A 25 mL sample of 0.1 M ethyl acetate (C4H8O2) requires how many grams of NaOH for saponification? (Molar mass NaOH=40 g/mol)
Q1. A 25 mL sample of 0.1 M ethyl acetate (C4H8O2) requires how many grams of NaOH for saponification? (Molar mass NaOH=40 g/mol)
Answer: 0.40 g
Explanation: Moles ethyl acetate = 0.1 × 0.025 = 0.0025 mol. 1:1 ratio. Mass NaOH = 0.0025 × 40 = 0.1 g. Option B doubles for a 2:1 reaction.