The equivalent weight of $Na_2S_4O_6$ in the reaction $2Na_2S_2O_3 + I_2 \to Na_2S_4O_6 + 2NaI$ is :-

  • A
    $M$
  • B
    $\frac{M}{8}$
  • C
    $\frac{M}{0.5}$
  • D
    $\frac{M}{2}$

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