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

Explanation

To solve the problem step by step, we will follow the outlined process to calculate the current efficiency of the electrolysis of KCl in an alkaline medium.

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

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

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

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

### Step 2: Convert charge to Faraday
To convert the charge in coulombs to Faraday (F), use the relationship:
\[ \text{Charge in Faraday} = \frac{Q}{96500} \]
where 96500 C is the charge of one mole of electrons (Faraday's constant).

Now, calculate the charge in Faraday:
\[ \text{Charge in Faraday} = \frac{78800 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.816 \, \text{F} \]

### Step 3: Determine the amount of KClO3 produced theoretically
From the reaction, we know that 6 Faraday produces 1 mole of KClO3 (122.5 g). We can set up a proportion to find out how much KClO3 can be produced with the charge we calculated.

Using the relationship:
\[ \text{grams of KClO3} = \frac{122.5 \, \text{g}}{6 \, \text{F}} \times \text{Charge in Faraday} \]

Substituting the value of Charge in Faraday:
\[ \text{grams of KClO3} = \frac{122.5 \, \text{g}}{6} \times 0.816 \approx 16.67 \, \text{g} \]

### Step 4: Calculate the current efficiency
Current efficiency is calculated using the formula:
\[ \text{Current Efficiency} (\%) = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 \]

Given that the actual yield of KClO3 is 10 g:
\[ \text{Current Efficiency} (\%) = \left( \frac{10 \, \text{g}}{16.67 \, \text{g}} \right) \times 100 \approx 60 \% \]

### Step 5: Calculate \(\frac{\text{Current Efficiency}}{10}\)
Finally, we need to calculate \(\frac{\text{Current Efficiency}}{10}\):
\[ \frac{60}{10} = 6 \]

### Final Answer
The answer is:
\[ 6 \]