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

Options

A. (molecular weight)/2
B. (molecular weight)/6
C. (molecular weight)/3
D. same as molecular weight

Answer

B

Explanation

In iodometry, `K_(2)Cr_(2)O_(7)` liberates `I_(2)` from iodides (NaI or KI) which is titrated with `Na_(2)S_(2)O_(3)` solution.
`K_(2)Cr_(2)O_(7)+I^(-)+H^(+) to Cr^(3+)+I_(2)`
Here, one mole of `K_(2)Cr_(2)O_(7)` accepts 6 mole of electrons.
`therefore` Equivalents weight`=("molecular weight")/(6)`