基础化学人卫第版课后习题参考答案docx.docx

1、基础化学人卫第版课后习题参考答案docxChapter 1 Introduction习题3、(1) 5 位; (2) 3 位; (3) 2 位; (4) 1 位; (5) 2 位.XT0.1023 0. 10240.1%4、 RE0.10240.001T5、mol L 1JK 1 mol 1K JL1kgm 2 s 2 L 11000kg m 1 s 21000 Pa kPa或mol L 1 J K 1 mol 1 K J L 1 N m L 1 1000 N m 2 1000Pa kPan NaOHm( NaOH )0.010kg0.257、M ( NaOH )40.0g mol 1mol1

2、2)12m(Ca0.100 kgCa25.00 moln112M (2)40.0 g mol1Ca221m( 1 Na2 CO3 )0.100kgNa 2CO321.89moln112Na2 CO3 )106 gmol1M (228、 n( ZnCl 2 )m( ZnCl 2 )350g2.57molM ( ZnCl 2 )136.3gmol 1此溶液的物质的量浓度为：c( ZnCl 2 )n(ZnCl 2 )2.57mol3.48molL 1V739.5mL此溶液的质量摩尔浓度为：b(ZnCl 2 )n( ZnCl2 )2.57mol3.95molkg 1m(H 2 O)650gm( K)2

3、0 mg9、 c Kn( K) M(K) 39.0 g mol 15.1mmol L 1VV100 mLm(Cl)366mgc Cln(Cl) M (Cl)35.5gmol 1103mmol L 1VV100 mL10、 m(C6 H 12 O6 )50.0g L 1500mL25.0gm(C6 H 12 O6M (C6 H12O6 H 2O)198gmolH 2O) 25.0 g25.0gmolM (C6H12O6 )180g1127.5g即称取 27.5g C6H12O6H2O 晶体，溶于蒸馏水并稀释至 500mL。m(C 6 H 12O6 )25.0gn(C 6 H 12O 6 )M (

4、C6H 12O6)180 gmol1L1c(C 6 H 12O6 )VV500mL0.278mol溶液中水的质量为： 500mL1.00kgL 125.0g475g ，则葡萄糖的摩尔分数为：m(C6 H 12O6 )x(C 6H 12O6 )n(C 6H 12O6 )M (C6 H12O6 )n( H 2O)m(C 6H 12O6 )m( H 2O)n(C 6 H 12O6 )M (C6H 12O6 ) M (H 2O)25.0g180 gmol 10.0052525.0 g475g180g mol 118.0 g mol111、 m Na5.0gm NaClM (NaCl)58.5gmol5

5、.0g5.0gmolM (Na )23.0g1113gm( NaCl )13g1.4LV NaClL 1( NaCl ) 9.0 gm( I 2 )0.508g1 2.00 103mol12、 n I 2254gmolM (I2)n KI KMnO 42n( I 2 )22.00 10 3 mol 4.00 10 3 molExercises2.29.83729.240.018618 0.1932.0654.Assume the volume of the solution is 1 L, thus the mass of the solutionis:m 1.024 gmL 11L1024

6、gThe mass of NH4Cl is: m( NH 4Cl )1024 g 8.50%87.0gThe mass of solvent HO is:m(H2O) 1024 g (18.50%)937 g2m NH 4Cl87.0gb NH 4Cln NH 4ClM NH 4Cl53.5g mol 11.63mol1.74molkg 1m( H 2O)m(H 2O)937g0.937kgx NH 4Cln NH 4Cl1.63mol1.63mol0.0304n(H 2O)m( H 2 O)937 gn NH 4 Cl1.63mol1.63mol1M (H 2O)18.0g molc NH

7、4Cln NH 4Cl1.63mol1V1L1.63molL5. M (C3H8)44g mol 1 ;M (C4H 10 )58gmol 1Assume the total moles of the sample isx, thus,n(C3 H 8 ) M (C3 H 8 )n(C 4 H 10 )M (C 4 H 10 ) 58g0.43x 44g mol 10.57 x 58 gmol 158gx 1.1molThe mass of propane is: 0.431.1mol 44 g mol 121g ;The mass of butane is: 58g21g37g6.The m

8、ass of KAl(SO4)2 is:M KAl (SO4 )2258.2gmolm KAl (SO4 ) 212H 2O)118.6mgmolM (KAl (SO4 ) 2474.4g11 118.6mg 64.55mgm(KAl (SO4 ) 2 )64.55mgc( KAl ( SO4 ) 2 )n(KAl ( SO4 ) 2 )M ( KAl (SO4 ) 2 )258.2g mol 12.500 10 4 mol L 1VV1.000Lc(SO42 ) 2c(KAl (SO4 )2 )2 2.500 10 4 molL 15.000 10 4 mol L 1The density

9、of the solution is 1.00 gmL-1, thus the mass of the solution is 1.000 kg. Since the solution is very dilute, it can be considered that the mass of water(solvent) is approximately 1.000 kg,m( KAl (SO4 )2 )64.55mgn( KAl (SO4 ) 2 )M (KAl( SO4 )2 )258.2g mol1104mol kg1b(KAl (SO4 )2 )m( H 2O)1.000kg2.500

10、m( H 2 O)Chapter 2 Colligative Properties of Solutions习题100g1.x H 2O18.0g mol10.995100g10.0g18.0gmol 1342 gmol 1P P0 x H 2O2.34kPa0.9952.33kPa2.（1） n甲，溶质1.68g14.91 10 3 mol342g moln乙，溶质2.45g13.5510 3 mol n甲，溶质 ,溶剂的量相等690g molx甲， H 2O x乙， H 2O 。相同温度下，乙溶液蒸汽压高。（2）溶液乙中的溶剂可以转到甲溶液中。密闭容器中压强为 P，则 P甲 P P乙 ，甲

11、溶液中凝结速率大于蒸发速率， 溶质浓度降低；乙溶液中蒸发速率大于凝结速率， 浓度升高。直到两者浓度相等。4.9110 3 mol3.55 10 3 molmH O 3.22g（3），20.00g mH 2 O20.00gmH 2O23. Tb K bbBK bmBM BmAmB K b2.80g512Kg mol 128g mol1M B0.51K 100gTbmA方法一：mB1.86 103K g mol12.80g1.9KTf K f bB K f28g mol 1M B mA100g方法二：TfK f bBK fTb1.86K kg mol10.51K1.9KK b0.512K kg m

12、ol 1Tf0 CTfC1.9 C4. TbK bbB K bmBM BmAM BmB K b0.538g 512K g mol 1162g mol1TbmA0.17K 10.0g设尼古丁的分子式为：C5 H 7 N a162g mola81g mol11 2 ，分子式为： C10 H14 N2 。5. TbK bbB K bmBM BmAM BmB K b3.24g 5120K g mol 1256g mol 1TbmA1.62K 40.0g设此单质硫的分子式为： saa256g mol32g mol118 ，分子式为： s8 。6. 方法一：对于稀溶液， cB bBi.(1) (2) (4

13、) 均为水溶液，其中 (1) (2)为非电解质，浓度相等，凝固点下降值相等，凝固点 (1)(2)；(4)相对于 (1) (2) 来讲，凝固点下降更多，则凝固点 (1)(2) (4)ii.(3)为苯的溶液，纯苯的凝固点为 5.5 ，Tf K f C6 H 6 bB 5.12K kg mol 1 0.1mol kg 10.512K ，则其凝固点大于 0 ，为 (1) (2) (3) (4)中的最高值。排序： (3) (1)(2) (4)方法二：对于稀溶液， cB bB(1) T f K f H 2O bB 1.86K kg mol 1 0.1mol kg 1-1 蔗糖水溶液的凝固点为： 0.186

14、 0.1mol L(2) Tf K f H 2O bB1.86K kg mol10.11mol kg0.186K0.186K-1 甲醇水溶液的凝固点为： 0.186 0.1mol L(3) Tf K f C6 H 6 bB 5.12K kg mol 1 0.1mol kg 10.512K-10.1mol L甲醇苯溶液的凝固点为： 5.5 0.512 4.988 (4)Tf iK fH 2 O bB2 1.86Kkg mol 1 0.1mol kg 10.372K-10.1mol LNaCl 水溶液的凝固点为： 0.372 排序： (3) (1)(2) (4)7. (1)c C6 H 12O6R

15、T0.2molL 1RT(2)c Na2 CO31 c1 Na2CO310.2mol L 10.1molL 1222ic Na CO3RT3 0.1molL 1RT0.3mol L 1RT2(3)c Na3 PO41 c1 Na3 PO410.2molL 10.2 molL 13333ic Na PO4RT40.2 molL 1RT0.8 molL 1RT333(4)ic NaCl RT20.2molL 1RT0.4mol L 1 RT排序： (4) (2) (3) (1)8.ib NaClTf0.26K1 0.14molkg1K f1.86K kgmolcNaCl0.14molL 1140m

16、molL 1os(1)的说法正确。9.cB RTmBRTM BVm RT2.00g8.314Jmol 1 K1 298K41M BB6.91 10 gmolV0.717kPa 0.100L10. bTf0.52K0.28molkg1-1K f1.86Kkgmol 1Bcos 0.28 mol L 11 1 =cosRT = 0.28 mol L 8.314J K mol310K= 720 kPa11.由题可知，两种溶液的质量摩尔浓度相等。b 尿素bB 即：m尿素mBM 尿素 mA,1M B mA,2M BmBM 尿素 mA,142.8g60.05gmol 1200g343g mol1m尿素 m

17、A, 21.50g1000g12.bB0.020mol 20.010mol30.040mol30. 19mol1.9molkg 10.100kg0.100kgTbKb bB0.512K kgmol 11.9molkg 10.97KExercises1. (calculating process omitted). (a): The boiling point (b.p.) is 102.72C, the freezing point (f.p.) isC; (b):The9.89b.p. is 101.97 C,the f.p. isC.7.162.Tf 5.502.36 3.14(K )M

18、BmB K f4.00g 5.10Kkg mol 1118g mol1Tf mA3.14K 55.0g3.M BmB K f7.85g5.10K kgmol1mol1T fmA1.05K301g127gM (C5H4)64gmol 11127gmol 12Therefore, the molecular formula of this compound is C10H8.4. The molality of the EG solution is:bBmB62.01g651g4.19mol kg 1M Bm Amol 1 2505gTfK fbB1.86Kkg mol 14.19mol kg 1

19、7.79KThe freezing point of the solution is: (0 7.79) C7.79 C .(The calculation here has a relatively large error since the solution is not a dilute solution anymore.)5.10.0 mmHg corresponds to 10.0mmHg 101.3kPa 1.33kPa . 760mmHgcB RTnBRTmB RTmB RTVM BVM BVmB RT35.0 g 8.314 Jmol 1 K 1298K41M BV1.33kPa 1L6.52 10 gmol6. Assume the volume of the solution is 1 L, thus,The mass of NaCl is: 1.005g mL 1 1L 0.86% 8.6g .The molarity of NaCl solution is:nBnB.1cB8 6g0.15mol L, therefore,M BV.11LV58 5g molic B RT 20.15molL 1 8.314Jmol 1K 1(27337)K 7.7 102 kPaChapter 3