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 the molecular weight

Answer

B

Explanation

In iodometry, Kl reacts with `K_2Cr_2O_7` in acidic medium to give `I_2` which is titrated against `Na_2S_2O_3`.
`underset"1 M"(" "K_2Cr_2O_7) + 7H_2SO_4 + 6KI to 4K_2SO_4 + Cr_2(SO_4)_3+ underset"6 x 127 parts"(7H_2O + 3I_2)`
`underset"2 M"(" "2Na_2S_2O_3) +I_2 to underset"2 x 147 parts"(Na_2S_4O_6+2NaI)`
For 1 M `Na_2S_2O_3` , 127 parts of `I_2` is required which is obtained from 1/6 M of `K_2Cr_2O_7`. Hence, eq. wt. of `K_2Cr_2 O_7` is Mol. wt./2