When measured in a 1.00 cm cuvette, a 4.1x10M solution of species X exhibited absorbances of 0.1025 and 0.533 at 530 nm and 325 nm, respectively. A 8.20x10M solution of species Y gave absorbances of 0.230 and 0.508 at 530 nm and 325 nm, respectively. Both species were dissolved in the same solvent, and the solvent'sabsorbance was 0.000 and 0.000 at 530 nm and 325 nm, respectively, in a 1 cm cuvette.Calculate the concentrations of X and Y in an unknown solution that yielded absorbance data of 0,683 at 530 nm and 1.351 at 325 nm in a 1.0 cm cuvetteA. X= 4.20 X10MY=3.20 x10+MO B. X1.25 X10MY= 4.20 x10-5MOCX-3.75X10MY-1 4x10MOD.X-3.80 x 10 MY 2.10K10M

The amount of a substance that dissolves in a specified volume of solvent to yield a solution is…

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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

Chapter14: Applications Of Ultraviolet-visible Molecular Absorption Spectrometry

Section: Chapter Questions

Problem 14.9QAP

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When measured in a 1.00 cm cuvette, a 4.1x10M solution of species X exhibited absorbances of 0.1025 and 0.533 at 530 nm and 325 nm, respectively. A8.20x10 M solution of species Y gave absorbances of 0.230 and 0.508 at 530 nm and 325 nm, respectively. Both species were dissolved in the same solvent, and the solvent's absorbance was 0.000 and 0.000 at 530 nm and 325 nm, respectively, in a 1 cm cuvette. Calculate the concentrations of X and Y in an unknown solution that yielded absorbance data of 0.683 at 530 nm and 1.351 at 325 nm in a 1.0 em cuvete O A.X-4.20 x10 M Y=3.20 x104 M OBX-1.25 X10M Y420 x10-SM OCX-3.75x1oM Y-1 A10M OD.X-3.80 x 10M Y=2.10x10M
Furosemide (MM: 330.7 g/mol) is a diuretic derived from anthranilic acid, this diuretic is used in the treatment of kidney disease, hypertension, liver cirrhosis, and others. Furosemide in alkaline dissolution has an absorption spectrum with a maximum peak centered at 271 nm. The analysis of a furosemide tablet was performed by dissolving its content in a 0.1 mol L-1 NaOH solution and this dissolution completed for a volume of 100 mL, with 1.0 mL of this dissolution was transferred to 50.0 ml and made up to volume with the same solvent. It is dissolution has an absorbance value of 0.475 to 271 nm, with a 1.0 cuvette cm of optical path. At this wavelength, a series of dissolutions of a Standard substance of this drug has the following absorbance measurements (1.0 cm bucket): a) Demonstrate whether the experimental data are in accordance with Bee's law (b) Determine absorptivity and molar absorptivity (c) Determine the amount of drug in the tablet in mg and in ppm (d) Determine the…
2. Answer ALL parts. (a) The absorbance of compound A was measured in a 1.00 cm cuvette at 238 nm. A 3.96 x 10 M solution of compound A exhibited an absorbance of 0.624. A blank solution containing only the solvent had an absorbance of 0.029 at the same wavelength. Calculate the molar absorptivity of compoundA in m' mole". (i) The absorbance of an unknown solution of Compound A in the same solvent and cuvette was 0.375 at 238 nm. Find the concentration of A in the unknown. (iii) A concentrated solution of Compound A in the same solvent was diluted from an initial volume of 2.00ml to a final volume of 25 ml and then had an absorbance of 0.733 under the same conditions. What is the concentration of A in the concentrated solution? (iv) What type of cuvette was used in the above analysis? Justify your answer. (b) Name ONE advantage and ONE disadvantage of a linear photodiode array detector compared to a traditional photomultiplier based spectrometer. (c) Why is an FTIR spectrum digital in…
When measured in a 1-cm cell, an 8.50 x 10-5 M solution of species A exhibited absorbances of 0.129 and 0.764 at 475 and 700 nm, respectively. A 4.65 x 10-5 M solution of species B gave absorbances of 0.567 and 0.083 under the same circumstances. Calculate the concentrations of A and B in solutions that yielded the following absorbance data in a 1.25 cm cell : 0.502 at 475 nm and 0.912 at 700 nm. Pls answer need help.
What color would you expect for light transmitted through a solution with an absorption maximum at (a) 450; (b) 550; (c) 650 nm?
A 2.78 x 10-4 M solution of a compound has an absorbance of 0.469 at 520 nm in a 1.00 cm cell. The solvent's absorbance under the same conditions is 0.024. (a) What is the molar absorptivity of the unknown compound? 4.0 1687 X M-¹cm-1
Mercury(II) forms a 1:1 complex with triphenyltetrazolium chloride that exhibits an absorption maximum at 255 nm. The mercury(II) in a soil sample was extracted into an organic solvent containing an excess of TTC , and the resulting solution was diluted to 100.0 mL in a volumetric flask. Five-milliliter aliquots of the analyte solution were then transferred to six 25-mL volumetric flasks. A standard solution was then prepared that was 5*10^-6 M in . Volumes of the standard solution shown in the table were then pipetted into the volumetric flasks, and each solution was then diluted to 25.00 mL. The absorbance of each solution was measured at 255 nm in 1.00-cm quartz cells. Please solve and explain these questions: A.) Enter the given data pictured above into a spreadsheet and show the correct standard additions plot B.) Determine the slope and the intercept of the line C.)Determine the standard deviation of the slope and the intercept. D.) Calculate the concentration of Hg(II) in the…
3) A compound, X, with a molecular weight of 292.16 g/mol was dissolved in a 5.00 mL volumetric flask and diluted to the mark. A 1.00 mL aliquot of this was withdrawn, placed into a 10.00 mL volumetric flask and diluted to the mark. The absorbance at 340 nm was 0.427 in a 1.000 cm cuvette. The molar absorptivity of X at 340 nm is 6130 M-¹ cm-¹ a) Using Beer's Law, what is the [X] (concentration in molarity) for the diluted solution in the cuvet ? mmmmm b) Use the dilution equation (McVc = MdVa) to calculate [X] in the original 5.00 mL flask. c) How many milligrams of X was used in the original solution ? (Find total moles X using original concentration and original volume - and then use the molar mass to go from moles to grams and then milligrams)
1) Pd (II) with Au (III) methiomeprazine (C19 H24 N2 S2) complex. A mixture of their complexes is being analyzed. Pd (II) complex at 480 nm, The measurement results for Au (II) with a maximum peak at 635 nm are as follows. Of this mixture What is its concentration? (b = 1) Pd(II) ɛ Au(III)ɛ 2.96x 103 A 480 nm 3.55x 103 0.533 635 nm 5.64x 102 1.45x 104 0.590
2. Two substances, A and B, have the same molar extinction coefficient at 260 nm: 1.80 x 10 dm³mol'cm'. At 340 nm A no longer absorbs, and for B EB = 6.22 x 103 dm³mol'cm'. A solution of the two has an absorbance of 0.215 at 340 nm and 0.850 at 260 nm in a 1 cm thick cell. Calculate the concentrations of A and B.
A compound with a molecular weight of 229.61 g/mol was dissolved in 50.0 mL of water. 1.00 mL of this solution was placed in a 10.0 mL flask and diluted to the mark. The absorbance of this diluted solution at 510 nm was 0.472 in a 1.000 cm cuvet. The molar absorptivity of the compound, at 510 nm, is 6,310 M-1 cm-1. Calculate the concentration of the compound in the initial 50.0 mL solution. A. 1.50 x 10-5 M B. 7.48 x 10-5 M C. 7.48 x 10-4 M D. 7.48 x 10-6 M
4. The total absorbance of a solution is the sum of the absorbances of all different materials present. AT = Aa + Ap + Ac .. etc. At 427 nm, the molar absorbtivity of Yellow #5 is 2.73×10°M-1-cm1. At 427 nm the molar absorptivity of Red 40 is 7.49x103 M-1.cm-1. Calculate the total absorbance of a solution in which the concentration of Yellow #5 = 2.00x10-5 M and the concentration of Red 40 = 4.80x10-5 M, when measured at 427 nm in a 1.00 cm cell.

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