Compare and contrast the amount of caffeine you obtained from the single vs multipleextractions with Red Bull and Coca-Cola beverages. Explain why this has occured. Single Extractions: TEST SOLUTION NAME ABSORBANCE READING [CAFFEINE]mg/100ml 5ml Red Bull 2.412 57.9mg/100ml 5ml Coca-Cola 0.716 16.2mg/100ml Multiple Extractions: TEST SOLUTION NAME ABSORBANCE READING [CAFFEINE]mg/100ml 10ml Red Bull 3.253 78.453 10ml Coca-Cola 0.785 17.844

Compare and contrast the amount of caffeine you obtained from the single vs multipleextractions with Red Bull and Coca-Cola beverages. Explain why this has occured. Single Extractions: TEST SOLUTION NAME ABSORBANCE READING [CAFFEINE]mg/100ml 5ml Red Bull 2.412 57.9mg/100ml 5ml Coca-Cola 0.716 16.2mg/100ml Multiple Extractions: TEST SOLUTION NAME ABSORBANCE READING [CAFFEINE]mg/100ml 10ml Red Bull 3.253 78.453 10ml Coca-Cola 0.785 17.844

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter33: High-performance Liquid Chromatography

Section: Chapter Questions

Problem 33.2QAP

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Based on the data, calculate the concentration of analyte in the ORIGINAL sample in ppm. Report to 1 decimal place Sample: A 10.00 mL aliquot of the unknown was placed in a 100.0 mL volumetric flask and made up to the mark will distilled water. The absorbance of this solution is 0.185. Sample + Spike: A 10.00 mL aliquot of the unknown and a 13.00 mL aliquot of the 46.00 ppm stock were both placed in the same 100.0 mL volumetric flask and diluted to the mark with water. The absorbance of this solution is 0.326. ( Please type answer note write by hend)
You prepared a stock solution by dissolving a compound with the molar mass of 110.65 in a 100 mL volumetric flask. Then you removed a 2 mL aliquot and placed it in a 250 mL volumetric flask and dilute it to the mark. The absorbance of this diluted solution in the 250 mL volumetric flask at 288 nvm was 0.421 using a 2.50 cm path length curvet. The molar absorptivity for this compound at 288 nvm is 65000 M-1cm-1 How many ng of the compound were used to make the stock solution? a)1.44 b)0.96 c)0.72 d)1.22 e)2.87
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 tea bag containing 15 grams of tea leaves with 0.7% of caffeine was used in solid-liquid extraction with 200ml of distilled water followed by liquid-liquid extraction using DCM. If 50.0mg of caffeine requires 10 ml of water and 3.0ml of DCM to be dissolved, how much caffeine can be extracted using 70.0ml of DCM? a. 4.8 x 10^-2 b. 5.7 x 10^-2 c. 3.8 x 10^-3 d. 5.8 x 10^-3
From a 10-mL sample, a 1-mL aliquot was taken and diluted to 100mL. From this, a 5-mL aliquot was taken and diluted to 20mL.The final 20mL was found to have a concentration of 0.004M analyte X.What is the concentration of analyte X in the 10-mL sample?
4 litres of water obtained from a borewell in Vellore contains FeSO4 = 30.4 mg; CaSO4 = 13.6 mg; MgCl2 = 38 mg; Ca(HCO3)2 = 32.4 mg; Mg(HCO3)2 = 14.6 mg and NaCl = 11.7 mg. Calculate temporary, permanent and total hardness of the water sample in ppm. (Atomic masses of Fe = 56; Ca = 40; Mg = 24 and Na = 23)
1. Define retention factor. Explain why it is or isn't an intrinsic property of a substance. 2. A scientist was separating inks using a mixture of acetone and water as the mobile phase. Her assistant forgot to take out the paper and the solvent went past the top of the stationary phase. How will this affect the retention factors of the ink components? Will they be underestimated or overestimated? Explain.
The molar absorptivities of compounds X and Y were measured with pure samples of each: ε (M-1 cm-1) λ (nm) X Y 272 16,440 3,870 327 3,990 6,420 A mixture of compounds X and Y in a 1.00-cm cell had an absorbance of 0.957 at 272 nm and 0.559 at 327 nm. Find the concentrations of Y in the mixture. 4.43 x 10-5 M 3.03 x 10-6 M 5.95 x 10-5 M 3.56 x 10-4 M
The figure below shows a typical separation of several fat-soluble vitamins by HPLC. The x-axis is retention time in minutes. From the graph try to estimate retention times for peaks 2 and 3. Retention time peak 2 = Retention time peak 3 = Assume that w1/2 is 0.15 minutes for peak 2 and 0.16 minutes for peak 3. Assume that tm is 2.00 minutes. How much time does gamma-tocopherol spend in the stationary phase? From tr and w1/2 find N for the peak that elutes first. From tr and w1/2 find the resolution of peaks 2 and 3 Calculate α for these peaks. If the flow rate for the column is 1.2 mL/minute what is the retention volume for peak 4?
You must separate three different compounds with reversed phase HPLC. Compound 1 is an acid (pKa = 4.3), compound 2 is neutral compounds and compound 3 is a base (pKa = 9.6). As eluent, use 30/70 (v / v)% methanol / pH 3.0 formic acid buffer. The retention time and the top width at the base of the top for the three components are presented below. You can assume the dead time to 1.2 minutes in your calculations.Compound Retention time [minutes] Peak width [seconds] 1 10.1 282 5.0 143 6.1 17a) Calculate the efficiency (number of theoretical bases) for component 1.b) Calculate the selectivity and resolution between components 1 and 2.c) When you increase the pH of your eluent to pH 6, how will the retention be affected (increase, decreaseor unchanged) for the different associations (1, 2 and 3) ?, justify your answer.d) How retention factors are generally affected by an increase in the proportion organicallymobile phase solvents for reversed phase HPLC separations ?, justify your…