The volume of water to be added to `100 cm^(3)` of 0.5 `NH_(2)SO_(4)` to get decinormal concentration is

Options

A. `400 cm^(3)`
B. `500 cm^(3)`
C. `450 cm^(3)`
D. `100 cm^(3)`

Explanation

To solve the problem of determining the volume of water to be added to 100 cm³ of 0.5 N H₂SO₄ to achieve a decinormal (0.1 N) concentration, we can use the dilution formula, which is given by:

\[ N_1 V_1 = N_2 V_2 \]

Where:
- \( N_1 \) = initial normality (0.5 N)
- \( V_1 \) = initial volume (100 cm³)
- \( N_2 \) = final normality (0.1 N)
- \( V_2 \) = final volume (unknown)

### Step-by-Step Solution:

1. **Identify the known values**:
- Initial normality, \( N_1 = 0.5 \, \text{N} \)
- Initial volume, \( V_1 = 100 \, \text{cm}^3 \)
- Final normality, \( N_2 = 0.1 \, \text{N} \)

2. **Set up the equation using the dilution formula**:
\[ N_1 V_1 = N_2 V_2 \]
Plugging in the known values:
\[ 0.5 \, \text{N} \times 100 \, \text{cm}^3 = 0.1 \, \text{N} \times V_2 \]

3. **Calculate \( V_2 \)**:
\[ 50 = 0.1 V_2 \]
To find \( V_2 \), divide both sides by 0.1:
\[ V_2 = \frac{50}{0.1} = 500 \, \text{cm}^3 \]

4. **Determine the volume of water to be added**:
Since the final volume \( V_2 \) is 500 cm³ and the initial volume \( V_1 \) is 100 cm³, the volume of water to be added is:
\[ \text{Volume of water} = V_2 - V_1 = 500 \, \text{cm}^3 - 100 \, \text{cm}^3 = 400 \, \text{cm}^3 \]

### Final Answer:
The volume of water to be added is **400 cm³**.