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A professor found that there are 109 drops of glacial acetic acid in 2.00 mL. Calculate the number of drops of glacial acetic acid, HC2H3O2 required to make 155 ml of 0.25 M acetic acid solution

Can anyone walk me through the steps and provide an answer. Ive tried everything and it always ends up incorrect!


    The problem may well be the ambiguity in the molarity of glacial acetic acid. The molarity depends upon the density and percent composition.
    mLacid x M acid = 155 x 0.25
    IF we use 100% for the glacial acetic acid (I see 98-100% on the bottles) and density of 1.049 g/mL, then the calculated molarity of glacial acetic acid is 17.4M.)
    So you need (155 x 0.25/17.4) = 2.227 mL which you would round to 2.23 mL.
    Then 109 drops x (2.23 mL/2.00 mL) = ?
    drops. You can adjust the numbers depending upon the molarity of the glacial acetic acid you are using.


    Thank you so much!

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