`5.7 g` of bleaching powder was suspended in `500 mL` of water. `25 mL` of this suspension on treatment with `KI` and `HCl` liberated iodine which reacted with `24.35 mL "of" N//10Na_(2)S_(2)O_(3)`. Calculate `%` of available `Cl_(2)` in bleaching powder.

Explanation

To solve the problem, we need to calculate the percentage of available Cl₂ in the bleaching powder based on the reactions that occur when it is treated with KI and HCl, leading to the liberation of iodine (I₂), which then reacts with sodium thiosulfate (Na₂S₂O₃).

### Step 1: Write the reactions involved

1. **Bleaching Powder Reaction**:
\[ \text{Ca(OCl)₂} + 2 \text{KI} + 2 \text{HCl} \rightarrow \text{I}_2 + \text{CaCl}_2 + 2 \text{KCl} + 2 \text{H}_2\text{O} \]
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