For the reaction 3 Br 2-6 OH - 5 Br - BrO 3-3 H 2 OEquivalent weight of Br 2 is - M - mol wt- of -Br 2-A- M-fB-C- M -2- M -10D-

The correct option is C (M/2+M/10 ) 0Br_2→ 1Br^ n factor = 1 × 2 = 2 0Br_2→+5BrO^ _5 n factor = 5 × 2 = 10 eq wt = (M/2+M/10 ) ...

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Ion Electron Method

For the react...

Question

For the reaction 3 $B r_{2} + 6 O H^{-} \to 5 B r^{-} + B r O_{3}^{-} + 3 H_{2} O$
Equivalent weight of $B r_{2}$ is [M = mol wt of $B r_{2}$ ]

$\frac{M}{2}$

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$\frac{M}{10}$

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$(\frac{M}{2} + \frac{M}{10})$

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$\frac{M}{6}$

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For the reaction

$3 {B r}_{2}$ + $6 {O H}^{Θ}$ $\to$ $5 {B r}^{Θ}$ + $B r {O_{3}}^{Θ}$ + $3 H_{2} O$

Equivalent weight of ${B r}_{2}$ (Molecular weight M) is

B r_{2}

B r^{-}

{B r O_{3}}^{-}

3 B r_{2} + 6 O H^{-} \to 5 B r^{-} + {B r O_{3}}^{-} + 3 H_{2} O

B r_{2}

= M

3 B r_{2} + 6 O H^{⊖} \to 5 B r^{⊖} + B r O_{3}^{⊖} + 3 H_{2} O

B r_{2}

M

5 B r^{-} + B r O_{3}^{-} + 6 H^{+} \to 3 B r_{2} + 3 H_{2} O

B r_{2}

M

B r_{2}

3 B r_{2} + 6 O H^{-} \to 5 B r^{-} + B r O_{3}^{-} + 3 H_{2} O

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