posted by mathew
10 ml sample of vinegar an aqueous solution 0f acetic acid( HC2H3O2) is titrated with 0.5062 M and 16.58 ml is required to reach equivalence point what is the molarity of the acetic acid
b. if the density of vinegar is 1.006 g/cm3 what is the mass percent of acetic acid of vinegar
HAc + NaOH ==> NaAc + H2O
mols NaOH = M x L = estimated 0.008 but you need a more accurate answer for this and all of the other estimates that follow.
From the equation.
0.008 mols NaOH = 0.008 mols HAc
M = mols/L = 0.008/0.010L = about 0.8M
g HAc = mols x molar mass = approx 0.5g
mass of 10 mL = 1.006g/mL x 10 mL =?
mass% = g solute/g solution = about 5%?