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∆T b=K b⋅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∆T b=K b⋅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-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。

（提示：从溶液中的粒子数考虑。

）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：∏= c⋅R⋅Tc =∏/RT = [0.366/(8.314⨯293.15)] mol⋅L-1 = 1.50⨯10-4 mol⋅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：ΔT f = K f(H2O) ⨯b(C3H8O3)b(C3H8O3) =ΔT f / K f(H2O)=[3.00/1.86] mol⋅kg-1=1.61 mol⋅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⋅n HS-⋅(n-x)H+]x-⋅x H+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⋅n Ag+⋅(n-x)NO3-]x+⋅x NO3-1-14．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) ∆T f= K f(H2O)⨯b(C6H12O6)b(C6H12O6) = ∆T f/K f(H2O)=0.543K/1.86 K⋅kg⋅mol-1=0.292 mol⋅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：∆T f =T f-T f'=[278.66-(273.15+5.18)]K=0.33K∆T f = K f(苯) ⨯b(孕甾酮)= K f(苯)⋅m(孕甾酮)/[M(孕甾酮)⋅m(苯)]M(孕甾酮)= K f(苯)⋅m(孕甾酮)/[ΔT f⋅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-) = 0.029 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：∆T f = K f(H2O) ⨯b= [1.86 ⨯ (0.57 + 0.029 + 0.002 + 0.49 + 0.055 + 0.011 +0.011)]K= 2.17KT f = 273.15K –2.17K= 270.98KΔT b=K b(H2O)·b= [0.52 ⨯ (0.57 + 0.029 + 0.002 + 0.49 + 0.055 + 0.011 +0.011)]K= 0.61KT b = 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0.53mL 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。

1-18．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 78.59︒C. The boiling point of ethanol is 78.35︒C, and K b for ethanol is 1.20 K⋅kg.mol-1. What is the molecular formula of fructose?Solution: ∆T b=T b-T b'=[78.59 -78.35]K=0.24K∆T b= K b(ethanol) ⨯b(fructose)= K b(ethanol)⨯m(fructose)/M(fructose)⋅m(ethanol)M(fructose)= K b(ethanol)⨯m(fructose)/∆T b⋅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 : 6Molecular formula of fructose is C6H12O6.1-19．A 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. (K b=1.20K⋅kg⋅mol-1)Solution : If HgCl2 is not an electrolyte in ethanolb(HgCl2)=[9.41/(271.5⨯0.03275)] mol⋅kg-1=1.05mol⋅kg-1now, ∆T b= K b(ethanol)⨯b(HgCl2)b(HgCl2)= ∆T b/K b(ethanol)= [1.27/1.20] mol⋅kg-1=1.05 mol⋅kg-1Therefore, HgCl2 is not an electrolyte in ethanol.1-20.Calculate the percent by mass and the molality in terms of CuSO4for 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⨯159.6/249.68]g= 7.35gm(H2O) = [11.5 - 7.35]g = 4.15gPercent by mass: 7.35/(100.0+4.15)= 0.071b = [7.35/(159.6⨯0.1042) ] mol⋅kg-1= 0.442 mol⋅kg-11-21．The 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-1sodium 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.。