Calculating Mass of Subsample of Tobacco & Barium Hydroxide

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The discussion revolves around calculating the mass of tobacco in a 25 mL aliquot from a solution created by mixing 2.0830 grams of tobacco and 1 gram of barium hydroxide in a chloroform:toluene mixture. The initial assumption was to use a straightforward proportion (2.0830 x (25ml/100 mL)) to find the mass in the aliquot. However, it was pointed out that this method is incorrect because the addition of barium hydroxide increases the total volume of the solution, resulting in a final volume that exceeds 100 mL. Consequently, the mass of tobacco in the 25 mL sample cannot be accurately calculated without knowing the final volume of the solution. To obtain a precise measurement, the solution must be adjusted to exactly 100 mL with solvent.
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Hello,

If I get 2.0830 grams of tobacco and add 1 gram of barium hyrdoxide to it. Mix it in 100 mL of chloroform:toluene mixture, I essential have a 100 mL solution right? Now I take a 25 mL aliquote from that? what will be the mass in grams of tobaco in that 25 mL sample?

I thought simply 2.0830 x (25ml/100 mL) would do but my T.A. asked me to recalculate that because I did it wrong.

Thanks.
 
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If you add these things to 100 mL of the solvent, you would have slightly more than 100 mL, so taking 25 mL would yield sample that is below 1/4 of the total. To be sure you can do it this way you would have to fill up to 100 mL with solvent.

I don't see a way to correctly estimate the sample mass with the data given, as the final volume is not known and impossible to calculate.
 
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