Plz answer as soon aspossible

1.	Plz answer as soon aspossible
Answer» Let the MOLECULAR weight of KBrO₃ be 'M'. Oxidation state of Br in BrO₃⁻ : x + 3(-2) = -1 x - 6 = -1 x = +5 Oxidation state of Br in Br⁻ : x = -1 Thus, the oxidation state of Br in the GIVEN reaction, CHANGES from +5 to -1. → n-factor = change in oxidation state = 5 - (-1) = 6 Therefore Equivalent weight of KBrO₃ = M/n-factor = M/6 Hope it helps!!

Answer»

Let the MOLECULAR weight of KBrO₃ be 'M'.

Oxidation state of Br in BrO₃⁻ :

x + 3(-2) = -1

x - 6 = -1

x = +5

Oxidation state of Br in Br⁻ :

x = -1

Thus, the oxidation state of Br in the GIVEN reaction, CHANGES from +5 to -1.

→ n-factor = change in oxidation state = 5 - (-1) = 6

Therefore Equivalent weight of KBrO₃ = M/n-factor = M/6

Hope it helps!!

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