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The weight of sodium bromate required to...

The weight of sodium bromate required to prepare of solution for cell reaction,

A

`1.56` gm

B

`1.45` gm

C

`1.23` gm

D

`1.32` gm

Text Solution

Verified by Experts

The correct Answer is:
B
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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