The weight of sodium bromate required to prepare of solution for cell reaction,

Options

A. 1.56gm
B. 1.45gm
C. 1.23gm
D. 1.32gm

Explanation

To solve the problem of finding the weight of sodium bromate (NaBrO3) required to prepare a specific solution for a cell reaction, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Cell Reaction:**
The cell reaction involves the reduction of bromate ions (BrO3⁻). The half-reaction can be written as:
\[ 6H^+ + 6e^- \rightarrow Br^- + 3H_2O \]

2. **Determine the Oxidation States:**
- In NaBrO3, the oxidation state of bromine (Br) is +5.
- In the product (Br⁻), the oxidation state of bromine is -1.
- The change in oxidation state is from +5 to -1, which is a change of 6 units.

3. **Calculate the Equivalent Weight:**
The equivalent weight of sodium bromate can be calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{Valency Change}} \]
- The molar mass of NaBrO3 is approximately 151 g/mol (Na = 23, Br = 80, O3 = 48).
- The valency change is 6 (as calculated from the oxidation state change).
\[ \text{Equivalent Weight} = \frac{151}{6} \approx 25.17 \text{ g/equiv} \]

4. **Calculate the Mass Equivalent:**
The mass equivalent can be calculated using the formula:
\[ \text{Mass Equivalent} = \text{Volume (L)} \times \text{Normality (N)} \]
- Given volume = 85.5 mL = 0.0855 L
- Given normality = 0.672 N
\[ \text{Mass Equivalent} = 0.0855 \times 0.672 \approx 0.05745 \text{ g} \]

5. **Calculate the Weight of Sodium Bromate:**
Using the mass equivalent and equivalent weight, we can find the weight of sodium bromate:
\[ \text{Weight} = \text{Mass Equivalent} \times \text{Equivalent Weight} \]
Rearranging gives:
\[ \text{Weight} = \text{Mass Equivalent} \times \frac{\text{Equivalent Weight}}{1000} \]
Substituting the values:
\[ \text{Weight} = 57.45 \times \frac{151}{6} \approx 1.446 \text{ g} \]

### Final Answer:
The weight of sodium bromate required is approximately **1.446 grams**.

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