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

Explanation

To find 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 relationship between normality and volume strength
For hydrogen peroxide (H2O2):
- 1 normal (N) = 5.6 volume strength (VS)
- Therefore, 2 normal (N) = 11.2 volume strength (VS)

### Step 2: Assign values to the variables
Let:
- \( N_1 = 1 \, \text{N} \) (for 5.6 volume H2O2)
- \( N_2 = 2 \, \text{N} \) (for 11.2 volume H2O2)
- \( V_1 = 1 \, \text{L} \) (volume of 5.6 volume solution)
- \( V_2 = 1 \, \text{L} \) (volume of 11.2 volume solution)

### Step 3: Calculate the total volume of the mixed solution
The total volume \( V_3 \) of the mixed solution is:
\[ V_3 = V_1 + V_2 = 1 \, \text{L} + 1 \, \text{L} = 2 \, \text{L} \]

### Step 4: Use the normality equation
Using the formula for normality:
\[ N_1 V_1 + N_2 V_2 = N_3 V_3 \]
Substituting the known values:
\[ (1 \, \text{N} \times 1 \, \text{L}) + (2 \, \text{N} \times 1 \, \text{L}) = N_3 \times 2 \, \text{L} \]
This simplifies to:
\[ 1 + 2 = N_3 \times 2 \]
\[ 3 = N_3 \times 2 \]
Thus, solving for \( N_3 \):
\[ N_3 = \frac{3}{2} = 1.5 \, \text{N} \]

### Step 5: Calculate the volume strength of the mixed solution
To find the volume strength (VS) of the mixed solution, we use the relationship:
\[ \text{Volume Strength} = 5.6 \times \text{Normality} \]
Substituting \( N_3 \):
\[ \text{Volume Strength} = 5.6 \times 1.5 = 8.4 \, \text{volume} \]

### Final Answer
The normality of the mixed solution is **1.5 N**, and the volume strength is **8.4 volume**.