Fortification of food with iron is done using FeSO4.7H2O. The mass in grams of the FeSO4.7H2O requir — JEE Mains Chemistry Past Papers Chemistry Question
Question
Fortification of food with iron is done using FeSO4.7H2O. The mass in grams of the FeSO4.7H2O required to achieve 12 ppm of iron in 150 kg of wheat is __________ (Nearest integer) [Given: Molar mass of Fe, S and O respectively are 56, 32 and 16 g mol–1]
💡 Solution & Explanation
**Step 1: Calculate moles of Fe needed** 12 ppm means 12 g of Fe per 10⁶ g of wheat. Mass of Fe needed = (12 × 150,000 g) / 10⁶ = 1.8 g of Fe Moles of Fe = 1.8 g ÷ 56 g/mol = 0.0321 mol **Step 2: Determine moles of FeSO₄·7H₂O needed** Since each molecule of FeSO₄·7H₂O contains exactly one Fe atom: Moles of FeSO₄·7H₂O = 0.0321 mol **Step 3: Calculate molar mass of FeSO₄·7H₂O** FeSO₄·7H₂O = Fe + S + 4O + 7(H₂O) = 56 + 32 + (4 × 16) + 7(2 + 16) = 56 + 32 + 64 + 7(18) = 152 + 126 = 278 g/mol **Step 4: Calculate mass of FeSO₄·7H₂O required** Mass = moles × molar mass Mass = 0.0321 mol × 278 g/mol = 8.93 g Therefore, the answer is **9 g** (nearest integer)