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

Options

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

Answer

To solve the problem of determining the normality of a diluted solution of sulfuric acid (H₂SO₄), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Concentration**:
We start with a 5 M (molar) solution of H₂SO₄. Molarity (M) is defined as moles of solute per liter of solution.

2. **Understand the Dilution**:
The solution is diluted from a volume of 1 liter to a total volume of 100 liters. When a solution is diluted, the number of moles of solute remains constant.

3. **Calculate the Moles of H₂SO₄**:
The number of moles in the original solution can be calculated using the formula:
\[ \text{Moles of H₂SO₄} = \text{Molarity} \times \text{Volume} = 5 \, \text{M} \times 1 \, \text{L} = 5 \, \text{moles} \]

4. **Determine the New Molarity After Dilution**:
After dilution to 100 liters, the new molarity (M') can be calculated as:
\[ M' = \frac{\text{Moles of H₂SO₄}}{\text{New Volume}} = \frac{5 \, \text{moles}}{100 \, \text{L}} = 0.05 \, \text{M} \]

5. **Calculate the Normality**:
Normality (N) is related to molarity and the number of equivalents. For sulfuric acid (H₂SO₄), the n-factor is 2 because it can donate two protons (H⁺ ions) per molecule. The formula for normality is:
\[ \text{Normality} = \text{Molarity} \times \text{n-factor} \]
Thus, we have:
\[ \text{Normality} = 0.05 \, \text{M} \times 2 = 0.10 \, \text{N} \]

6. **Final Answer**:
The normality of the diluted solution of H₂SO₄ is **0.10 N**.