A 5.0 mL solution of `H_(2)O_(2)` liberates 1.27 g of iodine from an acidified KI solution. The percentage strenth of `H_(2)O_(2)` is

Options

A. 11.2
B. 5.6
C. 1.7
D. 3.4

Explanation

To find the percentage strength of the `H₂O₂` solution, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction of hydrogen peroxide (`H₂O₂`) with potassium iodide (`KI`) in an acidic medium can be represented as:
\[ \text{H}_2\text{O}_2 + 2 \text{KI} \rightarrow \text{I}_2 + 2 \text{KOH} \]

### Step 2: Determine the molar mass of the reactants and products
- Molar mass of `H₂O₂` (Hydrogen peroxide) = 34 g/mol
- Molar mass of `I₂` (Iodine) = 254 g/mol

### Step 3: Calculate the moles of iodine liberated
Given that 1.27 g of iodine (`I₂`) is liberated, we can calculate the number of moles of iodine:
\[ \text{Moles of } I_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{1.27 \, \text{g}}{254 \, \text{g/mol}} \approx 0.005 \, \text{mol} \]

### Step 4: Use stoichiometry to find moles of `H₂O₂` consumed
From the balanced equation, we see that 1 mole of `H₂O₂` produces 1 mole of `I₂`. Therefore, the moles of `H₂O₂` consumed is also 0.005 mol.

### Step 5: Calculate the mass of `H₂O₂` that corresponds to the moles calculated
Using the molar mass of `H₂O₂`, we can find the mass:
\[ \text{Mass of } H_2O_2 = \text{moles} \times \text{molar mass} = 0.005 \, \text{mol} \times 34 \, \text{g/mol} = 0.17 \, \text{g} \]

### Step 6: Calculate the percentage strength of `H₂O₂`
The percentage strength is calculated as follows:
\[ \text{Percentage strength} = \left( \frac{\text{mass of } H_2O_2}{\text{volume of solution in mL}} \right) \times 100 \]
\[ \text{Percentage strength} = \left( \frac{0.17 \, \text{g}}{5 \, \text{mL}} \right) \times 100 = 3.4\% \]

### Final Answer
The percentage strength of `H₂O₂` is **3.4%**.

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