基础化学(人卫第8版)课后习题参考答案.pdf

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1、Chapter1Introduction习题3、(1)5位;(2)3位;(3)2位;(4)1位;(5)2位.XT0.10230.10244、RE0.0010.1%T0.10245、111122112molLJKmolKJLkgmsL1000kgms1000PakPa或111112molLJKmolKJLNmL1000Nm1000PakPam(NaOH)0.010kg7、nNaOH10.25molM(N

2、aOH)40.0gmol12m(Ca)1220.100kgnCa5.00mol21211M(Ca)40.0gmol221m(NaCO)1230.100kg2nNa2CO31.89mol2111M(NaCO)106gmol2322m(ZnCl)350g28、n(ZnCl)2.57mol21M(ZnCl)136.3gmol2此溶液的物质的量浓度为：n(ZnCl2)2.57mol1c(ZnCl)3.48molL2V739.5mL此溶液的质量摩尔浓

3、度为：n(ZnCl2)2.57mol1b(ZnCl)3.95molkg2m(HO)650g2m(K)20mg1n(K)M(K)39.0gmol19、cK5.1mmolLVV100mLm(Cl)366mg1n(Cl)M(Cl)35.5gmol1cCl103mmolLVV100mL10、1m(CHO)50.0gL500mL25.0g61261M(C6H12O6H2O)198gmolm(CHOHO)25.0g25.0g27.5g

4、612621M(C6H12O6)180gmol即称取27.5gC6H12O6·H2O晶体，溶于蒸馏水并稀释至500mL。m(C6H12O6)25.0g1n(C6H12O6)M(C6H12O6)180gmol1c(CHO)0.278molL6126VV500mL1溶液中水的质量为：500mL1.00kgL25.0g475g，则葡萄糖的摩尔分数为：m(CHO)6126n(CHO)M(CHO)61266126x(CHO)6126n(C6H12O6)n(H2O)m(C6H12O6)m(

5、H2O)M(CHO)M(HO)6126225.0g1180gmol0.0052525.0g475g11180gmol18.0gmol11、mNa5.0g1M(NaCl)58.5gmolmNaCl5.0g5.0g13g1M(Na)23.0gmolm(NaCl)13gVNaCl1.4L1(NaCl)9.0gLm(I2)0.508g312、nI212.0010molM(I)254gmol233nKIKMnO42n(I2)2

6、2.0010mol4.0010molExercises29.83729.242.0.0186180.1932.0654.Assumethevolumeofthesolutionis1L,thusthemassofthesolutionis:1m1.024gmL1L1024gThemassofNH4Clis:m(NH4Cl)1024g8.50%87.0gThemassofsolventH2Ois:m(H2O)1024g(18.50%)937gmNH4Cl87.0g1

7、nNH4ClMNH4Cl53.5gmol1.63mol1bNH4Cl1.74molkgm(HO)m(HO)937g0.937kg22nNH4Cl1.63mol1.63molxNHCl0.03044nNH4Cln(H2O)m(H2O)937g1.63mol1.63mol1M(HO)18.0gmol2nNH4Cl1.63mol1cNH4Cl1.63molLV1L115.M(CH)44gmol;M(CH)58gmol38410Assumethe

8、totalmolesofthesampleisx,thus,n(C3H8)M(C3H8)n(C4H10)M(C4H10)58g110.43x44gmol0.57x58gmol58gx1.1mol1Themassofpropaneis:0.431.1mol44gmol21g;Themassofbutaneis:58g21g