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 when a current of 9.65 amperes is passed for 1000 seconds, we can use Faraday's laws of electrolysis. Here’s a step-by-step solution:

### Step 1: Calculate Total Charge (Q)
First, we need to calculate the total charge (Q) passed through the solution using the formula:
\[ Q = I \times t \]
where:
- \( I \) = current in amperes (9.65 A)
- \( t \) = time in seconds (1000 s)

Substituting the values:
\[ Q = 9.65 \, \text{A} \times 1000 \, \text{s} = 9650 \, \text{C} \]

### Step 2: Calculate Moles of Electrons (n)
Next, we calculate the number of moles of electrons using Faraday's constant (\( F \)), which is approximately 96500 C/mol. The formula is:
\[ n = \frac{Q}{F} \]

Substituting the values:
\[ n = \frac{9650 \, \text{C}}{96500 \, \text{C/mol}} = 0.1 \, \text{mol} \]

### Step 3: Determine Moles of NaOH Produced
In the electrolysis of NaCl, the formation of NaOH can be represented by the following half-reaction:
\[ \text{Na}^+ + \text{OH}^- \rightarrow \text{NaOH} \]

From the balanced equation, we see that 1 mole of NaOH is produced for every mole of electrons transferred. Therefore, the moles of NaOH produced will be equal to the moles of electrons:
\[ \text{Moles of NaOH} = n = 0.1 \, \text{mol} \]

### Step 4: Calculate Mass of NaOH
To find the mass of NaOH produced, we use the molar mass of NaOH, which is approximately 40 g/mol:
\[ \text{Mass of NaOH} = \text{Moles of NaOH} \times \text{Molar Mass of NaOH} \]
\[ \text{Mass of NaOH} = 0.1 \, \text{mol} \times 40 \, \text{g/mol} = 4 \, \text{g} \]

### Final Answer
The amount of NaOH formed during electrolysis is **4 grams**.

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