When electrolysis of `KCl` is doen in alkaline medium, `10g` of `KClO_(3)` is produced as follows `:`
`Cl^(c-)+6overset(c-)(O)H rarr ClO_(3) +3H_(2)O+6e^(-)`
A current of `2A` is passed for `10.941 ` hours. Calculate the `((Percent a g e current efficiency)/(10))` used in the process.
`(Mw` of `KClO_(3)=122.5)`

Explanation

To solve the problem, we need to calculate the percentage current efficiency of the electrolysis process and then divide that by 10 as requested. Here's a step-by-step breakdown of the solution:

### Step 1: Calculate the total charge passed during electrolysis
The total charge (Q) can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I \) is the current in amperes (A)
- \( t \) is the time in seconds (s)

Given:
- \( I = 2 \, \text{A} \)
- \( t = 10.941 \, \text{hours} \)

First, convert hours to seconds:
\[ t = 10.941 \, \text{hours} \times 3600 \, \text{s/hour} = 39400 \, \text{s} \]

Now calculate \( Q \):
\[ Q = 2 \, \text{A} \times 39400 \, \text{s} = 78800 \, \text{C} \]

### Step 2: Calculate the moles of \( KClO_3 \) produced
Using the formula:
\[ n = \frac{w}{M} \]
where:
- \( n \) is the number of moles
- \( w \) is the mass of the substance (in grams)
- \( M \) is the molar mass (in g/mol)

Given:
- \( w = 10 \, \text{g} \)
- \( M \) of \( KClO_3 = 122.5 \, \text{g/mol} \)

Calculate the moles of \( KClO_3 \):
\[ n = \frac{10 \, \text{g}}{122.5 \, \text{g/mol}} = 0.08163 \, \text{mol} \]

### Step 3: Calculate the total charge required to produce the moles of \( KClO_3 \)
From the reaction, we see that 6 moles of electrons are required to produce 1 mole of \( KClO_3 \). Therefore, for 0.08163 moles of \( KClO_3 \):
\[ \text{Moles of electrons} = 6 \times 0.08163 = 0.48978 \, \text{mol} \]

Now, calculate the total charge using Faraday's law:
\[ Q_{required} = n \times F \]
where \( F \) (Faraday's constant) is approximately \( 96500 \, \text{C/mol} \):
\[ Q_{required} = 0.48978 \, \text{mol} \times 96500 \, \text{C/mol} = 47200.5 \, \text{C} \]

### Step 4: Calculate the current efficiency
Current efficiency (\( \eta \)) can be calculated using:
\[ \eta = \frac{Q_{actual}}{Q_{required}} \times 100 \]
Where:
- \( Q_{actual} \) is the total charge passed (78800 C)
- \( Q_{required} \) is the charge required (47200.5 C)

Now substituting the values:
\[ \eta = \frac{78800 \, \text{C}}{47200.5 \, \text{C}} \times 100 = 166.88\% \]

### Step 5: Calculate the percentage current efficiency divided by 10
Now, we need to find:
\[ \frac{\eta}{10} = \frac{166.88}{10} = 16.688 \]

### Final Answer
The final answer is approximately:
\[ 16.69 \]