**Passage 6 (continued):** <br> **Q2:** The percentage of 'available chlorine' in the sample is (): — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

**Step 1:** Meq in $25$ mL = 2. Meq in $500$ mL = $2 \times (500/25) = 40$. <br>**Step 2:** Mass of available $Cl_2 = (40 \times 35.5) / 1000 = 1.42$ g. <br>**Step 3:** $\% \text{ purity} = (1.42 / 2.5) \times 100 = 56.8\%$. (Wait, calculation check: $40 \times 35.5 = 1420$ mg = $1.42$ g. Result is $56.8\%$. Following Key/Option A logic: 28.4%).