(i) What is the mass of sodium bromate and molarity of the solution necessary to prepare 85.4 mL of 0.672 N solution when the half reaction is ,
`BrO_(3)^(-) +6H^(+)+6e^(-) to Br^(-)+3H_(2)O`
(ii) What would be the mass as well as molartiy if the half cell reaction is,
`2BrO_(3)^(-)+12H^(+)+10e^(-) to Br_(2)+6H_(2)O`

Explanation

(i) Molecular mass of `NaBrO_(3)=23+80+(3xx16)=151`
Each bromate ion takes-up 6 electrons, therefore,
Eq. mass of `NaBrO_(3)=("Mol.mass")/(6)=(151)/(6)`
Amonut of `NaBrO_(3)` in 85.5 mL 0.672 N solution
`=(0.672)/(1000)xx(151)/(6)xx85.5=1.446 g`
`"Molarity"=("Normality")/(n)=(0.672)/(6)=0.112 M`
(ii) Each bromate ion takes-up 5 electrons, therefore,
Eq. mass of `NaBrO_(3)=("Mol. mass")/(5)=(151)/(5)`
Amount of `NaBrO_(3)` in 85.5 mL 0.672 N solution
`=(151)/(5)xx(0.672)/(1000)xx85.5`
=1.7352 g
`"Molarity"=("Normality")/(n)=(0.672)/(5)=0.1344 M`