The volume strength of 8.9 M H O solution calculated at 273 K and 1 atm is ……….. (R = 0.0821 L atm K — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

**Step 1: Understand Volume Strength** Volume strength of H₂O₂ is defined as the volume of O₂ gas (at STP: 273 K and 1 atm) produced per unit volume of H₂O₂ solution. **Step 2: Write the Decomposition Reaction** 2H₂O₂ → 2H₂O + O₂↑ From stoichiometry: 2 moles of H₂O₂ produce 1 mole of O₂ **Step 3: Calculate Moles of O₂ from Given Molarity** For 8.9 M H₂O₂ solution: - 1 L of solution contains 8.9 moles of H₂O₂ - From the equation: 8.9 moles H₂O₂ produces (8.9/2) = 4.45 moles of O₂ **Step 4: Calculate Volume of O₂ at STP** Using ideal gas law: PV = nRT V = nRT/P = (4.45 × 0.0821 × 273)/1 V = 99.99 ≈ 100 L **Step 5: Determine Volume Strength** Volume strength = Volume of O₂ (L) / Volume of solution (L) = 100 L / 1 L = **100** Therefore, the answer is 100.00.