100 mL of 0.01 M XO(4)^(-) is reduced to...

100 mL of 0.01 `M XO_(4)^(-)` is reduced to `X^(n+)` by 100 mL of 0.05 M `Fe^(2+)` in acidic medium. Thus oxidation state of X in `X^(n+)` is ______________

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The correct Answer is:

2

`100xx0.05xx1=100xx0.01xx(7-n)n=2`

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