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1.0gmL1，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) = 1000mL1.0gmL10.030 = 30g
m( H2O) = 1000mL1.0gmL1(10.030) = 9.7102g
n( H2O2) = 30g/34gmoL1=0.88mol
扬州慢n( H2O) = 970g/18g.mol1=54mol
b( H2O2)= 0.88mol /0.97kg = 0.91molkg1
c( H2O2)= 0.88mol/1L = 0.88molL1
x( H2O2) = 0.88/(0.88.+54) = 0.016
喝什么茶补肾1-4．计算5.0%的蔗糖(C12H22O11)水溶液与5.0%的葡萄糖(C小猫的特征6H12O6)水溶液的沸点。
对联哪边是上联
解： b(C12H22O11)=5.0g/(342g.mol10.095kg)=0.15molkg网上预报名1
b(C6H12O6)=5.0g/(180g.mol10.095kg)=0.29molkg1
蔗糖溶液沸点上升
Tb=Kbb(C12H22O11)= 0.52Kkgmol10.15molkg1=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=Kbb(C6H12O6)= 0.52Kkgmol10.29molkg1=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: 0.1molkg1 C12H22O11溶液，0.1molkg1 CH3COOH溶液，0.1molkg1 KCl溶液。
(2)渗透压：0.1molL1 C6H12O6溶液，0.1molL1CaCl2溶液，0.1molL1 KCl溶液，1molL1 CaCl2溶液。（提示：从溶液中的粒子数考虑。）
解：凝固点从高到低：
0.1molkg1 C12H22O11溶液>0.1molkg1 CH3COOH溶液>0.1molkg1 KCl溶液
渗透压从小到大：
0.1molL1 C6H12O6溶液<0.1molL1 KCl溶液<0.1molL1 CaCl2 溶液<1molL1CaCl2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： = cRT
c =/RT = [0.366/(8.314293.15)] molL1 = 1.50104 molL1
500103L1.50104molL1 = 5.0g/M
M = 6.7104gmol1
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(C3H勇敢说爱8O3)多少克?
解： ΔTf = Kf(H2O) b(C3H8O3)
b(C3H8O3) =ΔTf / Kf(H2O)
=[3.00/1.86] molkg1
=1.61 molkg1
m(C3H8O3)=1.610.50092.09g
=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)mnHS(nx)H+]xxH+
1-9．将10.0mL0.01molL1的KCl溶液和100mL0.05mo1L1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+(nx)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/K晓出净慈寺送林子方古诗f(H2O)
=0.543K/1.86 Kkgmol1
=0.292 molkg1
w=0.292180/(0.292180+1000)
= 0.0499
(2) = cRT
= 0.292molL18.314kPaLmol1K1(273.15+37)K
=753kPa
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： 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无穷等比数列(苯)]
M(孕甾酮)= Kf(苯)m(孕甾酮)/[ΔTfm(苯)]
=[5.120.100/(0.330.00500)]gmol1
=3.1102gmol1
CHO =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3.1102gmol1，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) = 0.57molkg1、b(SO42) = 0.029 molkg1、b(HCO3) = 0.002 molkg1、 b(Na+) = 0.49 molkg1、b(Mg2+) = 0.055 molkg1、 b(K+) = 0.011 molkg1 和 b(Ca2+) = 0.011 molkg1，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
= 2.17K
Tf = 273.15K –2.17K
= 270.98K
ΔTb=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 mo1L1的KCl溶液，在第二支试管中加入1.25mL 0.050 mo1L1的Na2SO4溶液，在第三支试管中加入0.74mL 0.0033 mo1L1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 mo1L1)
第二支试管聚沉值： 0.0501.251000/(20+1.25) = 2.9(m mo1L1)
第三支试管聚沉值： 0.00330.741000/(20+0.74)=0.12(m mo1L1)
溶胶带正电。
1-18．The sugar fructo contains 40.0% C, 6.7% H and 53.3% O by mass. A solution of 11.7 g of fructo 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.mol1. What is the molecular formula of fructo?
Solution: Tb=TbTb=[78.59 78.35]K=0.24K
Tb= Kb(ethanol) b(fructo)= Kb(ethanol)m(fructo)/M(fructo)m(ethanol)
M(fructo)= Kb(ethanol)m(fructo)/Tbm(ethanol)
=[1.2011.7/(0.240.325)]gmol1 =180gmol1
C:H:O =18040%/12.011 : 1806.75%/1.008 : 18053.32%/16.00
= 6 : 12 : 6
Molecular formula of fructo 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 1.27C. Is HgCI2 an electrolyte in ethanol? Show your calculations. (Kb=1.20Kkgmol1)
Solution : If HgCl2 is not an electrolyte in ethanol
b(HgCl2)=[9.41/(271.50.03275)] molkg1
=1.05molkg1
now, Tb= Kb(ethanol)b(HgCl2)
b(HgCl2)= Tb/Kb(ethanol)
= [1.27/1.20] molkg1
=1.05 molkg1
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 CuSO45H2O in 0.1000kg of water. Remember to consider the water relead from the hydrate.
