In the neutralization of `Na_(2)S_(2)O_(3)` using `K_(2)Cr_(2)O_(7)` by idometry, 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

`Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)rarr2Cr+7H_(2)O` since `K_(2)Cr_(2)O_(7)` accepts 6 electrons for its redcution to `Cr^(3+)` ions
`therefore` Eq wt=Mol wt/ 6