In the reaction
`Br_(2) +Na_(2) CO_(3) rarr NaBr + NaBrO_(3) + CO_(2)`
The equiv. wt. of `NaBrO_(3)` is

Options

A. `( "Mol. Wt. ")/( 1) `
B. `( "Mol. Wt. ")/( 10)`
C. `( "Mol. wt. ")/( 5)`
D. `( "Mol. wt.")/(4)`

Explanation

To find the equivalent weight of NaBrO₃ in the reaction:

\[ \text{Br}_2 + \text{Na}_2\text{CO}_3 \rightarrow \text{NaBr} + \text{NaBrO}_3 + \text{CO}_2 \]

we will follow these steps:

### Step 1: Determine the oxidation states of bromine in the reactants and products.
- In Br₂, the oxidation state of bromine (Br) is 0.
- In NaBrO₃, we need to find the oxidation state of Br. Let's denote it as \( x \). The oxidation states of sodium (Na) and oxygen (O) are +1 and -2, respectively.

### Step 2: Set up the equation for the oxidation state of Br in NaBrO₃.
The formula for NaBrO₃ can be expressed as:
\[ \text{Na} (+1) + \text{Br} (x) + 3 \times \text{O} (-2) = 0 \]
This gives us the equation:
\[ 1 + x - 6 = 0 \]
Solving for \( x \):
\[ x = +5 \]

### Step 3: Calculate the change in oxidation state.
- The change in oxidation state for bromine from 0 in Br₂ to +5 in NaBrO₃ is:
\[ \text{Change} = 5 - 0 = 5 \]

### Step 4: Determine the molecular weight of NaBrO₃.
- The molecular weight of NaBrO₃ can be calculated as follows:
- Na: 23 g/mol
- Br: 80 g/mol
- O (3 atoms): 3 × 16 g/mol = 48 g/mol
- Therefore, the molecular weight of NaBrO₃ is:
\[ 23 + 80 + 48 = 151 \text{ g/mol} \]

### Step 5: Calculate the equivalent weight of NaBrO₃.
- The equivalent weight is given by the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n} \]
Where \( n \) is the change in oxidation number, which we found to be 5.
- Thus, the equivalent weight of NaBrO₃ is:
\[ \text{Equivalent weight} = \frac{151 \text{ g/mol}}{5} = 30.2 \text{ g/equiv} \]

### Final Answer:
The equivalent weight of NaBrO₃ is **30.2 g/equiv**.

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