The normality and volume strength of a solution made by mixing `1.0 L` each of `5.6` volume and `11.2` volume `H_2O_2` solution are:

Options

A. `1N, 5.6 vol`
B. `1.5N, 5.6 vol`
C. `1.5 N, 8.4 vol`
D. `1 N, 8.4 vol`

Explanation

To solve the problem of finding the normality and volume strength of a solution made by mixing 1.0 L each of 5.6 volume and 11.2 volume H₂O₂ solutions, we can follow these steps:

### Step 1: Understand the Volume Strength
Volume strength of a solution indicates how many volumes of oxygen are released by 1 volume of the solution. The formula relating volume strength (VS) and normality (N) is:
\[ \text{Volume Strength} = 5.6 \times \text{Normality} \]

### Step 2: Determine the Normality of Each Solution
1. **5.6 Volume H₂O₂ Solution**:
- By definition, a 5.6 volume H₂O₂ solution corresponds to 1 normal (N₁ = 1 N).

2. **11.2 Volume H₂O₂ Solution**:
- To find the normality of the 11.2 volume solution (N₂), we can use the ratio of the volumes:
\[ N_2 = \left( \frac{11.2}{5.6} \right) \times N_1 = \left( \frac{11.2}{5.6} \right) \times 1 \text{ N} = 2 \text{ N} \]

### Step 3: Calculate the Final Normality of the Mixture
Using the formula for the normality of mixed solutions:
\[ N_f = \frac{N_1 V_1 + N_2 V_2}{V_1 + V_2} \]
Where:
- \(N_1 = 1 \text{ N}\), \(V_1 = 1 \text{ L}\)
- \(N_2 = 2 \text{ N}\), \(V_2 = 1 \text{ L}\)

Substituting the values:
\[ N_f = \frac{(1 \text{ N} \times 1 \text{ L}) + (2 \text{ N} \times 1 \text{ L})}{1 \text{ L} + 1 \text{ L}} = \frac{1 + 2}{2} = \frac{3}{2} = 1.5 \text{ N} \]

### Step 4: Calculate the Volume Strength of the Final Solution
Using the normality found in the previous step:
\[ \text{Volume Strength} = 5.6 \times N_f = 5.6 \times 1.5 = 8.4 \text{ volume} \]

### Final Result
The normality and volume strength of the mixed solution are:
- Normality = 1.5 N
- Volume Strength = 8.4 volume