Answer:
7.68%
Step-by-step explanation:
Let's consider the neutralization reaction between acetic acid and sodium hydroxide.
HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O
25.56 mL of 0.200 M NaOH reacted. The reacting moles of NaOH are:
0.02556 L × 0.200 mol/L = 5.11 × 10⁻³ mol
The molar ratio of HC₂H₃O₂ to NaOH is 1:1. The reacting moles of HC₂H₃O₂ are 5.11 × 10⁻³ moles.
The molar mass of acetic acid is 60.05 g/mol. The mass corresponding to 5.11 × 10⁻³ moles is:
5.11 × 10⁻³ mol × 60.05 g/mol = 0.307 g
0.307 g of acetic acid are in 4.00 g of vinegar. The mass percent of HC₂H₃O₂ is:
0.307 g/4.00 g × 100% = 7.68%