1.5 gm sample of bleaching power was suspended in water. If was treated with `CH_(3)COOH` followed by the addition of excess of Kl. The liberated iodine required 150 mL of `(M)/(10)` hypo solution for complete titration. The percentage of available chlorine in the sample is

Answer

To find the percentage of available chlorine in the bleaching powder sample, we can follow these steps:

### Step 1: Understand the Reaction
Bleaching powder (calcium oxychloride, \( \text{Ca(OCl)}_2 \)) reacts with water to release chlorine gas (\( \text{Cl}_2 \)). The chlorine gas then reacts with potassium iodide (\( \text{KI} \)) to liberate iodine (\( \text{I}_2 \)).

### Step 2: Write the Relevant Reactions
1. The reaction of bleaching powder with water:
\[ \text{Ca(OCl)}_2 + 2 \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{Cl}_2 \]

2. The reaction of chlorine with potassium iodide:
\[ \text{Cl}_2 + 2 \text{KI} \rightarrow 2 \text{KCl} + \text{I}_2 \]

3. The titration reaction of iodine with sodium thiosulfate (\( \text{Na}_2\text{S}_2\text{O}_3 \)):
\[ \text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 \]

### Step 3: Calculate Moles of Sodium Thiosulfate Used
Given that 150 mL of \( \frac{M}{10} \) (0.1 M) sodium thiosulfate solution was used:
\[ \text{Moles of } \text{Na}_2\text{S}_2\text{O}_3 = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.150 \, \text{L} = 0.015 \, \text{mol} \]

### Step 4: Relate Moles of Iodine to Moles of Chlorine
From the titration reaction, 1 mole of \( \text{I}_2 \) reacts with 2 moles of \( \text{Na}_2\text{S}_2\text{O}_3 \):
\[ \text{Moles of } I_2 = \frac{0.015}{2} = 0.0075 \, \text{mol} \]
Since 1 mole of \( \text{Cl}_2 \) produces 1 mole of \( \text{I}_2 \), the moles of chlorine gas produced is also 0.0075 mol.

### Step 5: Calculate the Mass of Available Chlorine
The molar mass of \( \text{Cl}_2 \) is approximately 71 g/mol (35.5 g/mol for each chlorine atom):
\[ \text{Mass of } \text{Cl}_2 = \text{Moles} \times \text{Molar Mass} = 0.0075 \, \text{mol} \times 71 \, \text{g/mol} = 0.5325 \, \text{g} \]

### Step 6: Calculate the Percentage of Available Chlorine
To find the percentage of available chlorine in the 1.5 g sample:
\[ \text{Percentage of available chlorine} = \left( \frac{\text{Mass of } \text{Cl}_2}{\text{Sample mass}} \right) \times 100 = \left( \frac{0.5325 \, \text{g}}{1.5 \, \text{g}} \right) \times 100 = 35.5\% \]

### Final Answer
The percentage of available chlorine in the bleaching powder sample is **35.5%**.

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