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If one wished to obtain 0.15 moles of glacial acetic acid, how many milliliters of glacial acetic acid should be obtained? Enter only the number with two significant figures.

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Answer:

8.6 mL

Explanation:

Glacial acetic acid (or anhydrous acetic acid) can be thought of as pure acetic acid, and has a density of 1.05 g/cm³ (or g/mL).

To answer this problem first we convert moles of acetic acid (CH₃CO₂H) to grams, using its molar mass (60 g/mol):

  • 0.15 mol acetic acid * 60 g/mol = 9.0 g acetic acid.

Now we convert grams of acetic acid to mL, using its density:

  • 9.0 g ÷ 1.05 g/mL = 8.6 mL