Here is a sample calculation:

Data:

Mass of 50.0 mL vinegar: 51.0 g
Number of mL of vinegar to reach endpoint: 23.0 mL

Calculations:

Density of vinegar:

51.0 g/ 50.0 mL = 1.02 g/mL

Number of grams of vinegar added:

(1.02 g/mL) x (23.0 mL) = 23.5 g

Mass of C₂H₄O₂ added:

(23.5 g) x (0.0500) = 1.18 g

Molecular Mass of C₂H₄O₂ = 60.0 amu

Number of moles of C₂H₄O₂ added:

(1.18 g/1) x (1 mole/60.0 grams) = 0.0197 moles

Number of moles of ammonia that exist in solution:

(0.0197 moles C₂H₄O₂ /1)x(1 mole ammonia/1 mole C₂H₄O₂) =
0.0197 moles ammonia

Concentration of ammonia:

0.0197 moles ammonia / 0.0100 L = 1.97 M