Question

. A 10.00-ml aliquot of vinegar requires 16.95 mL of the 0.4874 M standa point of the titration. Calculate the molarity of HC,H,O, in vinegar. 4 M standardized NaOH solution to reach the end w swer to Problem #2, calculate the "percent by 3. Assume the density of the vinegar is 1.005 g/mL. Using your answer to Problem #2, calculate th mass" of HC,H,O2 (molar mass = 60.05 g/mol) in vinegar.

Answer 1

no of moles of NaOH = molarity * volume in L

                                   = 0.4874*0.01695   = 0.00826143 moles

HC2H3O2(aq) + NaOH(aq) -------------> NaC2H3O2(aq) + H2O(l)

1 mole of NaOH react with 1 mole of CH3COOH

0.00826143 moles of NaOH react with 0.00826143 moles of CH3COOH

no of moles of HC2H3O2   = 0.00826143 moles

molarity of HC2H3O2   = no of moles /volume in L

                                   = 0.00826143/0.01   = 0.826143M

molarity of HC2H3O2 in vinegar = 0.826143M

3.

mass of HC2H3O2   = no of moles * gram molar mass

                               = 0.00826143*60.05 = 0.496g

mass of vinegar = volume * density

                          = 10*1.005 = 10.05g

percent by mass = 0.496*100/10.05   = 4.94% >>>>answer