Question

A student needs to prepare 100.0 mL of vitamin C solution according to the directions in Part 1 of the experiment. S/he weighs out 0.0471 g vitamin C (molar mass = 176.124 g/mol) and dissolves it in water to give 100.0 ml solution. S/he pipets 10.00 mL of the solution into a beaker and adds 20 mL of water and 10 mL of buffer. (a) How many mg of vitamin C are in the beaker? 4.9 4.71 mg (b) Suppose 22.9 mL of DCP are required to reach the endpoint of the titration. How many mg of vitamin C react with each mL of DCP? X mg

Answer 1

a) 4.71 mg

we dissolve 0.0471 g (47.1 mg) vitamin C in water to form 100 ml solution.

​​​​​​(since, 1 g = 1000 mg)

ie. 100 ml solution contain 47.1 mg Vitamin C

now we take 10 ml of this vitamin C solution in beaker.

since, 100 ml solution = 47.1 mg vitamin C.

therefore,

10 ml solution =
10(ml)x 47.1 mg) = 4.71 (mg) 100 ml)
vitamun C.

Therefore, 4.71 mg vitamin C are present in beaker.

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b) 0.2056 mg

22.9 ml of DCP required to end point of the titration.

i.e. 22.9 ml DCP = 4.71 mg vitamin C

therefore,

1 ml of DCP =
1(ml) x 4.71(mg) = 0.2057mg) 22.9(ml)
vitamin C.

Therefore, 0.2057 mg of vitamin C react with each ml of DCP.

i.e. 0.2057 mg/ml