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Chemistry

Pure acetic acid, known as glacial acetic acid, is a liquid with a density of 1.049 g/mL at 25°C. Calculate the molarity of a solution of acetic acid made by dissolving 35.00 mL of glacial acetic acid at 25°C in enough water to make 400.0 mL of solution.

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Matt

  1. Gms of HOAc = (35.0ml)(1.049g/ml) = 36.715gms HOAc
    moles HOAc = (36.715 g)/(60.05 g/mol) = 0.6114 mole
    Molarity = moles/Liters
    = (0.6114-mole/0.400-L)
    =1.53-Molar

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    Olereb48

  2. First determine the M of the glacial acetic acid.
    1000 mL x 1.049 g/mL x 1.00 x (1/molar mass acetic acid) = ?M

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    DrBob222

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