(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

(i) What is the mass of sodium bromate and molarity of the solution ne

1.	(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
Answer» Solution :(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`