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? B. Calculate the concentration of the reducing sugar in the 50.0 mL sample in mg/mL. Apply Beer-Lambert’s law: A=bc

Macroscale and Microscale Organic Experiments

7th Edition

ISBN:9781305577190

Author:Kenneth L. Williamson, Katherine M. Masters

Publisher:Kenneth L. Williamson, Katherine M. Masters

Chapter7: Extraction

Section: Chapter Questions

Problem 1Q

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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=ɛbc. Computations: Answer:
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. What is the reducing sugar concentration (mg/mL) in the juice concentrate? The equation of the line was plotted to be: y = 1.6656x - 0.0885 0.578 mg/mL 577.9 mg/mL 57.7 mg/mL 5.78 mg/mL
Below is the procedure described for determination of paracetamol according to BP: "Weigh and powder 20 tablets. Add a quantity of powder equivalent to about 0.35 g paracetamol to 50 ml of 0.1 M NaOH, dilute with water to 200 ml and shake for 15 min. Add sufficient water to make up 250 ml. Filter the solution and dilute 5 ml of the filtrate to 100 ml with water. Dilute further 10 ml to 100 ml. Measure the absorbance at 257 nm. The A," following is correct? 715 at 257 nm". Which of the O a. An analyst can take 0.37g of the powdered tablets instead of 0.35. O b. An analyst can replace NaOH with HCI OC. An analyst can skip the shaking step O d. B andC O e. All of the above
Below is the procedure described for determination of paracetamol according to BP: "Weigh and powder 20 tablets. Add a quantity of powder equivalent to about 0.35 g paracetamol to 50 ml of 0.1 M NaOH, dilute with water to 200 ml and shake for 15 min. Add sufficient water to make up 250 ml. Filter the solution and dilute 5 ml of the filtrate to 100 ml with water. Dilute further 10 ml to 100 ml. Measure the absorbance at 257 nm. The A = 715 at 257 nm". Which of the following is correct? O a. An analyst can take 0.37g of the powdered tablets instead of 0.35. O b. An analyst can replace NAOH with HCI C. An analyst can skip the shaking step d. B and C e. All of the above
Below is the procedure described for determination of paracetamol according to BP: "Weigh and powder 20 tablets. Add a quantity of powder equivalent to about 0.35 g paracetamol to 50 ml of 0.1 M NaOH, dilute with water to 200 ml and shake for 15 min. Add sufficient water to make up 250 ml. Filter the solution and dilute 5 ml of the filtrate to 100 ml with water. Dilute further 10 ml to 100 ml. Measure the absorbance at 257 nm. The A, = 715 at 257 nm". Which of the following is correct? O a. An analyst can take 0.37g of the powdered tablets instead of 0.35. O b. An analyst can replace NaOH with HCl O c. An analyst can skip the shaking step O d. B andC O e. All of the above
Your HRP sample had a concentration of 2.0 mg/mL and you added 10 uL to a tube containing 10 uL of 2X Sample Buffer. You mixed this well and then heated at 75C for 5 min. Then you added 10 uL of this solution to a well of a gel. How much protein did you add to that well?
During an assay of paracetamol by UV an analyst prepares a series of calibration solutions containing approximately 3-15 mcg/mL paracetamol as follows: 140.1 mg of pure paracetamol was added to a 200 mL volumetric flask. 0.1 M NaOH (50 mL) was added, the mixture well shaken and made up to volume with deionized water 10 mL of the above solution was diluted to 100 mL 1)What is the concentration of the final diluted solution in mcg/mL?(1 decimal) 2) What volume of the diluted solution is required to prepare 100 mL of a 8 mcg/mL standard solution? (1 decimal)
Tonic water (20.0 mL) was added into each of two volumetric flasks (100 mL). The first flask was made up to volume with deionised water and produced an absorbance of 0.200. To the second flask was added an aliquot (20.0 mL) of a quinine standard solution (20.0 ppm) and after being made up to volume this produced an absorbance of 0.400. What is the concentration of quinine in the tonic water? a. 8.00 ppm b. 800 ppm С. 40.0 ppm d. 400 ppm е. 20.0 ppm
Assume that 8.7 mg of commercial 2,4 DNPH were dissolved in 100 mL acetonitrile solution. 300 μL of this solution were diluted to 10 mL and the absorbance at 360 nm was found to be 0.120 nm. Calculate the % 2,4 DNPH in the commercial product.
A 5.00-mL aliquot of a solution that contains 3.05 ppm Ni2+ is treated with an appropriate excess of 2,3-quinoxalinedithiol and diluted to 50.0 mL. The molar absorptivity of a Ni2+- 2,3-quinoxalinedithiol solution at 510 nm is 5520 L mol-1 cm-1. What is the absorbance of the above diluted Ni2+- 2,3-quinoxalinedithiol solution at 510 nm in a 1.00-cm cell?
A 10.00mL sample of alcoholic ethyl acetate was diluted to 100.00 mL. 20.00 mL was aliquoted and mixed with 40.00 mL of 0.04672 M KOH. The resulting mixture was heated for 2 hours. CH3COOC2H5 + OH- → CH3COO- + C2H5OH After cooling, the excess OH was back titrated with 3.41 mL of 0.05042 M H2SO4. Answer the following: Calculate the number of moles of OH- that reacted with ethyl acetate. Calculate the number of moles of ethyl acetate in the 20.00 mL solution. What is the mass of ethyl acetate (FW=88.11 g/mol) in the original 10.00 mL sample?
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