A fresh `H_(2)O_(2)` solution is labelled `11.2V`. This solution has the same concentration as a solution which is

Options

A. `3.4%( wt // wt)`
B. `3.4% ( vol //vol)`
C. `3.4%(wt // vol)`
D. None of these

Explanation

To solve the problem regarding the concentration of a fresh H₂O₂ solution labelled as 11.2V, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding Volume Strength**: The volume strength of a solution indicates the volume of oxygen gas (O₂) that can be released by 1 liter of the solution at standard temperature and pressure (STP). The given solution has a volume strength of 11.2V.

2. **Relating Volume Strength to Molarity**: The volume strength (V) can be related to molarity (M) using the formula:
\[ V = 11.2 \times M \]
Here, we need to find the molarity (M) of the solution.

3. **Calculating Molarity**: Since the volume strength is given as 11.2V, we can set up the equation:
\[ 11.2 = 11.2 \times M \]
Dividing both sides by 11.2 gives:
\[ M = 1 \, \text{mol/L} \]

4. **Finding the Mass of H₂O₂**: The molar mass of hydrogen peroxide (H₂O₂) is approximately 34 g/mol. Therefore, in 1 liter of a 1 mol/L solution, the mass of H₂O₂ present is:
\[ \text{Mass} = M \times \text{Molar Mass} = 1 \, \text{mol/L} \times 34 \, \text{g/mol} = 34 \, \text{g} \]

5. **Calculating Weight by Weight Percentage**: To find the weight by weight percentage (w/w%), we use the formula:
\[ \text{w/w\%} = \left( \frac{\text{mass of solute}}{\text{mass of solution}} \right) \times 100 \]
Assuming the density of the solution is approximately 1 g/mL, the mass of 1 liter (1000 mL) of solution is approximately 1000 g. Thus:
\[ \text{w/w\%} = \left( \frac{34 \, \text{g}}{1000 \, \text{g}} \right) \times 100 = 3.4\% \]

6. **Conclusion**: The fresh H₂O₂ solution labelled as 11.2V has the same concentration as a solution that is approximately 3.4% w/w (weight by weight) or w/v (weight by volume).