18 " mL of " 1.0 M `Br_(2)` solution undergoes complete disproportionation in basic medium to `Br^(The hardness of water in terms of )` and `BrO_(3)^(ɵ)`. Then the resulting solution required 45 " mL of " `As^(3+)` solution to reduce `BrO_(3)^(ɵ)` to `Br^(ɵ)`. `As^(3+)` is oxidised to `As^(5+)` which statements are correct?

Options

A. `Ew(Br_(2))=(M)/(10)`
B. `Ew(Br_(2))=(5M)/(3)`
C. Molarity of `As^(+3)=0.4M`
D. Molarity of `As^(3+)=0.2M`

Explanation

To solve the problem step by step, we will analyze the disproportionation of bromine in basic medium and the subsequent reaction with arsenic.

### Step 1: Understand the Disproportionation Reaction
Bromine (Br₂) undergoes disproportionation in a basic medium to form bromide ions (Br⁻) and bromate ions (BrO₃⁻). The balanced reaction can be written as:
\[ 6 \text{Br}_2 \rightarrow 10 \text{Br}^- + 2 \text{BrO}_3^- \]

### Step 2: Calculate Millimoles of Br₂
Given that we have 18 mL of a 1.0 M Br₂ solution, we can calculate the millimoles of Br₂:
\[ \text{Millimoles of Br}_2 = \text{Volume (mL)} \times \text{Molarity (M)} = 18 \, \text{mL} \times 1.0 \, \text{mmol/mL} = 18 \, \text{mmol} \]

### Step 3: Determine Millimoles of BrO₃⁻
From the balanced reaction, we know that 6 moles of Br₂ produce 2 moles of BrO₃⁻. Therefore, we can find the millimoles of BrO₃⁻ produced:
\[ \text{Millimoles of BrO}_3^- = \frac{2}{6} \times \text{Millimoles of Br}_2 = \frac{2}{6} \times 18 \, \text{mmol} = 6 \, \text{mmol} \]

### Step 4: Calculate Milliequivalents of BrO₃⁻
The n-factor for BrO₃⁻ when it is reduced to Br⁻ is 6 (since it gains 6 electrons). Thus, the milliequivalents of BrO₃⁻ can be calculated as:
\[ \text{Milliequivalents of BrO}_3^- = \text{Millimoles} \times \text{n-factor} = 6 \, \text{mmol} \times 6 = 36 \, \text{meq} \]

### Step 5: Relate Milliequivalents of As³⁺
The problem states that 45 mL of As³⁺ solution is required to reduce BrO₃⁻ to Br⁻. The n-factor for As³⁺ when it is oxidized to As⁵⁺ is 2. Therefore, we can express the milliequivalents of As³⁺ as:
\[ \text{Milliequivalents of As}^{3+} = \text{Volume (mL)} \times \text{Molarity (M)} \times \text{n-factor} \]
Let the molarity of As³⁺ be \( x \):
\[ \text{Milliequivalents of As}^{3+} = 45 \, \text{mL} \times x \times 2 \]

### Step 6: Set Up the Equation
Since the milliequivalents of BrO₃⁻ must equal the milliequivalents of As³⁺, we can set up the equation:
\[ 36 = 45 \times x \times 2 \]

### Step 7: Solve for Molarity \( x \)
Now we solve for \( x \):
\[ 36 = 90x \implies x = \frac{36}{90} = 0.4 \, \text{M} \]

### Conclusion
The molarity of the As³⁺ solution is 0.4 M.