Dilute one litre 1 molar `H_2SO_4` solution by 5 litre water , the normality of that solution is

Options

A. 0.2 N
B. 5N
C. 10 N
D. 0.33N

Answer

To find the normality of the diluted solution, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Normality of the Initial Solution:**
- We have a 1 M (molar) solution of \( H_2SO_4 \).
- The normality (N) is calculated using the formula:
\[ \text{Normality} = \text{Molarity} \times n \text{ (n factor)} \]
- For \( H_2SO_4 \), the n factor is 2 because it can donate 2 protons (H⁺ ions).
- Therefore, the normality of the initial solution is:
\[ N_1 = 1 \, \text{M} \times 2 = 2 \, \text{N} \]

2. **Calculate the Volume of the Diluted Solution:**
- We are diluting 1 L of the \( H_2SO_4 \) solution with 5 L of water.
- The total volume of the diluted solution (V₂) is:
\[ V_2 = 1 \, \text{L} + 5 \, \text{L} = 6 \, \text{L} \]

3. **Use the Dilution Formula:**
- The dilution formula states that:
\[ N_1 \times V_1 = N_2 \times V_2 \]
- Where:
- \( N_1 \) = normality of the initial solution (2 N)
- \( V_1 \) = volume of the initial solution (1 L)
- \( N_2 \) = normality of the diluted solution (unknown)
- \( V_2 \) = volume of the diluted solution (6 L)

4. **Substitute the Known Values:**
- Plugging in the values we have:
\[ 2 \, \text{N} \times 1 \, \text{L} = N_2 \times 6 \, \text{L} \]

5. **Solve for \( N_2 \):**
- Rearranging the equation to find \( N_2 \):
\[ N_2 = \frac{2 \, \text{N} \times 1 \, \text{L}}{6 \, \text{L}} = \frac{2}{6} \, \text{N} = \frac{1}{3} \, \text{N} \approx 0.33 \, \text{N} \]

### Final Answer:
The normality of the diluted solution is approximately **0.33 N**.