Question

plz answer as soon aspossible

Answer 1

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

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