How many milligrams of Vitamin C were in the original tablet?

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The discussion revolves around calculating the amount of ascorbic acid in a vitamin C tablet using titration data. The titration involved a 0.0151M triiodide solution, and the stoichiometry of the reaction is one-to-one. The calculations show that 9.40 mL of titrant corresponds to 1.41 x 10^-4 moles of ascorbic acid in a 25 mL aliquot, leading to a total of 248.3 mg in the original tablet after scaling up to the full 250 mL solution. Concerns about the appropriate use of the M1V1=M2V2 equation and significant figures were also discussed, emphasizing the importance of accuracy in intermediate calculations. The final result should reflect three significant figures based on the provided data.
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Homework Statement



You ground your vitamin C tablet and dissolved it in 250. mL of water in a volumetric flask. You used a 25.0 mL aliquot of this solution for your titration. If it takes 9.40 milliliters of your 0.0151M titrant, how many milligrams of ascorbic acid (MW 176.12) were in the original tablet?

Homework Equations



M1V1=M2V2

The Attempt at a Solution



0.015 M x 0.0940 L = M 2 V2

I'm not sure I'm using the right equation.
 
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If you are not told what the titrant is, question doesn't make sense. And M1V1=M2V2 is not a "relevant equation"when it comes to titrations, it is a specific case which may, or may not work.

Compare

http://www.titrations.info/titration-calculation

for a very basic example of the titration where this equation doesn't work.
 
The titrant of this experiment is triiodide, but it didn't have that info in the question.
 
What is the reaction? Or more precisely - what is stoichiometry of the reaction? In what molar ratio do ascorbic acid react with iodine (or triiodide, it is the same in this context)?
 
Reaction: C6H8O6 + I3- +H2O ↔ C6H6O6 +3I- +2H3O+
Stoichiometry is one to one.

So 9.40 mL of 0.015 M titrant contains 1.41x10^-4 mol I3-.
Since it's a 1:1, 25mL aliquot contains 1.41x10^-4 moles vitamin C.
There are 10 25mL in 250 mL so 10 x 1.41x10^-4 moles = 1.41 x 10^-3 moles
1.41x10^-3 x 176.12 = 0.2483292g = 248.3 mg.

Are my sig figs correct?
 
Don't round down intermediate results - that is, round them down for reporting, but use as many digits as you have for further calculations.

Your final result is slightly off (I suppose that's because of this rounding down). Both volume and concentration of the titrant are given with three significant digits, so your final answer should have three digits as well.
 
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