`KMnO_4` reacts with `Na_2S_2O_3` in acidic, strongly basic and aqueous (neutral) media. `100mL` of `LMnO_4` reacts with 100 mL of 0.1 M `Na_2S_2O_3` in acidic, basic and neutral media.
Q. The molarity (M) of `KMnO_4` in aqueous medium is

Options

A. 0.8M
B. 0.08M
C. 0.26M
D. 0.026M

Explanation

To solve the problem of finding the molarity of `KMnO_4` in aqueous medium when it reacts with `Na_2S_2O_3`, we can follow these steps:

### Step 1: Understand the Reaction
In aqueous (neutral) medium, `KMnO_4` (which contains `Mn^7+`) is reduced to `Mn^4+`, while `Na_2S_2O_3` (thiosulfate) is oxidized to `HSO_4^-`.

### Step 2: Write the Equivalence Relationship
The number of equivalents of `KMnO_4` will equal the number of equivalents of `Na_2S_2O_3` in the reaction. The formula for equivalents is given by:
\[ N_1V_1 = N_2V_2 \]
Where:
- \( N_1 \) = normality of `KMnO_4`
- \( V_1 \) = volume of `KMnO_4` (in liters)
- \( N_2 \) = normality of `Na_2S_2O_3`
- \( V_2 \) = volume of `Na_2S_2O_3` (in liters)

### Step 3: Determine the n-factor
- For `KMnO_4` in neutral medium, the n-factor (number of electrons transferred) is 3 (since `Mn^7+` to `Mn^4+` involves a change of 3 electrons).
- For `Na_2S_2O_3`, the n-factor is 8 (since thiosulfate ion is oxidized to sulfate).

### Step 4: Set Up the Equation
Given:
- Volume of `KMnO_4` (V1) = 100 mL = 0.1 L
- Volume of `Na_2S_2O_3` (V2) = 100 mL = 0.1 L
- Molarity of `Na_2S_2O_3` (M2) = 0.1 M

Using the n-factors:
\[ N_1 = M_1 \times n_1 \]
\[ N_2 = M_2 \times n_2 \]

Substituting the known values:
\[ N_1 = M_1 \times 3 \]
\[ N_2 = 0.1 \times 8 \]

### Step 5: Substitute into the Equivalence Equation
Now substituting into the equivalence relationship:
\[ (M_1 \times 3) \times 0.1 = (0.1 \times 8) \times 0.1 \]

### Step 6: Simplify and Solve for M1
\[ M_1 \times 3 \times 0.1 = 0.8 \times 0.1 \]
\[ M_1 \times 0.3 = 0.08 \]
\[ M_1 = \frac{0.08}{0.3} = \frac{8}{30} = \frac{4}{15} \approx 0.267 \text{ M} \]

### Final Answer
The molarity of `KMnO_4` in aqueous medium is approximately **0.267 M**.

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