A 0.1097 gm sample of `As_(2)O_(3)`required 26.10 mL of `Kmno_(4)` solution for its titration. The molarity of solution is :

Options

A. 0.02
B. 0.04
C. 0.018
D. 0.3

Explanation

To find the molarity of the KMnO4 solution used in the titration with As2O3, we can follow these steps:

### Step 1: Determine the molar mass of As2O3
The molar mass of As2O3 can be calculated as follows:
- Arsenic (As) has an atomic mass of approximately 74.92 g/mol.
- Oxygen (O) has an atomic mass of approximately 16.00 g/mol.

The molar mass of As2O3 = (2 × 74.92) + (3 × 16.00) = 149.84 + 48.00 = 197.84 g/mol.

### Step 2: Calculate the number of moles of As2O3
Using the mass of the sample (0.1097 g) and the molar mass (197.84 g/mol), we can find the number of moles of As2O3:
\[ \text{Moles of As2O3} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.1097 \text{ g}}{197.84 \text{ g/mol}} \approx 0.000554 \text{ moles}. \]

### Step 3: Determine the n-factor for As2O3
In the reaction, As2O3 is oxidized from As(III) to As(V). The change in oxidation state for each arsenic atom is 2 (from +3 to +5), and since there are 2 arsenic atoms, the total change is:
\[ \text{n-factor for As2O3} = 2 \times 2 = 4. \]

### Step 4: Calculate the equivalents of As2O3
Using the number of moles and the n-factor, we can find the equivalents of As2O3:
\[ \text{Equivalents of As2O3} = \text{moles} \times \text{n-factor} = 0.000554 \text{ moles} \times 4 = 0.002216 \text{ equivalents}. \]

### Step 5: Determine the n-factor for KMnO4
In the reaction, KMnO4 is reduced from Mn(VII) to Mn(II). The change in oxidation state is:
\[ \text{n-factor for KMnO4} = 7 - 2 = 5. \]

### Step 6: Calculate the equivalents of KMnO4
According to the equivalence principle, the equivalents of As2O3 will equal the equivalents of KMnO4:
\[ \text{Equivalents of KMnO4} = \text{Equivalents of As2O3} = 0.002216 \text{ equivalents}. \]

### Step 7: Calculate the molarity of KMnO4
We know the volume of KMnO4 used is 26.10 mL, which we convert to liters:
\[ \text{Volume in liters} = \frac{26.10 \text{ mL}}{1000} = 0.02610 \text{ L}. \]
Now, we can use the formula for molarity:
\[ \text{Molarity (M)} = \frac{\text{Equivalents}}{\text{Volume in L}} = \frac{0.002216 \text{ equivalents}}{0.02610 \text{ L}} \approx 0.0847 \text{ M}. \]

### Final Answer:
The molarity of the KMnO4 solution is approximately **0.0847 M**.

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