Don't used Ai solution 17-Substances A and B have retention times of 3.20 and 5.40min, respectively, on a 20 cm column. The peak widths (atbase) for A and B are 0.80 and0.90 min, respectively.Calculate the column length that required to seperatesubstances A and B from each other when resolution of thecolumn increased to 3.00A) 26.88 cmBY 53.67 cmC) 17.80 cmD) 47.27 cmE) 37.15 cm OReqL= 41C-20.20H

Resolution and Column LengthThe relationship between resolution (Rs) and column length (L) can be…

Answered: 17-Substances A and B have retention…

Principles of Instrumental Analysis

7th Edition

ISBN:9781305577213

Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Chapter13: An Introduction To Ultraviolet-visible Molecular Absorption Spectrometry

Section: Chapter Questions

Problem 13.24QAP

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Consider the separation of components A and B on a GC column. The retention time for component A is 1.4 min and for component B is 2.1 min. The width at the base for the peak due toA is 0.38 min and for the peak due to B is 0.53 min. Calculate the resolution, R, and tell whetherthe two peaks are considered to be resolved and why.
A peak with a retention time of 407s has a width at half-height of 7.6 seconds. A neighboring peak is eluted 17 seconds later with w1/2 = 9.4 seconds. What is the difference in retention time required for an adequate resolution of 1.5?
Substances A and B have retention times of 12.75 and 14.28 min, respectively, on a 30.0-cm column. An unretained species passes through the column in 1.10 min. The peak widths (at base) for A and B are 1.11 and 1.27 min, respectively. What is the resolution?
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 peakwidths (at base) for 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.
Two components in an HPLC separation have retention times thatdiffer by 22 s. The first peak elutes in 10.5 min and the peakwidths are approximately equal. Use a spreadsheet to find theminimum number of theoretical plates needed to achieve thefollowing resolution, Rs, values: 0.50, 0.75, 0.90, 1.0, 1.10, 1.25,1.50, 1.75, 2.0, and 2.5. How would the results change if peak 2were twice as broad as peak 1?
A spectrophotometric method for the quantitative determination of the concentration ofPb2+ in blood yields an Ssamp of 0.133 for a 1 mL sample of blood that has been diluted to 6 mL. A second sample is spiked with 1 µL of a 1467 ppb Pb2+ standard and diluted to 6 mL, yielding an Sspike of 0.491. Determine the concentration of Pb2+ in the original sample of blood.
A peak with a retention time of 10 minutes has a width at half-height of 7.6 s neighboring peak is eluted 15 s later with W1⁄2 = 7.5 s. (a) Find out the resolution for these two components. (b) Is this resolution value good for quantitative analysis, why?
A student researcher performed a chromatographic separation of caffeine and aspartame. The retention time for caffeine, te, was found to be 200.8 s with a baseline peak width, we, of 16.1 s. The retention time for aspartame, ta, was 258.7 s with a baseline peak width, wa, of 20.6 s. The retention time for the unretained solvent methanol was 44.2 s. Calculate the average plate height, H, in micrometers for this separation, given that it was performed on a 22.1 cm long column. H = Calculate the resolution, R, for this separation using the widths of the peaks. R = 88.2 R₂5N = 3.16 Calculate the resolution if the number of theoretical plates were to increase by a factor of 2.5. 3.533 Incorrect μm
Two substances A and B have retention times of 16.40 and 17.63 min respectively on 30.0 cm column. An unsuccessful species passed over the column in 1.30 min. The widths (at the base) of the peaks ofA and B are respectively 1.11 min and 1.21 min: Calculate: a) The Resolution Rs of the column. b) The average number (Nav of theoretical plates in the column c) The equivalent height (H) d) length of column required to achieve a resolution of 1.5, e) time required to elute substance B on the longer column.
3. The lead in a swab sample, lead standards, together with a blank were made up in 5.00 mL volumetric flasks using 0.2 % HNO3. 20 μL aliquots of these solutions were injected into the spectrophotometer and the absorbance measured at 217 nm. The following results were obtained. lead / ppb Absorbance 0 0.0591 10.00 0.0858 50.00 0.1926 100.0 0.3260 150.0 0.4594 200.0 0.5929 Swab sample 0.3700 (a) Determine the amount of lead in the swab sample in μg.
A 2.54 x 10-M sample of compound X in a solvent has an absorbance of 0.370 at 528 nm in a 1.000 cm cuvette. The solvent alone has an absorbance of 0.058. What is the molar absorptivity of compound X? € = M-1.cm- A different of compound X in the same solvent has an absorbance of 0.475 at 528 nm when measured with the same cuvette. What is the concentration of this second sample? mol X [X] = L * TOOLS |x10
Substances A and B have retention times of 6.40 and 7.20 min, respectively, on 15.0 cm liquid chromatography (LC) column. The mobile phase peak is seen at 0.60 min. The peak widths (at base) for A and B are 1.10 and 1.15 min, respectively. Calculate: a) Column capacity factor, K b) Selectivity factor, a c) Resolution of substances A and B, R d) Average number of theoretical plates, Neve e) Theoretical plate height, H f) Under the same experimental conditions, length (2) of column required to achieve a resolution of 1.5, g) Time (t.) required to elute substance B on the column that gives Rs = 1.5

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