`2.5 cm^(3)` of `0.2 M H_(2)SO_(4)` solution is diluted to 0.5 `dm^(3)` . Find normality of the diluted solution .

Options

A. 0.2 N
B. 0.02 N
C. 0.002 N
D. 0.04 N

Explanation

To solve the problem of finding the normality of the diluted solution, we can follow these steps:

### Step 1: Identify and Convert Initial Volumes and Molarity
Given:
- Initial volume, \( V_1 = 2.5 \, \text{cm}^3 \)
- Initial molarity, \( M_1 = 0.2 \, \text{M} \)
- Final volume, \( V_2 = 0.5 \, \text{dm}^3 \)

First, convert the initial volume from cm³ to dm³:
\[ V_1 = 2.5 \, \text{cm}^3 = 2.5 \times 10^{-3} \, \text{dm}^3 \]

### Step 2: Calculate Initial Normality
For sulfuric acid (\( \text{H}_2\text{SO}_4 \)), the equivalent factor (n-factor) is 2 because it can donate 2 protons (\( \text{H}^+ \)).

Normality (\( N \)) is given by:
\[ N = M \times \text{n-factor} \]

So, the initial normality (\( N_1 \)) is:
\[ N_1 = 0.2 \, \text{M} \times 2 = 0.4 \, \text{N} \]

### Step 3: Apply Dilution Formula
The relationship between the initial and final states of a solution during dilution is given by:
\[ N_1 V_1 = N_2 V_2 \]

Substitute the known values:
\[ 0.4 \, \text{N} \times 2.5 \times 10^{-3} \, \text{dm}^3 = N_2 \times 0.5 \, \text{dm}^3 \]

### Step 4: Solve for Final Normality (\( N_2 \))
\[ N_2 = \frac{0.4 \times 2.5 \times 10^{-3}}{0.5} \]
\[ N_2 = \frac{1.0 \times 10^{-3}}{0.5} \]
\[ N_2 = 2.0 \times 10^{-3} \, \text{N} \]
\[ N_2 = 0.002 \, \text{N} \]

Thus, the normality of the diluted solution is \( 0.002 \, \text{N} \).