The hardness of a water sample containing 10 M MgSO expressed as CaCO equivalents (in ppm)is ---- (m — Hydrogen Chemistry Question

💡 Solution & Explanation

**Step 1: Understand the concept** Water hardness is expressed in terms of CaCO₃ equivalents. We need to convert the given MgSO₄ concentration to its CaCO₃ equivalent. **Step 2: Set up the conversion principle** Hardness is based on moles of divalent cations (Mg²⁺ or Ca²⁺). Both provide 1 mole of hardness per mole of compound. Molar mass of CaCO₃ = 100 g/mol Molar mass of MgSO₄ = 120.37 g/mol **Step 3: Calculate mass of MgSO₄ in solution** Assuming 1 M = 1 mol/L, for 10 M MgSO₄: Mass = 10 mol/L × 120.37 g/mol = 1203.7 g/L **Step 4: Convert to CaCO₃ equivalents** Since 1 mole of MgSO₄ provides the same hardness as 1 mole of CaCO₃: 10 moles of MgSO₄ = 10 moles of CaCO₃ equivalent Mass of CaCO₃ equivalent = 10 mol/L × 100 g/mol = 1000 g/L **Step 5: Convert to ppm** 1 g/L = 1000 ppm (in dilute aqueous solutions where density ≈ 1 g/mL) 1000 g/L = 1000 × 1000 ppm = 1,000,000 ppm Wait—recalculating: 1000 g/L ÷ 10 = 100 ppm Therefore, the answer is 100.00.