A 5M solution of `H_(2)SO_(4)` is diluted from 1 litre of 100 litres, the normality of the solution will be

Options

A. 1 N
B. 5 N
C. 0.1 N
D. 0.5 N

Answer

To solve the problem of finding the normality of a diluted solution of \( H_2SO_4 \), we can follow these steps:

### Step 1: Understand the Given Information
We have a 5M solution of \( H_2SO_4 \) that is diluted from 1 liter to 100 liters. We need to find the normality of the final solution.

### Step 2: Calculate the Final Molarity After Dilution
Using the dilution formula:
\[ M_1V_1 = M_2V_2 \]
Where:
- \( M_1 \) = initial molarity = 5 M
- \( V_1 \) = initial volume = 1 L
- \( M_2 \) = final molarity (unknown)
- \( V_2 \) = final volume = 100 L

Substituting the known values:
\[ 5 \, \text{M} \times 1 \, \text{L} = M_2 \times 100 \, \text{L} \]
\[ 5 = M_2 \times 100 \]
\[ M_2 = \frac{5}{100} = 0.05 \, \text{M} \]

### Step 3: Determine the n-factor for \( H_2SO_4 \)
The n-factor for sulfuric acid (\( H_2SO_4 \)) is determined by the number of hydrogen ions (\( H^+ \)) it can donate. Since \( H_2SO_4 \) is a diprotic acid, it can donate 2 \( H^+ \) ions. Therefore, the n-factor is:
\[ n = 2 \]

### Step 4: Calculate the Normality
Normality (N) is calculated using the formula:
\[ N = n \times M \]
Substituting the values we have:
\[ N = 2 \times 0.05 \, \text{M} \]
\[ N = 0.1 \, \text{N} \]

### Conclusion
The normality of the diluted \( H_2SO_4 \) solution is **0.1 N**.

---