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大学仪器分析经典考试真题3

课程名称:仪器分析 命题教师: 适用班级:

1、满分100分。要求卷面整洁、字迹工整、无错别字。

2、考生必须将姓名、班级、学号完整、准确、清楚地填写在试卷规定的地方,否则视为废卷。

3、考生必须在签到单上签到,若出现遗漏,后果自负。

4、如有答题纸,答案请全部写在答题纸上,否则不给分;考完请将试卷

和答题卷分别一同交回,否则不给分。 试 题 一、单选题(每题3分,共30分) 1、可见光波长范围是在( C ) A.10-200nm; B.200-400nm; C.400-750nm D.2.5-5.0μm 2、吸光度读数在(B) 范围内,测量较准确。 A .0~1 B .0.15~0.7 C .0~0.8 D .0.15~1.5 3、对于电位滴定法,下面哪种说法是错误的(D) A.在酸碱滴定中,常用pH 玻璃电极为指示电极,饱和甘汞电极为参比电极 B.弱酸、弱碱以及多元酸(碱)不能用电位滴定法测定 C.电位滴定法具有灵敏度高、准确度高、应用范围广等特点 D.在酸碱滴定中,应用电位法指示滴定终点比用指示剂法指示终点的灵敏度高得多 4、一般气相色谱法适用于(C )

A. 任何气体的测定,

B. 任何有机物和无机物的分离测定,

C.无腐蚀性气体与在气化温度下可以气化的液体的分离与测定,

D. 无腐蚀性气体与挥发的液体和固体的分离与测定。

5、在气相色谱分析中,影响组份之间分离程度的最大因素是(B)

A. 进样量;

B. 柱温;

C. 载体粒度;

D. 气化室温度。

6、分离有机胺时,最好选用的气相色谱柱固定液为(D)

A.非极性固定液;B. 高沸点固定液;C. 混合固定液;D. 氢键型固定液。

7、色谱法不具有下列哪种特性?( D)

A.分析速度快

B.灵敏度高

C.适用于混合物的分析

D.组分快速定性

8、下列说法错误的是(B)

A.醋酸电位滴定是通过测量滴定过程中电池电动势的变化来确定滴定终点

B.滴定终点位于滴定曲线斜率最小处

C.电位滴定中,在化学计量点附近应该每加入0.1~0.2mL滴定剂就测量一次电动势

D.除非要研究整个滴定过程,一般电位滴定只需准确测量和记录化学计量点前后1~2mL的电动势变化即可

9、在液相色谱分析中,提高色谱柱柱效的最有效的途径是(A)

A. 减小填料粒度;

B. 适当升高柱温;

C. 降低流动相的流速;

D. 降低流动相的粘度。

10、标准甘汞电极的外玻璃管中装的是(A)

A.1.0mol/LKCl溶液

B. 0.1mol/LKCl溶液

C. 0.1mol/LHCl溶液

D.纯水

二、简述题(共20分)

1、测量吸光度时,应该如何选择参比溶液?(6分)

答:当试液及显示剂均无色时,可用蒸馏水作参比溶液。(2分)如果显色剂为无色,而被测试液中存在其他有色离子,可采用不加显色剂的被测试液作为参比溶液。(2分)如果显色剂和被测试液均有颜色,可将一份试液加入适量掩蔽剂,将被测组份掩蔽起来,使之不再与显色剂作用,而显色剂及其他试剂按试剂测定方法加入,以此作为参比溶液,这样可以消除显色剂和一些共存组份的干扰。(2分)

2、色谱分析法的特点有哪些?(6分)

答:1)高选择性;(1.5分)2)高分离效能;(1.5分)3)高灵敏度;(1分);4)分析速度

快;(1分)5)应用范围广。(1分) 3、确定电位滴定终点的方法有哪些?(8分) 答:①E-V 曲线法:根据实验数据测得的E 代表电动势作纵轴,V (滴定体积)作横轴,在S 形滴定曲线上,作两条与滴定曲线相切的平行直线,两平行线的等分线与曲线的交点为曲线的拐点,对应的体积即滴定终点是所需的体积。(2分) ②ΔE/ΔV-V 曲线法: 曲线的最高点对应于滴定终点。(2分) ③二级微商法:基于ΔE/ΔV-V 曲线的最高点正是二级微商Δ2E/ΔV 2等于零处。可通过 绘制二级微商曲线图,或通过计算求得终点。 ④直线法.(2分) 三、计算题(共50分) 1、用一般分光光度法测量0.00100 mo l ·L -1 Zn 标准溶液和含Zn 的试液,分别测得A=0.700和A=1.000, 两种溶液的透光率相差多少?如果采用0.00100 mol ·L -1标准溶液作为参比溶液,试液的吸光度为多大?示差分光光度法与一般的分光光度法相比较,读数标尺放大了多少倍?(14分) 解: 由T A lg -=得A T 101= 标准溶液的透光率:%201011017.0===S A S T (1分) 含Zn 试液的透光率: %101011011===X A X T (1分) 两种溶液的透光率相差: %10%10%20=-=-=?X S T T T (3分) 以0.00100mol ·L -1标准溶液为参比溶液时, 试液的透光率为: %50%

20%10===S X T T T 相对 (2分)

这时, 试液的吸光度: 301.05.0lg lg =-=-=相对相对T A 或30.07.00.1=-=-=S X A A A 相对 (3分) 透光率标尺放大倍数 = 5%10%50==X T T 相对

或 = 5%20%100%100==S T (4分) 2、在一定色谱条件下分析只有二氯乙烷、二溴乙烷、四异基铅的样品,得到其峰面积和校正因子如下:

计算三种物质的百分含量。(12分) 解: 二氯乙烷:50.10.150.1'=?=二氯乙烷二氯乙烷F A ; (1分) 二溴乙烷:67.165.101.1'=?=二溴乙烷二溴乙烷F A ;(1分) 四异基铅:94.475.182.2'=?=四异基铅四异基铅F A (1分) 11.894.467.150.1'=++=∑i i F A (3分) %9.60%%6.20%%5.18%10011.850.1%===?=四异基铅二溴乙烷二氯乙烷;;P P P (3x2分)

3、在2.0 m 长的色谱柱上测得环己烷的保留时间为30 min,惰性组份的流出时间是1.5min,峰底宽为5.0min ,求有效塔板数及有效塔板高度。(12分) 解:环己烷的调整保留时间t R ’=30-1.5=28.5min (2分)

有效塔板数n 有效=16(t R ’/Y )2=16(28.5/5.0)2=520 (5分)

有效塔板高度H 有效=L/n 有效=2.0/520=0.385cm (5分)

4、晶体膜氯电极对C r O 42-的电位选择性系数为2×10-3。当氯电极用于测定pH 为

6的0.01mol/L 铬酸钾溶液中的5×10-4mol/L 氯离子时,估计方法的相对误差有多大?(12分)

解:%测量误差=分)

分)5(4010010501

.01025(1004

2/13/=????=??--i

z z j ij a a K j i

即相对误差有40% (2分)

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