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:
(i) `BrO_(3)^(-)rarr6H^(+)+6e^(-)rarrBr^(-)+3H_(2)O`

Explanation

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

### Step 1: Determine the n-factor
The half-cell reaction provided is:
\[ \text{BrO}_3^{-} + 6 \text{H}^{+} + 6 \text{e}^{-} \rightarrow \text{Br}^{-} + 3 \text{H}_2\text{O} \]

In this reaction, the oxidation state of bromine changes from +5 in bromate (\( \text{BrO}_3^{-} \)) to -1 in bromide (\( \text{Br}^{-} \)). The change in oxidation state is:
\[ +5 \rightarrow -1 \]
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