In basic medium `CrO_(4)^(2-)` reacts with `S_(2)O_(3)^(2-)` resulting in the formation of `Cr_(OH)_(4)^(ɵ)` and `SO_(4)^(2-)` How many " mL of " 0.1 M `Na_(2)CrO_(4)` is required to react with 40 " mL of " 0.25 M `Na_(2)S_(2)O_(3)`?

Options

A. `240.2mL`
B. `24.02mL`
C. `266.65mL`
D. `26.67`mL

Answer

C

Explanation

`S_(2)O_(3)^(2-)to2SO_(4)^(2-)+8e^(-)`
`3e^(-)+CrO_(4)^(2-)to[Cr(OH)_(4)]^(ɵ)`
`CrO_(4)^(2-)-=S_(2)O_(3)^(2-)`
`mEq=mEq`
`Vxx0.1xx3-=40 mL xx0.25xx8`
`V=266.67 mL`