Equivalent mass of `KMnO_(4)` in acidic basic and netural are in the ratio, of:

Options

A. `3:5:15`
B. `5:3:1`
C. `5:1:3`
D. `3:15:5`

Answer

To find the ratio of the equivalent mass of KMnO₄ in acidic, basic, and neutral conditions, we will follow these steps:

### Step 1: Determine the Molecular Weight of KMnO₄
The molecular weight of KMnO₄ is given as 158 g/mol.

### Step 2: Identify the Valence Factor (n) in Different Conditions
- In acidic medium, the valence factor (n) is 5.
- In basic medium, the valence factor (n) is 1.
- In neutral medium, the valence factor (n) is 3.

### Step 3: Calculate the Equivalent Mass in Each Condition
The formula for equivalent mass is given by:

\[ \text{Equivalent Mass} = \frac{\text{Molecular Weight}}{\text{Valence Factor}} \]

- **In Acidic Medium:**
\[ \text{Equivalent Mass (Acidic)} = \frac{158}{5} = 31.6 \]

- **In Basic Medium:**
\[ \text{Equivalent Mass (Basic)} = \frac{158}{1} = 158 \]

- **In Neutral Medium:**
\[ \text{Equivalent Mass (Neutral)} = \frac{158}{3} \approx 52.67 \]

### Step 4: Establish the Ratio of Equivalent Masses
Now we can establish the ratio of the equivalent masses in acidic, basic, and neutral conditions:

\[ \text{Ratio} = \text{Equivalent Mass (Acidic)} : \text{Equivalent Mass (Basic)} : \text{Equivalent Mass (Neutral)} = 31.6 : 158 : 52.67 \]

To simplify this ratio, we can multiply each term by a common factor to eliminate decimals:

\[ \text{Ratio} = 31.6 \times 3 : 158 \times 3 : 52.67 \times 3 = 94.8 : 474 : 158 \]

Now, we can simplify this ratio further:

\[ \text{Ratio} \approx 3 : 15 : 5 \]

### Final Answer
Thus, the ratio of the equivalent mass of KMnO₄ in acidic, basic, and neutral conditions is approximately **3 : 15 : 5**.