.00= 21 : 29 : 2所以孕甾酮的相对分子质量是3.1102gmol-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-) = 0.57molkg-1、b(SO42-) = 0.029 molkg-1、b(HCO3-) = 0.002 molkg-1、 b(Na+) = 0.49 molkg-1、b(Mg2+) = 0.055 molkg-1、 b(K+) = 0.011 mo
9、lkg-1 和 b(Ca2+) = 0.011 molkg-1，请计算海水的近似凝固点和沸点。 解： DTf = 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= 270.98KTb=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
10、少在第一支试管加入0.53mL 4.0 mo1L-1的KCl溶液，在第二支试管中加入1.25mL 0.050 mo1L-1的Na2SO4溶液，在第三支试管中加入0.74mL 0.0033 mo1L-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： 4.00.531000/(20.00+0.53) =1.0102 (m mo1L-1)第二支试管聚沉值： 0.0501.251000/(20+1.25) = 2.9(m mo1L-1)第三支试管聚沉值： 0.00330.741000/(20+0.74)=0.12(m mo1L-1)溶胶带正电。1-18
11、The sugar fructose contains 40.0% C, 6.7% H and 53.3% O by mass. A solution of 11.7 g of fructose in 325 g of ethanol has a boiling point of 78.59C. The boiling point of ethanol is 78.35C, and Kb for ethanol is 1.20 Kkg.mol-1. What is the molecular formula of fructose? Solution: DTb=Tb-Tb=78.59 -78.
12、35K=0.24KDTb= Kb(ethanol) b(fructose)= Kb(ethanol)m(fructose)/M(fructose)m(ethanol)M(fructose)= Kb(ethanol)m(fructose)/DTbm(ethanol)=1.2011.7/(0.240.325)gmol-1 =180gmol-1C:H:O =18040%/12.011 : 1806.75%/1.008 : 18053.32%/16.00= 6 : 12 : 6Molecular formula of fructose is C6H12O6.1-19A sample of HgCl2
13、weighing 9.41 g is dissolved in 32.75 g of ethanol, C2H5OH. The boiling-point elevation of the solution is 1.27C. Is HgCI2 an electrolyte in ethanol? Show your calculations. (Kb=1.20Kkgmol-1) Solution : If HgCl2 is not an electrolyte in ethanolb(HgCl2)=9.41/(271.50.03275) molkg-1=1.05molkg-1now, DTb
14、= Kb(ethanol)b(HgCl2)b(HgCl2)= DTb/Kb(ethanol)= 1.27/1.20 molkg-1=1.05 molkg-1Therefore, HgCl2 is not an electrolyte in ethanol.1-20.Calculate the percent by mass and the molality in terms of CuSO4 for a solution prepared by dissolving 11.5g of CuSO45H2O in 0.1000kg of water. Remember to consider th
15、e water released from the hydrate. Solution : m(CuSO4) = 11.5159.6/249.68g= 7.35gm(H2O) = 11.5 - 7.35g = 4.15gPercent by mass: 7.35/(100.0+4.15)= 0.071b = 7.35/(159.60.1042) molkg-1= 0.442 molkg-11-21The cell walls of red and white blood cells are semipermeable membranes. The concentration of solute particles in the blood is about 0.6 mo1L-1. What happens to blood cells that are place in pure water? In a 1 mo1L-1 sodium chloride solution?Solution : In pure water: Water will entry the cell.In a 1 mo1L-1 sodium chloride solution: The cell will lose water.
