无机及分析化学第二版答案.docx
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无机及分析化学第二版答案
无机及分析化学第二版答案
【篇一:
无机及分析化学答案(第二版)第一章】
txt>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
?
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=373.30k
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?
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?
t
c=?
/rt=[0.366/(8.314?
293.15)]mol?
l?
1=1.50?
10?
4mol?
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?
=[3.00/1.86]mol?
kg?
1
=1.61mol?
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)?
tf=kf(h2o)?
b(c6h12o6)
b(c6h12o6)=?
tf/kf(h2o)
=0.543k/1.86k?
kg?
mol?
1
=0.292mol?
kg?
1
w=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
=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:
?
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)]
=[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?
)=0.029mol?
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
=2.17k
tf=273.15k–2.17k
=270.98k
=[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0.53ml4.0mo1?
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?
1
c:
h:
o=180?
40%/12.011:
180?
6.75%/1.008:
180?
53.32%/16.00
=6:
12:
6
molecularformulaoffructoseisc6h12o6.
1-19.asampleofhgcl2weighing9.41gisdissolvedin32.75gofethanol,c2h5oh.theboiling-pointelevationofthesolutionis1.27?
c.ishgci2anelectrolyteinethanol?
showyourcalculations.(kb=1.20k?
kg?
mol?
1)
solution:
ifhgcl2isnotanelectrolyteinethanol
b(hgcl2)=[9.41/(271.5?
0.03275)]mol?
kg?
1
=1.05mol?
kg?
1
now,?
tb=kb(ethanol)?
b(hgcl2)
b(hgcl2)=?
tb/kb(ethanol)
=[1.27/1.20]mol?
kg?
1
=1.05mol?
kg?
1
therefore,hgcl2isnotanelectrolyteinethanol.
1-20.calculatethepercentbymassandthemolalityintermsofcuso4forasolutionpreparedbydissolving11.5gofcuso4?
5h2oin0.1000kgofwater.remembertoconsiderthewaterreleasedfromthehydrate.
solution:
m(cuso4)=[11.5?
159.6/249.68]g=7.35g
m(h2o)=[11.5?
7.35]g=4.15g
percentbymass:
7.35/(100.0+4.15)=0.071
b=[7.35/(159.6?
0.1042)]mol?
kg?
1=0.442mol?
kg?
1
1-21.thecellwallsofredandwhitebloodcellsaresemipermeablemembranes.theconcentrationofsoluteparticlesinthebloodisabout0.6mo1?
l?
1.whathappenstobloo