**Passage 3 (continued):** <br> **Q2:** The percentage of Copper in the ore is ( of ): — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

**Step 1:** From $2Cu^{2+} \equiv I_2$, moles $Cu = 2 \times \text{moles } I_2 = 2 \times 0.001 = 0.002$. <br>**Step 2:** Mass $Cu = 0.002 \times 63.5 = 0.127$ g. <br>**Step 3:** $\%$ in $0.5$ g sample = $(0.127 / 0.5) \times 100 = 25.4\%$.