The equivalent weight of $Na_2S_2O_3$ (Gram molecular weight $= M$) in the given reaction is $I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$.

  • A
    $M/2$
  • B
    $M$
  • C
    $2M$
  • D
    $M/4$

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