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assuming the density of a 5% acetic acid solution is 1.0 g/mL, determine the volume of the acetic acid solution necessary to neutralize 25.0mL of 0.10 M Na)H. Also record this calculation on your Report Sheet.

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    5% acetic acid solution means 5 g acetic acid in 100 g soln.
    If the density of the soln is 1 g/mL, then the soln has a volume of 100 mL and the acetic solution is 50 g/1000 mL.
    Convert 50 g CH3COOH to moles and you will have moles/L = M.

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