Dilute one litre 1 molar `H_2SO_4` solution by 5 litre water , the normality of that solution is

Options

A. 0.2 N
B. 5N
C. 10 N
D. 0.33N

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 initial solution
We start with a 1 liter solution of 1 molar \( H_2SO_4 \).

### Step 2: Determine the n-factor for \( H_2SO_4 \)
The n-factor for \( H_2SO_4 \) (sulfuric acid) is 2 because it can release 2 \( H^+ \) ions per molecule when dissolved in water.

### Step 3: Calculate the normality of the initial solution
Normality (N) is calculated using the formula:
\[ \text{Normality} = \text{Molarity} \times \text{n-factor} \]
For our solution:
\[ \text{Normality} = 1 \, \text{M} \times 2 = 2 \, \text{N} \]

### Step 4: Determine the final volume after dilution
We are diluting the 1 liter of \( H_2SO_4 \) solution by adding 5 liters of water. Therefore, the final volume \( V_2 \) is:
\[ V_2 = 1 \, \text{L} + 5 \, \text{L} = 6 \, \text{L} \]

### Step 5: Use the dilution formula to find the new normality
We can use the dilution formula:
\[ N_1 V_1 = N_2 V_2 \]
Where:
- \( N_1 \) is the initial normality (2 N)
- \( V_1 \) is the initial volume (1 L)
- \( N_2 \) is the final normality (what we need to find)
- \( V_2 \) is the final volume (6 L)

Rearranging the formula to find \( N_2 \):
\[ N_2 = \frac{N_1 V_1}{V_2} \]
Substituting the known values:
\[ N_2 = \frac{2 \, \text{N} \times 1 \, \text{L}}{6 \, \text{L}} = \frac{2}{6} = 0.33 \, \text{N} \]

### Final Answer
The normality of the diluted \( H_2SO_4 \) solution is \( 0.33 \, \text{N} \).

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