HC2H3O2(aq) + NaOH(aq)-> H2O(l) + NaC2H3O2
no of moles of NaOH = molarity of NaOH * volume in L
                           Â
= 0.1031*0.0349 = 0.003598 moles
From balanced equation 1 mole of CH3COOH react with 1 mole of
NaOH
no of moles of NaOH = no of moles of CH3COOH
no of moles of CH3COOH = 0.003598 moles
2.52 ml vinegar contains 0.003598 moles
mass of CH3COOH = no of moles * G.M.Wt
                         Â
= 0.003598*60 = 0.2159 gm
                         Â
= 0.2159*1000/2.52 = 85.67g/L
1 US quart = 0.95L
                           Â
= 85.67*0.95 =81.38gm acetic acid ina 1qt sample of vinegar
