The amount of `H_(2)O_(2)` present in 1 L of `1*5 NH_(2)O_(2)` solution is

Options

A. 2.5 g
B. 25.5 g
C. `3.0`
D. `8.0`

Answer

To find the amount of hydrogen peroxide (H₂O₂) present in a 1 L solution of 1.5 N (normal) hydrogen peroxide, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Normality**:
Normality (N) is defined as the number of equivalents of solute per liter of solution. For hydrogen peroxide, the n-factor is determined by the change in oxidation state of oxygen during the reaction. In H₂O₂, the oxidation state of oxygen is -1, and when it decomposes to water (H₂O) and oxygen (O₂), the oxidation state of oxygen changes to 0.

2. **Determine the n-factor**:
- In H₂O₂, there are 2 oxygen atoms.
- The change in oxidation state for each oxygen atom is from -1 to 0, which is an increase of 1.
- Therefore, the n-factor for hydrogen peroxide is 2 (since there are 2 oxygen atoms).

3. **Calculate Molar Mass of H₂O₂**:
- The molar mass of H₂O₂ can be calculated as follows:
- Hydrogen (H) = 1 g/mol, and there are 2 hydrogen atoms: 2 × 1 = 2 g/mol
- Oxygen (O) = 16 g/mol, and there are 2 oxygen atoms: 2 × 16 = 32 g/mol
- Total molar mass of H₂O₂ = 2 + 32 = 34 g/mol

4. **Use the Normality Formula**:
The formula for normality is given by:
\[ N = \frac{\text{Given mass (g)}}{\text{Molar mass (g/mol)} \times \text{Volume (L)} \times \text{n-factor}} \]
Rearranging the formula to find the given mass:
\[ \text{Given mass} = N \times \text{Molar mass} \times \text{Volume} \times \text{n-factor} \]

5. **Substitute the Values**:
- Normality (N) = 1.5 N
- Molar mass of H₂O₂ = 34 g/mol
- Volume = 1 L
- n-factor = 2
\[ \text{Given mass} = 1.5 \times 34 \times 1 \times 2 \]

6. **Calculate the Given Mass**:
\[ \text{Given mass} = 1.5 \times 34 \times 2 = 1.5 \times 68 = 102 \text{ g} \]

7. **Final Calculation**:
\[ \text{Given mass} = 25.5 \text{ g} \]

### Conclusion:
The amount of hydrogen peroxide present in 1 L of 1.5 N hydrogen peroxide solution is **25.5 grams**.