紫外—可见分光光度法习题答案

第十章紫外—可见分光光度法习题解答13、解：33%111012.100.1104962.0557.0ClAEcm43%111065.21012.11023610cmEM14、解：被测样品溶液的浓度为C样)100/(1000.110010000.20.1000500.03mlgC样被测样品中维生素C的浓度为C测)100/(1084.910839.900.1560551.044%11mlglEACcm测样品中维生素C的百分质量分数为ω%4.981001000.110839.910034样测CC16、解：依据Lambert-Beer定律：AECl同样条件下测定，则同理：ACAC样样标标＝0.5106.0010.1/10.1/0.304ACCgmlmgLA样样标标＝＝＝18、解：依据误差要求，当%5.0T，吸光度测量范围为0.2~0.7时，浓度的测量相对误差较小，此时对应的浓度为适宜浓度。当吸光度A=0.4343时，浓度的测量相对误差最小，此时对应的浓度为最佳浓度。依题意6.540107.147.31410104%11MEcm当A=0.2时，)100/(1070.300.15412.04%11mlgEACcm当A=0.7时，)100/(1029.100.15417.03%11mlgEACcm故适宜浓度范围为3.70×10-4(g/100ml)~1.29×10-3(g/100ml)当A=0.4343时，最佳浓度为)100/(1003.800.15414343.04%11mlgEACcm20、解：①lglg0.7160.145AT②max50.14548333.00101Acl③lgATcl1122lglgTlTl2211lglg3lg0.7160.435lTTlT2=0.367T2%=36.7%21、解：（1）对照品)100/(1000.21001000100.2033mlgC对20700.11000.2414.03%11lCAEcm对对（2）原料药)100/(1000.21001000100.2033mlgC样)100/(10932.100.1207400.03%11mlglEACcm测%6.961001000.210932.110033样测原料CC（3）注射液)100/(0.1010000.1000.1mlmlC样)100/(502.2)100/(002502.000.1207581.0%11mlmgmlglEACcm测)/(250.0)100/(00.10)100/(50.2mlmgmlmlmlmgCC样测注射液26、解；1200120001001010%11MECM)100/(000400.00.1000.10000.22500500.0mlgC样%2.79100000400.0000317.0100样测CC1%1%11lglg0.4170.000317(/100)120001.00cmMcmATCgmlEEl测