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

meq of `NaBrO_(3)=N xx V_(mL)=85.5 xx 0.672`
=57.456
(i) meq of `NaBrO_(3) =w/E xx 1000 =57.456 (or) w/(151//6) xx 1000=57.456 rArr w=1.446g`
But, `"Molarity"=("Normality")/("Valency factor")=(0.672)/(6)=0.112M`
(ii) Solve as part (i) value of valency factro is 5 w=1.735g, M=0.1344M