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

In the above chemical reaction, the equivalent mass of $K B r O_{3}$ is:

167

No worries! We‘ve got your back. Try BYJU‘S free classes today!

27.83

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

55.67

No worries! We‘ve got your back. Try BYJU‘S free classes today!

83.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D 27.83
$a N_{2} H_{4} + b B r O_{3}^{-} ⟶ a N_{2} + b B r^{-} + 6 H_{2} O$
+5 -1
Each Br gains 6 electrons in this reaction.

Therefore, the equivalent weight of $K B r O_{3} = \frac{Molar mass of K B r O_{3}}{No of electron gain by Br} = \frac{167}{6} = 27.83$ .

Hence, the correct option is $(B)$ .

Suggest Corrections

Similar questions

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

N_{2} H_{4}

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

a

b

2 K^{+} B r O_{3}^{-} + 12 H^{+} + 10 e^{-} \to B r_{2} + 6 H_{2} O + 2 K^{+}

K B r O_{3}

=

K B r O_{3}

2 K^{+} + B r O^{3 -} + 12 H^{+} + 10 e^{-} \to B r_{2} + 6 H_{2} O + 2 K

K B r O_{3}

N H_{3}

4 N H_{3} + 5 O_{2} ⟶ 4 N O + 6 H_{2} O

2 N H_{3} ⟶ N_{2} + 3 H_{2}

Join BYJU'S Learning Program