1.

The weight of sodium bromate required to prepare 85.5 ml" of " 0.672N solution for cell reaction , BrO_(3)^(-) + 6H^(+) + 6e to Br^(-) + 3H_(2)O, is

Answer»

`1.56` gm
`1.45` gm
`1.23` gm
`1.32` gm

SOLUTION :Meq. of `NaBrO_(3) = 85.5 xx 0.672 = 57.456`
Let weight of `NaBrO_(3) = W`
`:. W/M_(NaBrO_(3)) xx 6 xx 1000 = 57 .456 `(EQUIVALENT weight = M/6) of n-factor = 6
` :. M/151 xx 6 xx 1000 = 57.456`
` :. W = 1.45 ` gm


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