36 " mL of " 0.5 M `Br_(2)` solution when made alkaline undergoes complete disproportionation into `Br^(ɵ)` and `BrO_(3)^(ɵ)`. The resulting solution requried 45 " mL of " As (III) solution to reduce `BrO_(3)^(ɵ)` to `Br^(ɵ)`. Given that as (III) is oxidised to As (V), what is the molarity of As (III) solution?

Options

A. 0.2
B. 0.1
C. 0.4
D. 0.5

Explanation

To solve the problem step by step, we will follow the stoichiometric principles and calculations based on the information provided.

### Step 1: Calculate the moles of `Br2`
Given:
- Volume of `Br2` solution = 36 mL = 0.036 L
- Molarity of `Br2` = 0.5 M

Using the formula for moles:
\[ \text{Moles of } Br_2 = \text{Molarity} \times \text{Volume} = 0.5 \, \text{mol/L} \times 0.036 \, \text{L} = 0.018 \, \text{mol} \]

### Step 2: Determine the moles of `BrO3^-` produced
From the stoichiometry of the disproportionation reaction:
\[ 6 \, Br_2 \rightarrow 2 \, BrO_3^- + 4 \, Br^- \]
This means:
- 6 moles of `Br2` produce 2 moles of `BrO3^-`.

Using the ratio:
\[ \text{Moles of } BrO_3^- = \left( \frac{2}{6} \right) \times \text{Moles of } Br_2 = \left( \frac{2}{6} \right) \times 0.018 \, \text{mol} = 0.006 \, \text{mol} \]

### Step 3: Calculate the moles of `As(III)` required to reduce `BrO3^-`
The reaction between `BrO3^-` and `As(III)` can be represented as:
\[ BrO_3^- + 3 \, As^{3+} \rightarrow Br^- + 3 \, As^{5+} \]
From the stoichiometry:
- 1 mole of `BrO3^-` requires 3 moles of `As(III)`.

Therefore:
\[ \text{Moles of } As^{3+} = 3 \times \text{Moles of } BrO_3^- = 3 \times 0.006 \, \text{mol} = 0.018 \, \text{mol} \]

### Step 4: Calculate the molarity of `As(III)` solution
Given:
- Volume of `As(III)` solution = 45 mL = 0.045 L

Using the formula for molarity:
\[ \text{Molarity of } As^{3+} = \frac{\text{Moles of } As^{3+}}{\text{Volume}} = \frac{0.018 \, \text{mol}}{0.045 \, \text{L}} = 0.4 \, \text{M} \]

### Final Answer
The molarity of the `As(III)` solution is **0.4 M**.

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