A constant current was flown for 2 hour through a solution of KI At the end of experiment liberated lodlne consumed 21.75 mL of 0.0831 M solution of sodium thiosulphate followihg the reaction: `I_(2)+2S_(2)O_(3)^(2-) to 2I^(-) +S_(4)O_(6)^(2-)`. What was the aveagre rate of current flow in ampere?

Explanation

To solve the problem step by step, we need to calculate the average rate of current flow in amperes based on the given information about iodine liberation and sodium thiosulphate consumption.

### Step 1: Calculate the number of moles of sodium thiosulphate (Na2S2O3)

Given:
- Volume of Na2S2O3 solution = 21.75 mL = 0.02175 L (convert mL to L by dividing by 1000)
- Molarity of Na2S2O3 = 0.0831 M

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