Find out `%` strength of 11.2V `H_(2)O_(2)`

Options

A. `34%`
B. `3.4%`
C. `1.7%`
D. `13.8%`

Answer

To find the percentage strength of 11.2V H₂O₂, we can follow these steps:

### Step 1: Understand the Concept of Volume Strength
Volume strength (V) of a solution indicates the volume of oxygen gas (in liters) that can be produced by 1 liter of the solution at standard conditions. For H₂O₂, 11.2V means that 1 liter of this solution can produce 11.2 liters of oxygen gas.

### Step 2: Use the Formula for Molarity
The formula for molarity (M) in terms of volume strength is given by:
\[ \text{Molarity} = \frac{\text{Volume Strength}}{11.2} \]
Substituting the given volume strength:
\[ \text{Molarity} = \frac{11.2}{11.2} = 1 \, \text{M} \]

### Step 3: Calculate the Molar Mass of H₂O₂
The molar mass of H₂O₂ can be calculated as follows:
- Hydrogen (H) = 1 g/mol, and there are 2 H atoms: \(2 \times 1 = 2\)
- Oxygen (O) = 16 g/mol, and there are 2 O atoms: \(2 \times 16 = 32\)

Thus, the molar mass of H₂O₂ is:
\[ 2 + 32 = 34 \, \text{g/mol} \]

### Step 4: Relate Molarity to Weight/Volume Concentration
The formula for molarity can also be expressed as:
\[ \text{Molarity} = \frac{\text{Weight (g)}}{\text{Volume (L)} \times \text{Molar Mass (g/mol)}} \]
Rearranging gives:
\[ \text{Weight} = \text{Molarity} \times \text{Volume (L)} \times \text{Molar Mass} \]
For 1 liter of solution:
\[ \text{Weight} = 1 \, \text{mol/L} \times 1 \, \text{L} \times 34 \, \text{g/mol} = 34 \, \text{g} \]

### Step 5: Calculate Weight/Volume Percentage
Weight/volume percentage (W/V %) is given by:
\[ \text{W/V \%} = \left( \frac{\text{Weight of solute (g)}}{\text{Volume of solution (mL)}} \right) \times 100 \]
Substituting the values:
\[ \text{W/V \%} = \left( \frac{34 \, \text{g}}{1000 \, \text{mL}} \right) \times 100 = 3.4\% \]

### Conclusion
The percentage strength of 11.2V H₂O₂ is **3.4%**.