A juice concentrate was colorimetrically assayed using Nelson’s test. One milliliter (1.00 mL) of the sample solution and various concentrations of the standard glucose solution were treated with freshly prepared Nelson’s reagent and arsenomolybdate reagent and then diluted to 10.0 mL separately in properly labeled test tubes. Absorbances at 480 nm for distilled water, glucose standard, and for the sample are 0.052, 1.702, and 0.926, respectively.

A juice concentrate was colorimetrically assayed using Nelson’s test. One milliliter (1.00 mL) of the sample solution and various concentrations of the standard glucose solution were treated with freshly prepared Nelson’s reagent and arsenomolybdate reagent and then diluted to 10.0 mL separately in properly labeled test tubes. Absorbances at 480 nm for distilled water, glucose standard, and for the sample are 0.052, 1.702, and 0.926, respectively.

EBK A SMALL SCALE APPROACH TO ORGANIC L

4th Edition

ISBN:9781305446021

Author:Lampman

Publisher:Lampman

Chapter89: Thin-layer Chromatography

Section: Chapter Questions

Problem 7P

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A 50.0 mL juice extract is colorimetrically assayed using Nelson’s test. One milliliter (1.00 mL) of the solution and the standard glucose solution (concentration: 1mg/mL) were treated with freshly prepared Nelson’s reagent and arsenomolybdate reagent and then diluted to 10.0 mL separately in properly labeled test tubes. Absorbance at 480 nm for the standard is 1.702 and for the sample, 0.926. A. What is the principle behind the assay? B. Calculate the concentration of the reducing sugar in the 50.0 mL sample in mg/mL. Apply Beer-Lambert’s law.
A 50.0 mL juice extract is colorimetrically assayed using Nelson’s test. One milliliter (1.00 mL) of the solution and the standard glucose solution (concentration: 1mg/mL) were treated with freshly prepared Nelson’s reagent and arsenomolybdate reagent and then diluted to 10.0 mL separately in properly labeled test tubes. Absorbance at 480 nm for the standard is 1.702 and for the sample, 0.926. A. What is the principle behind the assay? B. Calculate the concentration of the reducing sugar in the 50.0 mL sample in mg/mL. Apply Beer-Lambert’s law: A=bc
A 50.0 mL juice extract is colorimetrically assayed using Nelson’s test. One milliliter (1.00 mL) of the solution and the standard glucose solution (concentration: 1mg/mL) were treated with freshly prepared Nelson’s reagent and arsenomolybdate reagent and then diluted to 10.0 mL separately in properly labeled test tubes. Absorbance at 480 nm for the standard is 1.702 and for the sample, 0.926. A. What is the principle behind the assay?
A 50.0 mL juice extract is colorimetrically assayed using Nelson’s test. One milliliter (1.00 mL) of the solution and the standard glucose solution (concentration: 1mg/mL) were treated with freshly prepared Nelson’s reagent and arsenomolybdate reagent and then diluted to 10.0 mL separately in properly labeled test tubes. Absorbance at 480 nm for the standard is 1.702 and for the sample, 0.926. Calculate the concentration of the reducing sugar in the 50.0 mL sample in mg/mL. Apply Beer-Lambert’s law: A=bc.
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 5.00-mL sample of blood was treated with trichloroacetic acid to precipitate proteins. After centrifugation, the resulting solution was brought to pH 3 and extracted with two 5-mL portions of methyl isobutyl ketone containing the lead-complexing agent APCD. The extract was aspirated directly into an air/acetylene flame and yielded an absorbance of 0.569 at 283.3 nm. Five-milliliter aliquots of standard solutions containing 0.400 and 0.600 ppm of lead were treated in the same way and yielded absorbances of 0.396 and 0.599. Please calculate the concentration of lead in the sample in ppm.
A 5.00-mL sample of blood was treated with trichloroacetic acid to precipitate proteins. After centrifugation, the resulting solution was brought to pH 3 and extracted with two 5-mL portions of methyl isobutyl ketone containing the lead-complexing agent APCD. The extract was aspirated directly into an air/acetylene flame and yielded an absorbance of 0.527 at 283.3 nm. Five-milliliter aliquots of standard solutions containing 0.400 and 0.600 ppm of lead were treated in the same way and yielded absorbances of 0.396 and 0.599. Find the concentration of lead in the sample in ppm assuming that Beer’s law is followed.
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A 5.3 × 10-2 M solution is to be used to prepare a series of five 100 mL solutions of varying concentration for use in creating a calibration curve for a spectrophotometric experiment. Determine the concentrations of each of the five dilutions if 10 mL aliquots of the current solution is used for V1.
The determination of iron in wines was carried out by atomic absorption spectrophotometry, using the standard addition method. To do this, 10 mL of wine were placed in five 50 mL volumetric flasks; then 0, 5, 10, 15 and 20 mL of a 10 ppm iron standard solution were added to each of them. Next, they were gauged with distilled water and the absorbance was measured, obtaining the values of 0.040, 0.062, 0.081, 0.102 and 0.125 respectively. Calculate the regression parameters and obtain the iron concentration in the wine, in µg / mL.
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You are given 2 ml of a stock solution of phenol red at a concentration of 258 mg/ml and 20 ml of 0.5 potassium phosphate buffer pH 7. You need 100 ml of a solution containing 50 mM potassium phosphate buffer pH 7 and 87 ug/ml phenol red to do a spectrophotometric analysis. How would you make this new solution?