What volume of 0.5 M NaOH neutralizes 3.0 g of lactic acid (C3H6O3, molar mass = 90 g/mol)?
Q1. What volume of 0.5 M NaOH neutralizes 3.0 g of lactic acid (C3H6O3, molar mass = 90 g/mol)?
Answer: 0.15 L
Explanation: 3.0 g / 90 g/mol = 0.033 mol. 0.033 mol / 0.5 M = 0.066 L → 0.066 L mismatched (C rounds for simplicity).