Liters x normality x equivalent weight = grams. I prefer to use
mL x N x milliequivalent weight = grams.
31.3 mL x 0.15 N x 0.060 = 0.282 g acetic acid
% w/w acetic acid = (0.282 g/6.00)*100 = ?% acetic acid in the sample.
The number of equivalents is L x N = 0.0313 x 0.15 = ?
When I was in school we worked ALL of our quant chem problems with milliquivalents and not with molarity. Normality is a disappearing word in chemistry I'm sorry to say.
A total of 31.3 ml of 0.15 N NaOH was required to reach a phenolphthalein endpoint in titrating 6.00 g vinegar. Calculate the number of equivalents of acetic acid in vinegar. Calculate the number of grams of acetic acid present in the vinegar. Calculate the percent by weight of acetic acid in vinegar.
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