Please help me! Showing the work will be much appreciated! Thank you! Please ans

Question

Please help me! Showing the work will be much appreciated! Thank you!

Please answer the Pre-Lab questions listed. Thank you.

31 Experiment 5: Determination of Caffeine Exp 5: Quantitative Determination of Caffeine Content in "X" Objectives Design an experimental procedure to determine the concentration of caffeine in a commercial product using techniques leaned from two previous experiments Learn rules of scientific writing- prepare a lab report Introduction Beer's Law (Abs.-e*b*c, where Abs, absorbance value at wavelength absorptivity constant, b path length of cuvette (usually 1 cm), c - concentration) tells us that the absorbance of light by a compound at a certain wavelength is proportional to the concentration of that compound, Beer's law is a “limiting law" and only applies to quite dilute solutions. Therefore, when using Beer's law one wants to keep all the absorbance values between 0-1 absorbance units, ideally 0-0.7 abs units. As long as one knows the units they are working with, the concentration units can be anything and will dictate what the units on are. The overall objective is to use a class generated calibration curve to determine how the extraction of caffeine from a bag of tea is dependent on the extraction technique used by the tea drinker. Pre-Lab Questions 1. What is the reported molar absorptivity constant () at 270 nm for caffeine dissolved in water? List your reference source and make sure to include the units in your answer. 2. Convert the molar absorptivity into units of (mg/ml) cm1. Show your work.

Explanation / Answer

Ans. Molar absorptivity constant (e) of caffeine at 270 nm = 9900 M-1 cm-1

# Molarity = Moles of caffeine/ L

Molar mass of caffeine = 194.19 g/ mol

# Now,

            9900 M-1 cm-1 = 9900 (mol/ L)-1 / cm

                                    = (9900 L/ 1 mol) / cm                     ; [1 mol caffeine = 194.19 g]

                                    = (9900 L/ 194.19 g) / cm

                                    = (9900 L/ 194190 mg) / cm                       ; [1 g = 1000 mg]

                                    = (9900000 mL/ 194190 mg) / cm             ; [1 L = 1000 mL]

                                    = (50.981 mL/ mg) / cm

                                    = 50.981 (mg/ mL)-1 cm-1

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