What is the weight of sodium bromate and molarity of solution to prepare `85.5 mL` of `0.672N` solution when half cell reaction are:
(ii) `2BrO_(3)^(-)+12H^(+)+10e^(-)rarrBr_(2)+6H_(2)O`

Explanation

To solve the problem of finding the weight of sodium bromate (NaBrO₃) and the molarity of the solution needed to prepare 85.5 mL of a 0.672 N solution, we can follow these steps:

### Step 1: Determine the n-factor of bromate (BrO₃⁻)
The half-reaction provided is:
\[ 2 \text{BrO}_3^{-} + 12 \text{H}^{+} + 10 e^{-} \rightarrow \text{Br}_2 + 6 \text{H}_2\text{O} \]

In this reaction, each bromate ion (BrO₃⁻) is reduced to bromine (Br₂), which involves a change in oxidation state. The oxidation state of bromine in BrO₃⁻ is +5, and in Br₂ it is 0. Therefore, the change in oxidation state is:
\[ +5 \text{ to } 0 \]
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