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`
(ii) `2BrO_(3)^(-)+12H^(+)+10e^(-)rarrBr_(2)+6H_(2)O`

Explanation

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

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

To find the n-factor, we need to determine the change in oxidation state of bromine (Br):
- In \(\text{BrO}_3^{-}\), the oxidation state of Br is +5.
...