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 - bobpursley, 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.