Chem 115 Laboratory

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Experiment 4 - Pre-Lab Assessment Video
Vitamin Analysis
You have been given a jar of red fruit punch and been asked to determine the concentration of the red dye in the fruit punch. It is known that a 0.00010 M standard solution of this red dye absorbs green light (the complementary color) with an absorbance value of 0.77.
You decide to measure the absorbance of the fruit punch, but your colorimeter only works properly for absorbance values between 0.1 and 0.9 (which is common). As a result, the absorbance value for your fruit punch must be between 0.1 and 0.9 or your data will be worthless because you have either (i) exceeded the detection limit of your colorimeter or (ii) reached a point where concentration and absorbance are no longer directly proportional. Your fruit punch is dark red in color and you deduce that your fruit punch is too concentrated and will have a green light absorbance much greater than 0.9 (i.e. out of range of your colorimeter). Therefore, you will have to do a serial dilution. You take 5.0 mL of the fruit punch and dilute it to 100.0 mL using distilled water. You call this new solution Dilution #1. You then take 5.0 mL of Dilution #1 and dilute it to 100.0mL using distilled water; you call this Dilution #2. The solution in Dilution #2 absorbs green light with an absorbance value of 0.21.
1. Using the procedure you read about in the introduction section, calculate the concentration of red dye in the Dilution #2 solution.
2. Calculate the concentration of red dye in the Dilution #1 solution.
3. Calculate the concentration of red dye in the original fruit punch.