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Q.

A 60 ml sample solution of KI was electrolysed for 579 sec using a 5 amp constant current. The   produced, completely reacted with 0.20 M sodium thiosulphate solution and unreacted KI solution required 30 ml of 0.04 M acidic solution of KMnO4. If the original molarity of KI solution can be expressed in scientific notation as Y × 10-1. Calculate the value of Y.

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Detailed Solution

nF=579×596500=0.032I−→I2+2e−nF=2nI2nI2=00.015mol⇒nKI=0.03=30×10−3eI−=eKMNO4nI−=301000×.04×5=6×10−3[KI]=36×10−3×10360=0.6[KI]=6×10−1M
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