Calculating Mass of Subsample of Tobacco & Barium Hydroxide

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SUMMARY

The discussion centers on calculating the mass of tobacco in a 25 mL aliquot from a solution containing 2.0830 grams of tobacco and 1 gram of barium hydroxide mixed in a chloroform:toluene solvent. The initial assumption that the total volume remains 100 mL is incorrect, as the addition of solids increases the final volume slightly. Therefore, the mass of tobacco in the 25 mL sample cannot be accurately determined without knowing the exact final volume of the solution.

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waqaszeb
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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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