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 solve the problem step by step, we will follow the reactions and calculations as described in the video transcript.

### Step 1: Understanding the Reaction
Bleaching powder, which is primarily calcium oxychloride (Ca(OCl)₂), reacts with water to produce calcium hydroxide (Ca(OH)₂) and chlorine gas (Cl₂). The chlorine gas then reacts with excess potassium iodide (KI) to liberate iodine (I₂).

### Step 2: Reaction with Water
The reaction of bleaching powder with water can be represented as:
\[ \text{Ca(OCl)}_2 + 2 \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{Cl}_2 \uparrow \]

### Step 3: Reaction of Chlorine with KI
The liberated chlorine gas reacts with potassium iodide to produce potassium chloride and iodine:
\[ \text{Cl}_2 + 2 \text{KI} \rightarrow 2 \text{KCl} + \text{I}_2 \]

### Step 4: Titration of Iodine with Hypo Solution
The liberated iodine (I₂) is then titrated with hypo solution (sodium thiosulfate, Na₂S₂O₃). The reaction can be represented as:
\[ \text{I}_2 + 2 \text{S}_2\text{O}_3^{2-} \rightarrow 2 \text{I}^- + \text{S}_4\text{O}_6^{2-} \]

### Step 5: Calculate Moles of Hypo Solution
Given that the liberated iodine required 150 mL of \( \frac{M}{10} \) hypo solution for complete titration, we can calculate the moles of hypo solution used:
- Molarity (M) = \( \frac{1}{10} \) M
- Volume (V) = 150 mL = 0.150 L

Using the formula:
\[ \text{Moles of Na}_2\text{S}_2\text{O}_3 = \text{Molarity} \times \text{Volume} \]
\[ \text{Moles of Na}_2\text{S}_2\text{O}_3 = \frac{1}{10} \times 0.150 = 0.015 \text{ moles} \]

### Step 6: Relate Moles of Iodine to Moles of Hypo
From the stoichiometry of the reaction:
- 1 mole of I₂ reacts with 2 moles of Na₂S₂O₃.
Thus, the moles of I₂ can be calculated as:
\[ \text{Moles of I}_2 = \frac{0.015}{2} = 0.0075 \text{ moles} \]

### Step 7: Calculate Moles of Cl₂
From the reaction, we know that 1 mole of Cl₂ produces 1 mole of I₂. Therefore:
\[ \text{Moles of Cl}_2 = 0.0075 \text{ moles} \]

### Step 8: Calculate Mass of Available Chlorine
The molar mass of Cl₂ is approximately 71 g/mol. Thus, the mass of available chlorine can be calculated as:
\[ \text{Mass of Cl}_2 = \text{Moles} \times \text{Molar Mass} \]
\[ \text{Mass of Cl}_2 = 0.0075 \times 71 = 0.5325 \text{ g} \]

### Step 9: Calculate Percentage of Available Chlorine
To find the percentage of available chlorine in the 1.5 g sample of bleaching powder:
\[ \text{Percentage of Available Cl}_2 = \left( \frac{\text{Mass of Cl}_2}{\text{Sample Mass}} \right) \times 100 \]
\[ \text{Percentage of Available Cl}_2 = \left( \frac{0.5325}{1.5} \right) \times 100 = 35.5\% \]

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

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