Asked by Ashley
A student completed the standardization procedure & found that the value for her dropper was 0.133 mg Vitamin C/dp l2. The student then selected a juice & titrated it with the iodine from the dropper bottle. The sample turned blue with the addition of the 19th drop. How many mg of Vitamin C were in the juice sample? Show work & give answer in the tenth place.
Answers
Answered by
DrBob222
0.133 mg C/dp x 19 dps = ? mg vit C.
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