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

`"Cr"_(2)O_(7)^(2-) + 14 "H"^(+) + 6"e"^(-) rarr 2Cr^(3+)+7H_(2)O`
` 6l^(-) rarr 3l_(2) + 6 e^(-)`
`2"Na"_(2) S_(2) O_(3)+l_(2) rarr "Na"_(2) S_(4) O_(6) + 2 "Nal"`
`Cr_(2)O_(7)^(2-)equiv 6l` atom
`therefore " 1 l atom" = (Cr_(2)O_(7)^(2-))/6`
Equivalent weight of `K_(2)Cr_(2)O_(7)=("Molecular weight")/6`