1.

The weight of sodium bromate required to prepareofsolution for cell reaction,

Answer»

1.56gm
1.45gm
1.23gm
1.32gm

Solution :Meg .pf `NaBrO_(3)=85.5xxo.672=57.456`
LET WEIGHT of `NaBrO_(3)`=W
`therefore W/(M_(NaBrO_(3)))xx6xx1000=57.456("equivalent weight"=M//6`)of n-factor=6
Hence ,(B)is the correct anwer.


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