**Passage 6:** g of a bleaching powder sample was suspended in water and made to mL. mL of this susp — Redox Reactions and Volumetric Analysis Chemistry Question
Question
**Passage 6:** $2.5$ g of a bleaching powder sample was suspended in water and made to $500$ mL. $25$ mL of this suspension was treated with excess $KI$ and $CH_3COOH$. The liberated $I_2$ required $20$ mL of $0.1 N$ $Na_2S_2O_3$. <br> **Q1:** The milliequivalents of $Cl_2$ in $25$ mL suspension is:
Answer: A
💡 Solution & Explanation
**Step 1:** $Cl_2$ liberated from bleaching powder reacts with $I^-$ to give $I_2$. <br>**Step 2:** Meq $Cl_2 = \text{Meq } I_2 = \text{Meq } Na_2S_2O_3$. <br>**Step 3:** Meq Hypo = $20 \times 0.1 = 2$.
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