What is the weight of sodium bromate and molarity of solution to prepare `85.5 mL` of `0.672N` solution when half cell reaction are:
(i) `BrO_(3)^(-)rarr6H^(+)+6e^(-)rarrBr^(-)+3H_(2)O`
(ii) `2BrO_(3)^(-)+12H^(+)+10e^(-)rarrBr_(2)+6H_(2)O`

Explanation

(a). `BrO_(3)^(ɵ)+6H^(o+)+6e^(-)toBr^(ɵ)+3H_(2)O`
The oxidation number of Br chages from `+5` to `-1`
(Change of 6)`
n factors`=6`
Normality`=nxx` molarity
Molarity`=(0.672)/(6)=0.112M`
Mass of `NaBrO_(3)=` Molarity`xxV_(mL)xx10^(-3)` mol mass
`=0.112xx85.5xx10^(-3)xx151`
`=1.446g`
(b). `2BrO_(3)^(ɵ)+12H^(o+)+10e^(-)toBr_(2)+6H_(2)O`
The oxidation number of Br changes from `+5` to `0`.
(Change of 5)`
n factor`=5`
Normality `=nxx` molarity
molarity`=(0.672)/(6)0.1344M`
Mass of `NaBrO_(3)=` Molarity`xxV_(mL)xx10^(-3)` mol mass
`=0.1344xx85.5xx10^(-3)xx151`
`=1.735g`