无机及分析化学答案第一章.docx

无机及分析化学答案第一章第一章 物质的聚集状态 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 = 30g m( H2O) = 1000mL´1.0g×mL-1´(1-0.030) = 9.7´102g n( H2O2) = 30g/34g×moL-1=0.88mol n( H2O) = 970g/18g.×mol-1=54mol b( H2O2)= 0.88mol /0.97kg = 0.91mol×kg-1 c( H2O2)= 0.88mol/1L = 0.88mol×L-1 x( H2O2) = 0.88/(0.88.+54) = 0.016 1-4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-1 b(C6H12O6)=5.0g/(180g.×mol-1´0.095kg)=0.29mol×kg-1 蔗糖溶液沸点上升 DTb=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 DTb=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 = 373.30K 1-5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-1 C12H22O11溶液，0.1mol×kg-1 CH3COOH溶液，0.1mol×kg-1 KCl溶液。 (2)渗透压：0.1mol×L-1 C6H12O6溶液，0.1mol×L-1CaCl2溶液，0.1mol×L-1 KCl溶液，1mol×L-1 CaCl2溶液。 解：凝固点从高到低： 0.1mol×kg-1 C12H22O11溶液>0.1mol×kg-1 CH3COOH溶液>0.1mol×kg-1 KCl溶液 渗透压从小到大： 0.1mol×L-1 C6H12O6溶液<0.1mol×L-1 KCl溶液<0.1mol×L-1 CaCl2 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： P = c×R×T c =P/RT = 0.366/(8.314´293.15) mol×L-1 = 1.50´10-4 mol×L-1 500´10-3L´1.50´10-4mol×L-1 = 5.0g/M M = 6.7´104g×mol-1 1-7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： Tf = Kf(H2O) ´b(C3H8O3) b(C3H8O3) =Tf / Kf(H2O) =3.00/1.86 mol×kg-1 =1.61 mol×kg-1 m(C3H8O3)=1.61´0.500´92.09g =74.1g 1-8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： 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 K×kg×mol-1 =0.292 mol×kg-1 w=0.292´180/(0.292´180+1000) = 0.0499 (2) P = c×R×T = 0.292mol×L-1´8.314kPa×L×mol-1×K-1´(273.15+37)K =753kPa (AgCl)m×nAg+×(n-x)NO3-x+×xNO3- 1-15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=0.33K DTf = 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)/Tf×m(苯) =5.12´0.100/(0.33´0.00500)g×mol-1 =3.1´102g×mol-1 C: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-) = 0.029 mol×kg-1、b(HCO3-) = 0.002 mol×kg-1、 b(Na+) = 0.49 mol×kg-1、b(Mg2+) = 0.055 mol×kg-1、 b(K+) = 0.011 mol×kg-1 和 b(Ca2+) = 0.011 mol×kg-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 = 2.17K Tf = 273.15K 2.17K = 270.98K Tb=Kb(H2O)·b = 0.52 ´ (0.57 + 0.029 + 0.002 + 0.49 + 0.055 + 0.011 +0.011)K = 0.61K Tb = 373.15K + 0.61K = 373.76K 1-17三支试管中均放入20.00mL同种溶胶。欲使该溶胶聚沉，至少在第一支试管加入0.53mL 4.0 mo1×L-1的KCl溶液，在第二支试管中加入1.25mL 0.050 mo1×L-1的Na2SO4溶液，在第三支试管中加入0.74mL 0.0033 mo1×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 (m mo1×L-1) 第二支试管聚沉值： 0.050´1.25´1000/(20+1.25) = 2.9(m mo1×L-1) 第三支试管聚沉值： 0.0033´0.74´1000/(20+0.74)=0.12(m mo1×L-1) 溶胶带正电。 1-18The 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 78.59°C. The boiling point of ethanol is 78.35°C, and Kb for ethanol is 1.20 K×kg.mol-1. What is the molecular formula of fructose? Solution: DTb=Tb-Tb¢=78.59 -78.35K=0.24K DTb= Kb(ethanol) ´b(fructose)= Kb(ethanol)´m(fructose)/M(fructose)×m(ethanol) M(fructose)= Kb(ethanol)´m(fructose)/DTb×m(ethanol) =1.20´11.7/(0.24´0.325)g×mol-1 =180g×mol-1 C:H:O =180´40%/12.011 : 180´6.75%/1.008 : 180´53.32%/16.00 = 6 : 12 : 6 Molecular formula of fructose is C6H12O6. 1-19A sample of HgCl2 weighing 9.41 g is dissolved in 32.75 g of ethanol, C2H5OH. The boiling-point elevation of the solution is 1.27°C. Is HgCI2 an electrolyte in ethanol? Show your calculations. (Kb=1.20K×kg×mol-1) Solution : If HgCl2 is not an electrolyte in ethanol b(HgCl2)=9.41/(271.5´0.03275) mol×kg-1 =1.05mol×kg-1 now, DTb= Kb(ethanol)´b(HgCl2) b(HgCl2)= DTb/Kb(ethanol) = 1.27/1.20 mol×kg-1 =1.05 mol×kg-1 Therefore, 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 CuSO4×5H2O in 0.1000kg of water. Remember to consider the water released from the hydrate. Solution : m(CuSO4) = 11.5´159.6/249.68g= 7.35g m(H2O) = 11.5 - 7.35g = 4.15g Percent by mass: 7.35/(100.0+4.15)= 0.071 b = 7.35/(159.6´0.1042) mol×kg-1= 0.442 mol×kg-1 1-21The cell walls of red and white blood cells are semipermeable membranes. The concentration of solute particles in the blood is about 0.6 mo1×L-1. What happens to blood cells that are place in pure water? In a 1 mo1×L-1 sodium chloride solution? Solution : In pure water: Water will entry the cell. In a 1 mo1×L-1 sodium chloride solution: The cell will lose water.