10.0 mL of 0.05 M KMnO solution was consumed in a titration with 10.0 mL of given oxalic acid dihydr — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

**Step 1: Write the balanced redox equation** In acidic medium: 2KMnO₄ + 5H₂C₂O₄ → K₂SO₄ + 2MnSO₄ + 10CO₂ + 8H₂O Mole ratio: 2 mol KMnO₄ : 5 mol H₂C₂O₄ **Step 2: Calculate moles of KMnO₄** Moles of KMnO₄ = 0.05 M × 0.010 L = 0.0005 mol **Step 3: Calculate moles of oxalic acid using stoichiometry** From the ratio: 2 mol KMnO₄ reacts with 5 mol H₂C₂O₄ Moles of H₂C₂O₄ = (0.0005 × 5)/2 = 0.00125 mol **Step 4: Calculate mass of oxalic acid dihydrate** Molar mass of H₂C₂O₄·2H₂O = 126 g/mol Mass = 0.00125 mol × 126 g/mol = 0.1575 g **Step 5: Calculate strength in g/L** Volume of oxalic acid solution = 10.0 mL = 0.010 L Strength = 0.1575 g / 0.010 L = 15.75 g/L **Step 6: Express in the required form** 15.75 g/L = 1.575 × 10¹ g/L Rounding to nearest integer: **1575** (when expressed as × 10⁻² format) Therefore, the answer is 1575.00.