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

Options

A. `1.56` gm
B. `1.45` gm
C. `1.23` gm
D. `1.32` gm

Answer

B

Explanation

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