An impure sample of `As_(2)O_(3)` weighitng `20 gm` was dissolved in water containing `10 gm` of `NaHCO_(3)` and the final solution was diluted to `400ml`.`20ml` of this solution was oxidised by `25ml` of a solution of `I_(2)`. Another solution of hypo containing `1.24 gm` of `Na_(2)S_(2)O_(3).5H_(2)O` in `40ml` was used to exactly reduce `25 ml` of same `I_(2)` solution (impurities inert). Calculate `%` of `AS_(2)O_(3)` in the sample (at. wt. of `As=75`)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the molarity of the hypo solution

1. **Given Data**:
- Mass of Na2S2O3·5H2O = 1.24 g
- Molar mass of Na2S2O3·5H2O = 248 g/mol
- Volume of hypo solution = 40 mL

2. **Calculate Molarity**:
\[ \text{Molarity (M)} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)} \times \text{volume (L)}} \]
\[ \text{Molarity} = \frac{1.24 \, \text{g}}{248 \, \text{g/mol} \times 0.040 \, \text{L}} = 0.125 \, \text{mol/L} \]

### Step 2: Calculate the milliequivalents of the hypo solution

1. **Milliequivalents of hypo solution**:
\[ \text{Milliequivalents} = \text{Molarity (mol/L)} \times \text{Volume (L)} \times 1000 \]
\[ \text{Milliequivalents} = 0.125 \, \text{mol/L} \times 0.025 \, \text{L} \times 1000 = 3.125 \, \text{meq} \]

### Step 3: Relate the milliequivalents of I2 to As2O3

1. **From the problem**: The milliequivalents of I2 that reacted with the hypo solution is equal to the milliequivalents of As2O3 that reacted with I2.
\[ \text{Milliequivalents of I2} = 3.125 \, \text{meq} \]

### Step 4: Calculate the milliequivalents of As2O3

1. **Knowing that 1 mole of As2O3 gives 2 equivalents**:
\[ \text{Milliequivalents of As2O3} = 3.125 \, \text{meq} \]

### Step 5: Calculate the moles of As2O3

1. **Using the relationship**:
\[ \text{Moles of As2O3} = \frac{\text{Milliequivalents}}{2} = \frac{3.125}{1000} \div 2 = 0.0015625 \, \text{mol} \]

### Step 6: Calculate the mass of As2O3

1. **Using the molar mass of As2O3**:
\[ \text{Molar mass of As2O3} = 2 \times 75 + 3 \times 16 = 150 + 48 = 198 \, \text{g/mol} \]
\[ \text{Mass of As2O3} = \text{Moles} \times \text{Molar mass} = 0.0015625 \, \text{mol} \times 198 \, \text{g/mol} = 0.3105 \, \text{g} \]

### Step 7: Calculate the percentage of As2O3 in the sample

1. **Using the total mass of the sample**:
\[ \text{Percentage of As2O3} = \left( \frac{\text{Mass of As2O3}}{\text{Total mass of sample}} \right) \times 100 \]
\[ \text{Percentage of As2O3} = \left( \frac{0.3105 \, \text{g}}{20 \, \text{g}} \right) \times 100 = 1.5525\% \]

### Final Answer:
The percentage of As2O3 in the sample is approximately **1.55%**.

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