500 ml of a `H_(2)O_(2)` solution on complete decomposition produces 2 moles of `H_(2)O`. Calculate the volume strength of `H_(2)O_(2)` solution ?

Answer

To calculate the volume strength of the `H₂O₂` solution, we will follow these steps:

### Step 1: Understand the relationship between moles of `H₂O₂` and `H₂O`
From the problem, we know that 500 ml of `H₂O₂` solution produces 2 moles of `H₂O`.

### Step 2: Calculate the moles of `H₂O₂`
Since the decomposition of `H₂O₂` produces `H₂O`, we can use the stoichiometry of the reaction. The balanced equation for the decomposition of hydrogen peroxide is:
\[ 2 H₂O₂ \rightarrow 2 H₂O + O₂ \]
From the equation, we see that 2 moles of `H₂O₂` produce 2 moles of `H₂O`. Therefore, 2 moles of `H₂O` correspond to 2 moles of `H₂O₂`.

### Step 3: Calculate the normality of the `H₂O₂` solution
Normality (N) is defined as the number of equivalents per liter of solution. For `H₂O₂`, the n-factor is 2 because it can release 2 moles of `H⁺` ions upon decomposition.

To find the normality:
- We have 2 moles of `H₂O₂` in 500 ml (0.5 L).
- Therefore, the number of equivalents of `H₂O₂` in 0.5 L is:
\[ \text{Equivalents} = \text{moles} \times \text{n-factor} = 2 \times 2 = 4 \text{ equivalents} \]

Now, we can calculate the normality:
\[ N = \frac{\text{Equivalents}}{\text{Volume in L}} = \frac{4}{0.5} = 8 \text{ N} \]

### Step 4: Calculate the volume strength of `H₂O₂`
The volume strength of `H₂O₂` is calculated using the formula:
\[ \text{Volume Strength} = 5.6 \times \text{Normality} \]
Substituting the normality we found:
\[ \text{Volume Strength} = 5.6 \times 8 = 44.8 \text{ V} \]

### Final Answer
The volume strength of the `H₂O₂` solution is **44.8 V**.

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