In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromi — JEE Mains Chemistry Past Papers Chemistry Question
Question
In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is ___________ × 10–1%. (Nearest integer). (Given: Molar mass of Ag is 108 and Br is 80 g mol–1)
💡 Solution & Explanation
**Step 1: Calculate molar mass of AgBr** Molar mass of AgBr = 108 + 80 = 188 g/mol **Step 2: Find moles of AgBr formed** Moles of AgBr = 0.15 g ÷ 188 g/mol = 0.000798 mol **Step 3: Determine moles of Br in the compound** From the formula AgBr, 1 mole of AgBr contains 1 mole of Br Therefore, moles of Br = 0.000798 mol **Step 4: Calculate mass of Br** Mass of Br = moles × molar mass Mass of Br = 0.000798 mol × 80 g/mol = 0.0638 g **Step 5: Calculate percentage of Br** % of Br = (mass of Br / mass of compound) × 100 % of Br = (0.0638 g / 0.25 g) × 100 = 25.53% **Step 6: Express in the required form** The question asks for the answer × 10⁻¹ % Therefore: 25.53 × 10⁻¹ = 2.553 Rounding to the nearest integer: **3** (or **26** if the answer format means 25.53 ≈ 26, then 26 × 10⁻¹ = 2.6 ≈ 3) Therefore, the answer is **3** or **26** (depending on interpretation).