1 molarity of standard aa solution 055 grams 1 mol

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1) Molarity Of Standard AA solution (.055 grams)| 1 mol | = 3.125 x 10 -4 mols 176 g Final solution Volume = 100 mL = .1L [M] AA = ( mols solute ) Liters solution = ( 3.125 x 10 -4 mol ) .1 L = [.003125 M] AA 2) Molarity Of DCP Solution [M] DCP = ( mols solute ) Liters solution = ( 3.125 x 10 -5 mols ) .0264 L = [.001184 M] DCP 3) Mass Of Ascorbic Acid In Juice Sample Titrated (mg/mL) Avg. DCP solution Molarity = [1.23 x 10 -3 M]
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[1.23 x 10 -3 M]*(.011 L) = 1.35 x 10 -5 mols DCP required 1.35 x 10 -5 mols DCP | 1 mols AA | 176.13 g = 2.38 x 10 -3 g AA in titration 1 mol DCP | 1 mol AA 2.38 x 10 -3 g AA| 1000 mg | 1 = .476 mg/ml 1 g 5 mL Objective:
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To determine the different masses of ascorbic acid (vitamin C) contained in several solutions by using an oxidation-reduction ti- tration method using DCP. Procedure First the standardization of the DCP solution was performed. Reagent grade solid ascorbic acid was massed on an electric balance between (40-60 mg). Then the sample was taken and placed into a (100 mL) volumetric flask, and dissolved in (100 mL) distilled water. Then three (250 mL) erlenmeyer flasks were placed on the lab table to account for the three different titrations. Next, (10 mL) of the previ- ously prepared solution, (20 mL) distilled water, and (10 mL) pH3 buffer solution were combined and placed in each of the 250 mL erlenmeyer flasks. Next, the bur- et tube was filled with DCP and the meniscus calibrated to the (0 mL) line. After
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