In the chemical change,

$a N_{2} H_{4} + b {B r O}_{3}^{-} ⟶ a N_{2} + b B r^{-} + 6 H_{2} O$ ,

The equivalent mass of $N_{2} H_{4}$ in the above reaction is:

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10.6

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6.4

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Solution

The correct option is A 8
$a N_{2} H_{4} + b B r O_{3}^{-} ⟶ a N_{2} + b B r^{-} + 6 H_{2} O$
-2 0
As each N loses $2$ electrons, total number of electrons lost by $N_{2} H_{4} = 4$ .

Equivalent weight of $N_{2} H_{4} = \frac{M a s s o f N_{2} H_{2}}{N o . o f e l e c t r o n s l o s t} = \frac{32}{4} = 8$ .

Hence, the correct option is $(A)$

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Similar questions

a N_{2} H_{4} + b {B r O}_{3}^{-} ⟶ a N_{2} + b B r^{-} + 6 H_{2} O

K B r O_{3}

a N_{2} H_{4} + b {B r O}_{3}^{-} ⟶ a N_{2} + b B r^{-} + 6 H_{2} O

a

b

N_{2} H_{4}

N_{2} H_{4} ⟶ N_{2}

N_{2} H_{4}

N_{2} H_{4} ⟶ N_{2}

16

N_{2}

N_{2} H_{4} + O H^{-} \to N_{2} + H_{2} O + e^{-}

N_{2} H_{4}

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