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1、现代仪器分析课后作业6(色谱分析法 )参考答案专业年级班学号姓名1. Multiple Choice - circle correct answer(选择题 ) (1) The effect of Eddy diffusion on zone broadening ( 涡流扩散对谱带展宽的影响是) A) is independent of the diameter of packing particles. (与固定相粒径无关) B) is decreased with the increase of diameter of packing particles.( 随固定相粒径的增大而减小)
2、C) is increased with the increase of diameter of packing particles. ( 随固定相粒径的增大而增大) D) None of the above.(2) The plate theory in chromatography A) does not provide the information on the peak shapes and rate of movement. B) provides the information on the effects of mobile phase on resolution. C) te
3、lls one how to optimize the experimental parameters for a better separation. D) does not account for the mechanism causing peak broadening. E) none of above (3) The base peak(基峰 ) shown in Mass Spectrometry is A) The first peak eluted from column B) The highest peak C) Molecular ion peak D) None of
4、above(4) Which of the following statements regarding chromatography is correct? A) The higher the flow-rate of the mobile phase, the higher the effect of longitudinal diffusion. B) The higher the flow-rate of the mobile phase, the smaller the effect of longitudinal diffusion (纵向扩散 ). C) The faster t
5、he mobile phase moves, the smaller the band broadening caused by mass transfer in stationary phase D) None of above (5) Which of the following statements is correct regarding a chromatographic column with a certain length? A) The larger the theoretical plate height, the higher the chromatographic ef
6、ficiency. B) The larger the theoretical plate number, the higher the chromatographic efficiency. C) There is no relationship between theoretical plate number and chromatographic efficiency. D) None of above. (6) The Gas-solid chromatography is basically used for the separation of A) high molecular w
7、eight compounds. B) thermally unstable compounds. C) polar compounds D) All of the above E) none of the above (7) For the analysis of highly volatile compounds, A) a column with thick stationary film is preferred. B) a column with thin stationary film is preferred. C) All of the above. D) None of th
8、e above (8) In normal phase liquid chromatography (NPLC), A) weak mobile phase is a polar liquid. B) strong mobile phase is a polar liquid. C) All of the above. D) None of the above. (9) A diode-array spectrometer is especially valuable as a detector for HPLC because it A) is fully solid-state and e
9、asily interfaced to a computer. B) subtracts the interfering effects of co-eluting contaminants. C) provides a complete spectrum for each peak that elutes. D) can utilize a long path length for high sensitivity. E) None of the above. (10) Which type of GC detector is best suited for the selective de
10、tection of polychlorinated biphenyl residues ( 多氯联苯残基 )? A) FID. B) ECD. C) TCD. D) FPD. E) NPD. 2. Define the following ( 名词解释 ). (1) Adjusted retention time ( 调整保留时间)：扣除了死时间后的保留时间。(2) Eddy diffusion ( 涡流扩散 )：流动相中的组分分子在色谱柱中随载气或载液向前运行时，会碰到固定相的小颗粒，使前进受阻，改变前行方向而形成向垂直方向的流动，称为“涡流”，由此产生的色谱峰峰形变宽就称为涡流扩散。(3
11、) Longitudinal diffusion ( 纵向扩散 )： 待测组分分子在分离柱中由于浓差梯度形成造成的谱带展宽就称为纵向扩散。(4) Gradient elution ( 梯度洗脱 )：在分离过程中使两种或两种以上不同极性的溶剂按一定程序连续改变它们之间的比例，从而使流动相的强度、极性、pH值或离子强度相应地变化，达到提高分离效果，缩短分析时间的目的。(5) Stationary phase ：色谱分离柱中固定不动的相。(6) Mobile phase：色谱分离柱中携带待分离混合物流过固定相的相，如：气体、液体、超临界流体。3. What is van Deemter equati
12、on ( 范第姆特方程)? Explain each term of this equation and its effect on the efficiency of chromatographic separation ( 解释此方程的每一项及其对色谱分离效率的影响). Solution: van Deemter equation H=A+B/u+Cu A：涡流扩散项。A的大小与固定相的颗粒细度、均匀性和填充均匀性有关。B/u：分子扩散项。Cu：传质阻力项。从范第姆特方程可以看出，造成色谱峰扩展及柱效降低的主要原因是涡流扩散A、分子扩散B/u和传质阻力 Cu。各种因素相互制约，共同影响色谱
13、柱的分离效率。如：载气流速增大，分子扩散项影响降低，柱效提高，但此时传质阻力项影响增大，使柱效降低；升高柱温，有利于传质，但却加剧了分子扩散。因此，需选择最佳色谱条件才能使柱效达到最高。4. Consider a 50 cm column with a plate height of 1.5 mm that provides a theoretical plate number of 333 at a flow rate of 3 mL/min; VM=1.0 mL. (a) What are the solute retention time and retention volume whe
14、n k =1, 2, 5, and 10? (b) What is the baseline peak width for each of the foregoing values of k (每一个前面的 k 数值 ) ? Solution: min, n=333 According to the following equations, , then, tR=(1+k )tM, VR=(1+k )VMn=16 (tR/Y)2, then, Y=We can calculate the retention time, retention volume and baseline peak wi
15、dth for each of kk1 2 5 10 tR=(1+k )tM/min 2/3 1 2 11/3 VM=(1+k )VM/mL 2 3 6 11 Y=/min 0.15 0.22 0.44 0.80 5. Substances A and B have retention times of 16.40 and 17.63 min, respectively, on a 30.0 cm column. An unretained species passes through the column in 1.30 min. The peak widths (at base) for
16、A and B are 1.11 and 1.21 min, respectively. Calculate (a) the column resolution, (b) the average number of plates in the column, (C) the plate height, (d) the length of column required to achieve a resolution of 1.5, and (e) the time required to elute substance B on the column that gives an R value of 1.5 (Help hint: the retention time of substance B is proportional to its R2). Solution: (a) (b) nA=16 (tR(A)/YA)2=16 (16.4min/1.11min)2=3493, nB=16 (tR(B)/YB)2=16 (17.63min/
