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Question

In the reaction, I2+2Na2S2O32NaI+Na2S4O6, equivalent weight of Iodine is:

A
Molecularweight
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B
Molecularweight2
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C
Molecularweight4
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D
Molecularweight3
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Solution

The correct option is B Molecularweight2
Oxidation state of I changes from 0 to 1.
Therefore, change in oxidation state is 1.
Change, per mole of iodine molecule is 1×2 (as we have two atoms of I) = 2.
So, Eq. Wt is =molecular weight2.

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