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 sodium bromate = `85.5 xx 0.672 = 57.456`
(i) Meq. of `NaBrO_(3) = 57.456`
`w/E xx 1000 = 57.456`
`w/(151//6) xx 1000 = 57.456 (E_(NaBrO_(3)) = M/6 = 151/6)`
`w = 57.456 xx 151 /(6 xx 1000) = 1.446 g`
Also, molarity = normality/n-factor = 0.672/6 = 0.112 M
(ii) Similarly use n factor 5 in place of 6 in this problem,
Hence `w = 57.456 xx 151 /(5 xx 1000) = 1.735 g` and molarity = 0.672/5 = 0.1344 M