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Q.

The equivalent weight of iodine in the reaction (M is molecular weight)
2Na2S2O3 + I2 → 2NaI + Na2S4O6 is

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a

M

b

M/2

c

2M

d

M/3

 

answer is B.

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Detailed Solution

Oxidation state of iodine in I2 is zero and in NaI, it is -1. As there are two moles of NaI, therefore -1 becomes -2.

Equivalent weight is M/2.

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