基础化学第二章后习题答案

第二章后习题答案习题水在20℃时的饱和蒸气压为2.34kPa。若于100g水中溶有10.0g蔗糖(Mr解根据,甲溶液由1.68g蔗糖(Mr=342)和20.00g水组成，乙溶液由2.45g（Mr=690）的某非电解质和20.00g水组成。⑴ 在相同温度下，哪份溶液的蒸气压高？⑵ 将两份溶液放入同一个恒温密闭的钟罩里，时间足够长，两份溶液浓度会不会发生变化，为什么？⑶ 当达到系统蒸气压平衡时，转移的水的质量是多少？解(1)溶液乙的蒸气压下降小，故蒸气压高。（2）乙溶液浓度变浓,甲溶液浓度变稀。因为浓度不同的溶液置于同一密闭容器中，由于不同，P不同，蒸发与凝聚速度不同。乙溶液蒸气压高，溶剂蒸发速度大于甲溶液蒸发速度，所以溶液乙中溶剂可以转移到甲溶液。（3）设由乙溶液转移到甲溶液的水为x(g),当两者蒸气压相等时，则x=3.22g3. 将2.80g难挥发性物质溶于100g水中，该溶液在101.3kPa下，沸点为100.51℃。求该溶质的相对分子质量及此溶液的凝固点。(Kb=0.512K·kg·mol-1，Kf=1.86K·kg·mol-1解该溶液的凝固点为-1.85℃4. 烟草有害成分尼古丁的实验式是C5H7N，今将538mg尼古丁溶于10.0g水，所得溶液在101.3kPa下的沸点是100.17℃解尼古丁的分子式为：5.溶解3.24g硫于40.0g苯中，苯的凝固点降低1.62℃。求此溶液中硫分子是由几个Solution:（a）(b)2.If4.00gofacertainnonelectrolyteisdissolvedin55.0gofbenzene,theresultingsolutionfreezesat2.36℃.Calculatethemolecularweightofthenonelectrolyte.Solution:3.Theaverageosmoticpressureofseawaterisabout30.0atmat25℃.Calculatetheconcentration(molarity)ofanaqueoussolutionofurea(NH2CONH2)thatisisotonicwithseawater.Solution:4.Aquantityof7.85gofacompoundhavingtheempiricalformulaC5H4isdissolved301gofbenzene.Thefreezingpointofthesolutionis1.05℃belowthatofpurebenzene.Whatarethemolarmassandmolecularformulaofthiscompound?Solution:Sincetheformulamassofisandthemolarmassisfoundtobe,themolecularformulaofthecompoundis.5.Ethyleneglycol(EG)CH2(OH)CH2(OH),isacommonautomobileantifreeze.itischeap,water-soluble,andfairlynonvolatile(b.p.197℃).Calculatethefreezingpointofasolutioncontaining651gofthissubstancein2505gofwater.Wouldyoukeepthissubstanceinyourcarradiatorduringthesummer?Themolarmassofethyleneglycolis62.01g.Solution:M(EG)=62.01g·mol-1Becausethesolutionwillboilat102.15℃,itwouldbepreferabletoleavetheantifreezeinyourcarradiatorinsummertopreventthesolutionfromboiling.6.Asolutionispreparedbydissolving35.0gofhemoglobin(Hb)inenoughwatertomakeuponeliterinvolume.Iftheosmoticpressureofthesolutionisfoundtobe10.0mmHgat25℃Solution:theconcentrationofthesolution:7.A0.86percentbymasssolutionofNaCliscalled“physiologicalsaline”becauseitsosmoticpressureisequaltothatofthesolutioninbloodcell.Calculatetheosmot