基础化学标准答案第02

#/8此溶液中硫原子是由AT—fKf1.62K5.10K-kg-mol-1=0.318mol-kg-13.24gb 0.0400kgBx0.318mol-kg-1=255g-mol-18个硫原子组成。6,试比较下列溶液的凝固点的高低：（苯的凝固点为5.5℃,Kf=5.12K・kg・mol-1,水的 Kf1.86K・kg・mol-1)0.1mol・L-i蔗糖的水溶液；⑵ 0.1mol・L-1乙二醇的水溶液；序是：{C0s7.8.0.1mol・L-i乙二醇的苯溶液；⑷ 0.1mol・L-1氯化钠水溶液。对于非电解质溶液⑶口⑴口⑵口⑷AT=Kb，电解质溶液试排出在相同温度下，下列溶液渗透压由大到小的顺序：c(C6Hl2O6)=0.2mol・L-1；c(!NaPO)=0.2molL-1;3 3 4根据非电解质溶液今有一氯化钠溶液，测得凝固点为此溶液的渗透浓度为此溶液的渗透浓度为n=cRT,电解质溶液□⑶口⑴140mmol・L-1；70mmol・L-1；解由于NaCl在水溶液中可以电离出molL1Bmol-kg10.26K1.86K-kg-mol-1所以（ 1口正确，氯化钠溶液的渗透浓度应为9.100mL水溶液中含有2.00g白蛋白，子质量。解c（白蛋白）=AT=iKb，口相同浓度溶液的凝固点的大小口1c(—NaCO)=0.2mol-L-1；2 2 3⑷ c(NaCl)=0.2mol・L-1。n=icRT，渗透压大小顺序是：-0.26］，下列说法哪个正确，为什么？⑵此溶液的渗透浓度为280mmol・L-1；⑷此溶液的渗透浓度为7153mmol・L-112倍质点数目，该溶液的渗透浓度可认为0.140mol-kg-1140mmol・L-125℃时此溶液的渗透压力为0.717kPa求白蛋白的相对分0.717kPaRT 8,314kPa-L-mol-1-K-1x(273+25)K=2.89x10-4mol-L-1M□白蛋白)2.00g2.89x10-4mol-L-1x0.100L=6.92x104g-molt10,测得泪水的凝固点为-0.52］，口泪水的渗透浓度及371时的渗透压力。0.52K角牛b= =0.280mol-kg-1B1.86K-kg-mol-1泪水的渗透浓度为280mmol泪水的渗透浓度为280mmol•L-11n=0.28mol-L-1x8.314kPa-L-mol-1-K-1x(273+37)K=722kPa11,今有两种溶液，一为1.50g11,今有两种溶液，一为1.50g尿素(Mr=60,05)溶于 200g水中，另一为42.8g某非电解质溶于1000g水中，这两种溶液在同一温度下结冰，试求该非电解质的相对分子质量。解若口溶液在同一温度下结冰，则解若口溶液在同一温度下结冰，则b(尿素)=b□□□□□□□m/MAT=Kb=K rffBfV1.50g/60.05g-mol1.50g/60.05g-mol-1200g42.8g/M r1000gM=343g-mol-1

r12,10.100kg12,10.100kg的水中溶有0.020molNaCl,0.010molNa2sQ和0.040molMgCl2。假如它们在溶液中完全电离，计算该溶液的沸点升高值。0.020molb(NaCl)= =0.20mol-kg-10.100kgb(NaSO)=

2 4bb(NaSO)=

2 4b(MgCl)=20.010mol0.100kg0.040mol0.100kg=0.10mol-kg-1=0.40mol-kg-1bb(总)=2xb(NaCl)+3xb(NaSO)+3xb(MgCl)2 4 2=2x0.20mol•kg-1+3x0.10mol•kg-1+3x0.40mol-kg-1=1.9mol•kg-1=0.512K•kg•mol-1x1.9mol•kg-1=0.97KExercises1.Whatarethenormalfreezingpointsandboilingpointsofthefollowingsolution?(a)21.0gNaClin135mLofwater.(b)15.4gofureain66.7mLofwater.Solution:1a)c(NaCl)=21.0g/5&5g.mol-1=2.659mol-L-i0.135LAT=2x1,86K-kg-mol-1x2.659mol-L-i=9.89KfT=-9.89oCfAT=2x0.512K-kg-mol-1x2.659mol-L-i=2.72KbT=102.72oCb(b)c(CONH):154gz60.0g•mol-1=3.848mol•L-i24 0.0667LAT=1.86K-kg-mol-1x3.848mol-L-i=7.16KfT=—7.16oCfAT=0.512K•kg•mol-1x3.848mol.L-i=1.97KbT=101.97oCb2.If4.00gofacertainnonelectrolyteisdissolvedin55.0gofbenzene,theresultingsolutionfreezesat2.361.Calculatethemolecularweightofthenonelectrolyte.Solution:AT=To-T=5.5oC-2.36oC=3.14oCfffb= 314K =0.616mol•kg-1B 5.10K•kg•mol-1M),一如0550kg=n8g•mol-1r0.616mol•kg-13.Theaverageosmoticpressureofseawaterisabout30.0atmat25℃.Calculatetheconcentration(molarity)ofanaqueoussolutionofurea(NH2CONH2)thatisisotonicwithseawater.Solution:101.3kPa30.0atmx =3039kPa1atmc(HNCONH)=3039kPaRT―8.314kPa•L-K-1mol-1x298K=1.23mol•L-i4.Aquantityof7.85gofacompoundhavingtheempiricalformulaC5H4isdissolved301gofbenzene.Thefreezingpointofthesolutionis1.05belowthatofpurebenzene.Whatarethemolarmassandmolecularformulaofthiscompound?Solution:b=B1.05K5.10K-kg-mol-i=0.206mol-kg-i7.85g0.301kgx0.206mol-kg-i=127g-mol-iSincetheformulamassofCHis64g-mol-1andthemolarmassisfoundtobe127g-mol-15 4themolecularformulaofthecompoundisCH.10 85.Ethyleneglycol(EG)CH2(OH)CH2(OH),isacommonautomobileantifreeze.itischeap,water-soluble,andfairlynonvolatile(b.p.197口1Calculatethefreezingpointofasolutioncontaining651gofthissubstancein2505gofwater.Wouldyoukeepthissubstanceinyourcarradiatorduringthesummer?Themolarmassofethyleneglycolis62.01g.Solution:M(EG)=62.01g・mol-1 b(EG)=651gz62.01g.mol-1=4.19mol・kg-12.505kgAT=1.86K-kg-mol-1x4.19mol-kg-1=7.79KT=-7.79oCf fAT=0.512K-kg-mol-1x4.19mol-kg-1=2.15KT=102.15oCb bBecausethesolutionwillboilat102.15℃,itwouldbepreferabletoleavetheantifreezeinyourcarradiatorinsummertopreventthesolutionfromboiling.6.Asolutionispreparedbydissolving35.0gofhemoglobin(Hb)inenoughwatertomakeuponeliterinvolume.Iftheosmoticpressureofthesolutionisfoundtobe10.0mmHgat25℃,calculatethemolarmassofhemoglobin.Solution:theconcentrationofthesolution:c(Hb)=RTsTT101.3kPa10.mmHgx一 760mmHg =5.38x10-4mol-L-18.314kPa-L-mol-1-K-1x(273+25)KMM(Hb)= 35.0g5.38x10-4mol-L-1x1L=6.51x104g-mol-17.A0.86percentbymasssolutionofNaCliscalled“physiologicalsaline”becauseitsosmoticpressureisequaltothatofthesolutioninbloodcell.Calculatetheosmoticpressureofthissolutionatn