0.5 M of `H_(2)SO_(4)` is diluted from `1` litre to 10 litre, normality of resulting solution is

Options

A. `1N`
B. `0.1N`
C. `10N`
D. `11N`

Explanation

To find the normality of the resulting solution after diluting 0.5 M of \( H_2SO_4 \) from 1 liter to 10 liters, we can follow these steps:

### Step 1: Understand the relationship between Molarity and Normality
For sulfuric acid (\( H_2SO_4 \)), which is a diprotic acid, the normality is double the molarity. This is because each molecule of \( H_2SO_4 \) can donate two protons (H\(^+\)).

### Step 2: Calculate the initial normality
Given that the molarity (M) of the solution is 0.5 M, we can calculate the initial normality (N):
\[ N = 2 \times M = 2 \times 0.5 = 1 \, N \]

### Step 3: Use the dilution formula
When diluting a solution, we can use the dilution formula:
\[ N_1 V_1 = N_2 V_2 \]
Where:
- \( N_1 \) = initial normality = 1 N
- \( V_1 \) = initial volume = 1 L
- \( N_2 \) = final normality (what we are trying to find)
- \( V_2 \) = final volume = 10 L

### Step 4: Substitute the known values into the equation
Substituting the known values into the dilution formula:
\[ 1 \, N \times 1 \, L = N_2 \times 10 \, L \]

### Step 5: Solve for \( N_2 \)
Now, we can solve for \( N_2 \):
\[ N_2 = \frac{1 \, N \times 1 \, L}{10 \, L} = \frac{1}{10} \, N = 0.1 \, N \]

### Conclusion
The normality of the resulting solution after dilution is \( 0.1 \, N \).

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