A test for vitamin C (ascorbic acid, C6H8O6) is based on the reaction of the vitamin with iodine:

Found this online!

The balanced equation shows that the molar ratio between ascorbic acid and I2 is one to one.

Moles I2 = Molarity times volume in Liters

Moles I2 .0164 molar times .0118 liters = l.93 x l0^-4 moles I2 and also moles ascorbic acid in the sample.

Molarity of acid = moles acid over volume in liters = l.93 x l0^4 moles over .025 liters =7.7 x l0^-3rd molar acid

c 60mg = .060 grams. Moles vitamin C = .060 grams over 176 grams/mole = 3.4 x l0^-4 moles.

From previous work we know that 25 ml of juice contains l.93 x l0^-4 moles vitamin C

so we can scale up or down.


25 ml juice over l.93 x l0^-4 moles = 3.4 x l0^-4 moles vitamin C needed over X ml juice

Cross multiply and solve for X ml juice.

I get 44.2 ml. Then divide by 29.6 ml/fluid ounce = 1.49 fluid ounces if juice