A lead storage battery containing `5.0` L of 1N `H_(2)SO_(4)` solution is operated for `9.65xx10^(5)` s with a steady current of 100 mA. Assuming volume of the solution remaining constant, normality of `H_(2)SO_(4)` will

Options

A. Remain unchanges
B. Increases by `0.20`
C. Increases by unity
D. Decreases by `0.40`

Explanation

To solve the problem, we need to determine the normality of the `H2SO4` solution after operating a lead storage battery for a certain period of time with a given current. Here’s the step-by-step solution:

### Step 1: Calculate the Total Charge
The total charge (Q) can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I \) is the current in amperes,
- \( t \) is the time in seconds.

Given:
- Current \( I = 100 \, \text{mA} = 100 \times 10^{-3} \, \text{A} = 0.1 \, \text{A} \)
- Time \( t = 9.65 \times 10^5 \, \text{s} \)

Calculating the total charge:
\[ Q = 0.1 \, \text{A} \times 9.65 \times 10^5 \, \text{s} = 9.65 \times 10^4 \, \text{C} \]

### Step 2: Convert Charge to Moles of Electrons
Using Faraday's constant, which states that 1 mole of electrons carries approximately \( 96500 \, \text{C} \):
\[ \text{Moles of electrons} = \frac{Q}{F} = \frac{9.65 \times 10^4 \, \text{C}}{96500 \, \text{C/mol}} \approx 1 \, \text{mol} \]

### Step 3: Understand the Role of `H2SO4`
In a lead storage battery, `H2SO4` acts as an electrolyte and does not participate directly in the redox reactions. It facilitates the reactions but remains chemically unchanged throughout the process.

### Step 4: Conclusion on Normality
Since the concentration of `H2SO4` does not change during the operation of the battery, the normality of the `H2SO4` solution remains the same.

### Final Answer
The normality of the `H2SO4` solution remains unchanged.

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