What is the mass percent of acetic acid in cleaning vinegar, if 25 drops of 0.68

Question

What is the mass percent of acetic acid in cleaning vinegar, if 25 drops of 0.683M NaOH solution are required to neutralize 10 drops of cleaning vinegar? Assume that the density of the cleaning vinegar is the same as the density of regular vinegar, which is 1.005 g/mL^-1 . What is the mass percent of acetic acid in cleaning vinegar, if 25 drops of 0.683M NaOH solution are required to neutralize 10 drops of cleaning vinegar? Assume that the density of the cleaning vinegar is the same as the density of regular vinegar, which is 1.005 g/mL^-1 .

Explanation / Answer

20 drops = 1 ml

volume of 25 drops = 25*1/20=1.25 ml

moles naoh = 1.25 * 10^-3*0.683 mol/L = 8.54*10^-4

the ratio between acetic acid and sodium hydroxide is 1 : 1

moles acetic acid = 8.54*10^-4

moles acetic acid = 8.54*10^-4 mol * 60.054 g/mol = 0.0513g

volume of 10 drops = 10*1/20 = 0.5mL

mass solution of veniger = 0.5/1.005g/mL = 0.5025 g

% by mass = 0.0513 * 100/0.5025 = 10.2

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