0.25 g of an organic compound containing chlorine gave 0.40 g of silver chloride in Carius estimatio — Practical Organic Chemistry and Purification Chemistry Question

💡 Solution & Explanation

**Step 1: Find molar mass of AgCl** Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol **Step 2: Calculate moles of AgCl formed** Moles of AgCl = 0.40 g ÷ 143.5 g/mol = 0.00279 mol **Step 3: Determine moles of Cl in the compound** From the reaction: 1 mole of AgCl contains 1 mole of Cl Therefore, moles of Cl = 0.00279 mol **Step 4: Calculate mass of Cl in the compound** Mass of Cl = moles × molar mass Mass of Cl = 0.00279 mol × 35.5 g/mol = 0.099 g **Step 5: Calculate percentage of Cl** **Key formula:** $$\text{Percentage of Cl} = \frac{\text{Mass of Cl}}{\text{Mass of compound}} × 100$$ Percentage of Cl = (0.099 g ÷ 0.25 g) × 100 = 39.6% **Step 6: Round to nearest integer** 39.6% ≈ 40% Therefore, the answer is 40.00.