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. (a)`"molecular weight"//2`
B. (b)`"molecular weight"//6`
C. (c )`"molecular weight"//3`
D. (d)same as molecular weight

Answer

B

Explanation

In this reaction
`Cr_(2)O_(3)^(2-)` is changed into `Cr_(3)^(3+)` ion.
So, equivalent weight of
`K_(2)Cr_(2)O_(7)=("molecular wt".)/("decrease of oxidation number"xxn)`
`("where n= No. of atoms of" Cr)`
`=("molecular wt".)/(3xx2)=("molecular wt.")/6`