If 50 mL of 0.5 M oxalic acid is required to neutralise 25 mL of NaOH solution, the amount of NaOH in 50 mL of given NaOH solution is _____

Oxalic Acid NaOH Neutralisation Numerical | JEE Mains PYQ

Q. If 50 mL of 0.5 M oxalic acid is required to neutralise 25 mL of NaOH solution, the amount of NaOH in 50 mL of given NaOH solution is ______ g.

Explanation

Step 1: Write balanced chemical reaction

H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O

Step 2: Calculate moles of oxalic acid

Moles = M × V = 0.5 × 50/1000 = 0.025 mol

Step 3: Calculate moles of NaOH

From reaction, 1 mol oxalic acid reacts with 2 mol NaOH
Moles of NaOH = 2 × 0.025 = 0.05 mol (present in 25 mL)

Step 4: Calculate moles in 50 mL NaOH solution

Moles in 50 mL = 2 × 0.05 = 0.10 mol

Step 5: Calculate mass of NaOH

Molar mass of NaOH = 40 g mol⁻¹
Mass = 0.10 × 40 = 4 g

Correct Answer = 4 g

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