Bromine disproportionates in hot alkali: 3Br2 + 6OH- -> 5Br- + BrO3- + 3H2O (Br=80). The equivalent mass of Br2 in this reaction is:

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Bromine disproportionates in hot alkali: $3 Br X_{2} + 6 OH X^{-} 5 Br X^{-} + BrO X_{3} X^{-} + 3 H X_{2} O$ ( $Br = 80$ ). The equivalent mass of $Br X_{2}$ in this reaction is:

For

3

mol

Br X_{2}

, electrons gained

= 5

(five

Br X^{-}

formed), electrons lost

= 5

(

Br \to BrO X_{3} X^{-}

0 \to + 5

). Hence the n-factor of

Br X_{2}

5/3

and equivalent mass

= 160 / (5/3) = 96