Asked by Fai
A mixture of pure k2cr207 and pure kmn04 weighing 0.561g was treated with excess of ki in acidic medium. iodine liberated required 100ml of 0.15M of sodium thiosulphate solution for exact oxidation.
The right answer k2cr207 43.67%
kmn04 56.33%
K2cr207 =294
kmn04 = 158
x/294 + y/158 (0.561g-x) = 0.015
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The right answer k2cr207 43.67%
kmn04 56.33%
K2cr207 =294
kmn04 = 158
x/294 + y/158 (0.561g-x) = 0.015
Who check me step by step for me
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