A current of `9.65` ampere is passed through the aqueous solution `NaCI` using suitable electrodes for `1000s`. The amount of `NaOH` formed during electrolysis is

Options

A. `2.0g`
B. `4.0g`
C. `6.0g`
D. `8.0g`

Explanation

To find the amount of NaOH formed during the electrolysis of an aqueous NaCl solution, we can use Faraday's laws of electrolysis. Here’s a step-by-step solution:

### Step 1: Determine the parameters
We have:
- Current (I) = 9.65 A
- Time (t) = 1000 s
- Faraday's constant (F) = 96500 C/mol
- Molecular weight of NaOH = 40 g/mol
- Valency factor of NaOH = 1

### Step 2: Calculate the total charge (Q)
Using the formula:
\[ Q = I \times t \]
Substituting the values:
\[ Q = 9.65 \, \text{A} \times 1000 \, \text{s} = 9650 \, \text{C} \]

### Step 3: Calculate the number of equivalents of NaOH formed
Using Faraday's first law of electrolysis:
\[ \text{Number of equivalents} = \frac{Q}{F} \]
Substituting the values:
\[ \text{Number of equivalents} = \frac{9650 \, \text{C}}{96500 \, \text{C/mol}} = 0.1 \, \text{mol} \]

### Step 4: Calculate the amount of NaOH formed
Using the formula:
\[ \text{Weight of NaOH} = \text{Number of equivalents} \times \text{Equivalent weight of NaOH} \]
Since the equivalent weight of NaOH is equal to its molecular weight (40 g/mol) because its valency factor is 1:
\[ \text{Weight of NaOH} = 0.1 \, \text{mol} \times 40 \, \text{g/mol} = 4.0 \, \text{g} \]

### Final Answer
The amount of NaOH formed during electrolysis is **4.0 grams**.

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