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Title: A test for vitamin C (ascorbic acid, C6H8O6) is based on the reaction of the vitamin with iodine: Post by: e.margaret21 on Nov 24, 2013 A test for vitamin C (ascorbic acid, C6H8O6) is based on the reaction of the vitamin with iodine:
C6H8O6(aq) + I2(aq) → C6H6O6(aq) + HI(aq) (a) A 25.0 mL sample of juice requires 11.8 mL of a 0.0164 M I2 solution for reaction. how many moles of ascorbic acid are in the sample? (b) What is the molarity of the acid? (c) If a person wanted to consume the FDA recommended 60 mg of Vitamin C, how many ounces of juice would that person need to consume? (4 qts = 128 fluid oounces; 1L = 1.057 qt) Title: A test for vitamin C (ascorbic acid, C6H8O6) is based on the reaction of the vitamin with iodine: Post by: padre on Nov 24, 2013 Content hidden
Title: A test for vitamin C (ascorbic acid, C6H8O6) is based on the reaction of the vitamin with iodine: Post by: doubleu on Nov 24, 2013 a) moles of iodine = moles of ascorbic acid = ( 11.8 x0.0164/1000) = 0.00019352 moles ,
b) Molarity of acid = ( 0.00019352 x1000/25) = 0.00774 M, c) 60 mg vit C = 0.06/176.12 = 0.00034 moles vit C, vol of juice required = (25 x0.00034/0.00019352) = 44.01 ml , vol in ounces = ( 44.01 x1.057 x(128/4) )/1000 = 1.488 ounces |