If 50 mL of 0.5 M oxalic acid is required to neutralise 25 mL of NaOH — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

Equivalent of Oxalic acid = Equivalents of NaOH 50 × 0.5 × 2 = 25 × M × 1 MNaOH = 2M WNaOH in 50ml = 2 × 50 × 40 × 10–3 g = 4g