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1、食品仪器分析习题答案12页:6、7、空白平均值: 4.9 5 g/mL 的平均值: 9.7 9.7/4.9=1.98 当置信度为 95时:S/N=2 检出限 : (2/1.98)5=5.055.0 (g/mL) 当置信度为 99.7时: S/N=3 检出限 : (3/1.98)5=7.587.6 (g/mL)234页:9、解 cX cS =0.002mol/L VX =50 mL VS = 10mL ix =6.00 iS =18 10、解:注意此题根据公式计算出的浓度为稀释后溶液的浓度,必须还原为原溶液的浓度。 具体计算公式同上题: . 原子吸收:P 60页:12题: 解:计算法(1)设样品
2、中锌的浓度为cx,根据 A = Kc则:0.230 =K cx 0.453 = Kcx +(10.0500 / 25) 0.680 = Kcx +(20.0500 / 25) 得:K = 113.5 (L/mol)cx = 2.03 10-3 mol/L锌含量: X = (2.03 10-3) 25 10-3 65/1 = 3.29 10-3 (g/g)作图法:0.00 0.00200 0.004000.230 0.453 0.680从图上可知:cx = 2.00 10-3 mol/L锌含量: X = (2.00 10-3) 25 10-3 65/1 = 3.25 10-3 (g/g)13题:
3、(1)作图:浓度245810T/ %62.438.826.017.612.31/T1.6032.5773.8465.6828.130A0.2050.4110.5850.7540.910(2)T= 20.4% A= 0.690 c= 7.2 g/mL 12页:7题: 空白:4.5 4.9 5.1 4.7 5.5 4.3 5.3 =0.432c = 5 g/mL A = 9.7 Sc = 9.7/5 = 1.94 mL /gDc = 3/ Sc = 3 0.432 /1.94 = 0.67 g/mL261页习题答案7、解: (1) (2) (3) (4) 8、解: (1) (2) 乙柱效能高9、解:(1) (2) (3) 10、解: = 829 11、解: 302页7、解: