200 cc of x × 10 –3M potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr’s salt — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

**Step 1: Write the balanced redox equation** In acidic medium, potassium dichromate oxidizes Fe²⁺ (in Mohr's salt) to Fe³⁺: Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O Mole ratio: 1 mole Cr₂O₇²⁻ : 6 moles Fe²⁺ **Step 2: Calculate moles of Mohr's salt (Fe²⁺)** Moles of Fe²⁺ = Molarity × Volume (L) = 0.6 M × 0.750 L = 0.45 mol **Step 3: Use stoichiometry to find moles of K₂Cr₂O₇** From the equation: 1 mol Cr₂O₇²⁻ oxidizes 6 mol Fe²⁺ Moles of K₂Cr₂O₇ = 0.45 mol ÷ 6 = 0.075 mol **Step 4: Calculate molarity of K₂Cr₂O₇** Volume of K₂Cr₂O₇ = 200 cc = 0.200 L Molarity = moles ÷ volume (L) = 0.075 ÷ 0.200 = 0.375 M **Step 5: Express in scientific notation** 0.375 M = 3.75 × 10⁻¹ M = 37.5 × 10⁻² M If x × 10⁻³ M format is required: 0.375 M = 375 × 10⁻³ M, so **x = 375** Therefore, the answer is **x = 375** (or 0.375 M for K₂Cr₂O₇)