Equivalent weight of Baeyer's reagent is

Options

A. 158
B. 31.6
C. 52.6
D. All of these

Answer

To determine the equivalent weight of Baeyer's reagent, we follow these steps:

### Step 1: Identify the Chemical Composition
Baeyer's reagent is essentially a 1% alkaline solution of potassium permanganate (KMnO₄).

### Step 2: Determine the Molar Mass of KMnO₄
The molar mass of KMnO₄ can be calculated by adding the atomic masses of its constituent elements:
- Potassium (K): 39.1 g/mol
- Manganese (Mn): 54.9 g/mol
- Oxygen (O): 16.0 g/mol (there are 4 oxygen atoms)

Calculating the total:
\[ \text{Molar mass of KMnO}_4 = 39.1 + 54.9 + (4 \times 16.0) \]
\[ = 39.1 + 54.9 + 64.0 \]
\[ = 158.0 \, \text{g/mol} \]

### Step 3: Determine the M Factor
The M factor (or n factor) is the number of electrons transferred in the redox reaction. In the case of KMnO₄ being reduced to MnO₂, the manganese ion (Mn) undergoes a change in oxidation state from +7 in KMnO₄ to +4 in MnO₂. This means that 3 electrons are gained during the reduction process.

### Step 4: Calculate the Equivalent Weight
The equivalent weight can be calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{M Factor}} \]

Substituting the values we have:
\[ \text{Equivalent Weight} = \frac{158.0 \, \text{g/mol}}{3} \]
\[ = 52.67 \, \text{g/equiv} \]

### Conclusion
The equivalent weight of Baeyer's reagent (KMnO₄) is approximately **52.67 g/equiv**.

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