In the standardization of `Na_(2)S_(2)O_(3)` using `K_(2)Cr_(2)O_(7)` by iodometry, th equivalent weight of `K_(2)Cr_(2)O` is

Options

A. `("Molecular weight")//2`
B. `("Molecular weight")//6`
C. `("Molecular weight")//3`
D. Same as molecular weight

Answer

B

Explanation

In iodometry, `H_(2)Cr_(2)O_(7)` liberates `I_(2)` from iodides (Nal or Kl) which is titrated with `Na_(2)S_(2)O_(3)` solution `K_(2)Cr_(2)O_(7) + I^(-) + H^(+) rarr Cr^(3+) + I_(2)`
Here, one mole of `K_(2)Cr_(2)O_(7)` accepts 6 mole of electrons.
`:.` Equivalent weight `= ("Molecular weight")/(6)`