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 \( H_2SO_4 \), we will follow these steps:

### Step 1: Understand the initial conditions
We are given a 5 M (molar) solution of \( H_2SO_4 \) and we need to dilute 1 liter of this solution to a final volume of 100 liters.

### Step 2: Calculate the moles of \( H_2SO_4 \)
Using the formula for molarity:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
We can rearrange this to find the moles of \( H_2SO_4 \):
\[ \text{moles of } H_2SO_4 = \text{Molarity} \times \text{Volume} = 5 \, \text{M} \times 1 \, \text{L} = 5 \, \text{moles} \]

### Step 3: Calculate the new molarity after dilution
When we dilute the solution to 100 liters, the number of moles remains the same, but the volume changes. The new molarity can be calculated as:
\[ \text{New Molarity} = \frac{\text{moles of } H_2SO_4}{\text{new volume in liters}} = \frac{5 \, \text{moles}}{100 \, \text{L}} = 0.05 \, \text{M} \]

### Step 4: Determine the normality of the solution
Normality (N) is related to molarity (M) by the equation:
\[ \text{Normality} = \text{Molarity} \times \text{N factor} \]
For \( H_2SO_4 \), the N factor is 2 because it can donate 2 protons (H\(^+\)) in a reaction. Therefore:
\[ \text{Normality} = 0.05 \, \text{M} \times 2 = 0.1 \, \text{N} \]

### Conclusion
The normality of the diluted \( H_2SO_4 \) solution is \( 0.1 \, \text{N} \).

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