Homework Answers
one mole NaOH is neutralized by one mole CH3COOH.
25.47 mL of 0.1012 M NaOH = volume of solution in L * molarity = 0.02547 L * 0.1012 mole / L = 2.5776 * 10^-3 mole.
thus
25.00-mL portion of the dilute solution of acetic acid contains = 2.5776 * 10^-3 mole acetic acid.
250 ml solution contains = 2.5776 * 10^-3 mole * 250 ml / 25.00 ml = 2.5776 * 10^-2 mole acetic acid.
thus
50.00 ml sample of acetic acid contains 2.5776 * 10^-2 mole acetic acid.
mass of acetic acid = mole * molar mass = 2.5776 * 10^-2 mole * 60.05 g / mole = 1.548 g
thus
% (w/v) of acetic acid in the sample = 1.548 g * 100 / 50.00 ml = 3.096 % (answer
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