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e.margaret21 e.margaret21
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10 years ago
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)
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padrepadre
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10 years ago
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
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