色谱题解

第十九章经典液相色谱法题解1.to=100cm/s=1000s;tR=40min=2400s;R’=to/tR=1000/2400=0.4166666672.设wNaCl=x(g)，总摩尔数=03456.01023.091.7910.392567.046.3599.22cVxx01.1192567.003456.01023.001.119145.581x；x=0.158342389(g)NaCl:%68.61%1002567.01583.0;KBr:38.32%3.Rf=7.6/16.2=0.469135802;l=l0×Rf=14.3×0.47=6.72(cm)4.Rf(B)=9/18=0.5;Rr=Rf(A)/Rf(B)=1.5;Rf(A)=Rr×Rf(B)=1.5×0.5=0.755.1.8571434.16.257.083.06.59.62R;Rf(A)=0.43125169.6;Rf(B)=0.35166.56.lA=l0×Rf(A)=0.45l0;lB=l0×Rf(B)=0.63l0;l=lB-lA=0.18l0=2(cm);l0=1118.02(cm)7.0.8684211.151515133.01.05.011/11msfVKVR8.SN=.819679111.612712.49.11271)(100bbl第二十章气相色谱法题解1.1200mm/h=0.333mm/s对苯：2333.011.28054.5n883;HETP=(mm)27.28832000对甲苯：2333.091.212254.5n1.079×103;HETP=1.85(mm)10792000对乙苯：2333.040918154.5n1.203×103;HETP=1.66(mm)120320002.①4624289161171622nHETP=m)0.648789(m46243000②t’R1=14-1=13(min);t’R2=17-1=16(min)③409625616116162eff2n;HETPeff2=m)0.732422(m40963000④k1=13/1=13;k2=16/1=16⑤k2k11.2307693.①Vm=43.75×0.24=10.5(cm3)②t’R(苯)=1.41-0.24=1.17(min);t’R(甲苯)=2.67-0.24=2.43(min);t’R(乙苯)=4.18-0.24=3.94(min);t’R(异丙苯)=5.34-0.24=5.10(min)③根据式19.4ms/VKVk,RRttVVtVkVK'10.3/'/msmsm;K(苯)=3.10×t’R(苯)=3.10×1.17=3.63;K(甲苯)=3.10×t’R(甲苯)=3.10×2.43=7.54;K(乙苯)=3.10×t’R(乙苯)=3.10×3.94=12.22;K(异丙苯)=3.10×t’R(异丙苯)=3.10×5.10=15.82④(甲苯/苯)=k甲苯/k苯=t’R（甲苯）/t’R（苯）=2.43/1.17=2.08；同理(乙苯/甲苯)=3.94/2.43=1.62;(乙苯/甲苯)=5.10/3.94=1.294.自己可解5.①u1=200/18.2=10.98901cm/s;u2=200/8.0=25.0cm/s;u3=200/5.0=40.0cm/s②u1下：n=16×(2020/223)2=1312.844;H=2000/1312=1.523411(mm)u2下：n=16×(888.0/99.0)2=1287.287;H=2000/1287=1.553655(mm)u3下：n=16×(558.0/68.0)2=1077.384;H=2000/1077=1.856348(mm)③u1下：H=1.52=A+B/11+C×11u2下：H=1.55=A+B/25+C×25u3下：H=1.86=A+B/40+C×40解所得联立方程，得A=0.0576626cm;B=0.7026273cm2s-1;C=0.0027693s(注：这里的答案与书上的不一致，是书上计算的误差造成的)。6.WCACS/6021m=1896273(mvs/mg)=1.90109(mvs/g)Dm=2N/Sm=20.01/1.90109=1.051011g/s7.311232/211R2RWWttR6分离R=1.5，根据式20.33，LR2，为达到R=1.5，L须为原来的1/22即1/4，即0.75m8.R=1