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Posted by on Tuesday, January 25, 2011 at 1:47pm.

Assuming the density of acetic acid solution is 1.0 g/mL, determine the volume of the acetic acid solution necessary to neutralize 25.0 mL of 0.10 M NaOH. Record this calculation on your report sheet

  • CHM - , Tuesday, January 25, 2011 at 3:19pm

    moles acid needed= .025*.1 moles.

    grams acetic acid= molesacid*molmassAcid

    Now at this point, I am stupified on the density. The given density is the same as water. So how can it be acetic acid solution? Vinegar has a density of about 1.01g/ml, that is 5 percent acetic acid.

    I am not certain you can work this without better data.

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