10.00mL 5% acidity vinegar was used titrate with NaOH whosemolarity is 0.1030M.
Exp. # | Concentration of NaOH | Initial volume of NaOH | Final volume of NaOH | Volume of NaOH used for titration |
1 | 0.1030M | 5.50mL | 21.65mL | 16.15mL |
2 | 0.1030M | 3.60mL | 19.00mL | 15.40mL |
3 | 0.1030M | 19.00mL | 35.91mL | 16.91mL |
The vinegar was diluted to make 100.00mL final volume beforetrtration. If the molar mass of acetic acid is 60.05g/mole anddensity is 1.00 g/mL. Calculate the mass percent of acetic acid inthe original vinegar solution.
