Part B: Calculate the approximate amount of vinegar that reacts with 25 mL of 0.1 M NaOH.
| Trial 1 | Trial 2 | Trial 3 | |
|---|---|---|---|
| Mass of flask (g) | 151.080 | 152.200 | 149.551 |
| Mass of flask + vinegar sample (g) | 154.050 | 155.302 | 152.652 |
| Mass of vinegar sample (g) | |||
| Initial burette reading of NaOH (mL) | 0.00 | 0.00 | 0.00 |
| Final burette reading of NaOH (mL) | 24.81 | 25.15 | 24.70 |
| Volume of NaOH dispensed (mL) |
| Trial 1 | Trial 2 | Trial 3 | |
|---|---|---|---|
| (NaOH) (M), average from Part A | |||
| Moles of NaOH dispensed | |||
| Moles of acetic acid in vinegar sample | |||
| Mass of acetic acid in vinegar sample (g) | |||
| Percent by mass of acetic acid in vinegar sample (%) |
Average percent by mass of acetic acid in vinegar sample (%):
Standard deviation (percent by mass of acetic acid in vinegar sample (%)):