Question

Assuming the density of vinegar is 1.00 g/mL, determine the average percentage by mass of acetic acid in vinegar. Molarity of NaOH = .3 Volume of NaOH = 29.2 Mass of Vinegar = .205 mol

Answer 1

Dvnegar = 1 g/mL

average % of acetic acid in vinegar:

mol of NaOH = MV = 0.3*29.2 = 8.76 mmol of NaOH

mmol of acetic acid = mol of NaOH = 8.76 mmol of acid

so

MW of acetic acid = 60 g/mol

mass = mol*MW = (8.76*10^-3)(60) = 0.5256 g of actic acid

So.. for % mass:

% mass of acid = mass of acid / total mass * 100%

total mass = mass of vinegar... note that 0.205 mol is not a measurement of mass, so assume a sample of 10.205 g of sample...

so

% sample = 0.5256 /10.205 *100 = 5.15041 % (i.e. 5.15% is acetica acid in vinegar mix)