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

Explanation

In this reaction `Cr_(2)O_(7)^(2-)` is changed into `Cr^(+3)` ion
So, equivalent weight of
`K_(2)Cr_(2)O_(7)= ("mol ecular weight")/("decrese iof ox id a t ion number" xx n)` (where n = Number of atoms of Cr)
`=("mol ecular weight")/(3xx2)=("mol ecular weight")/6`