How many gram of `KMnO_(4)` are contained in 4 litre of 0.05 N solution . The `KMmO_(4)` is to be used as an oxidant in acidic medium ?

Options

A. 1.58 g
B. 15.8 g
C. 6.32 g
D. 31 .6 g

Answer

To find out how many grams of \( KMnO_4 \) are contained in 4 liters of a 0.05 N solution, we can follow these steps:

### Step 1: Determine the equivalent mass of \( KMnO_4 \)

The equivalent mass can be calculated using the formula:

\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{n} \]

Where:
- The molar mass of \( KMnO_4 \) is 158 g/mol.
- The \( n \) factor for \( KMnO_4 \) when it acts as an oxidant in acidic medium is 5 (as \( Mn^{+7} \) is reduced to \( Mn^{+2} \)).

Calculating the equivalent mass:

\[ \text{Equivalent mass} = \frac{158 \, \text{g/mol}}{5} = 31.6 \, \text{g/equiv} \]

### Step 2: Use the normality formula to find the mass

The formula for normality is given by:

\[ N = \frac{\text{mass (g)}}{\text{Equivalent mass (g/equiv)} \times \text{Volume (L)}} \]

Rearranging this formula to find the mass:

\[ \text{mass (g)} = N \times \text{Equivalent mass} \times \text{Volume (L)} \]

### Step 3: Substitute the known values into the equation

Given:
- Normality \( N = 0.05 \, N \)
- Equivalent mass \( = 31.6 \, \text{g/equiv} \)
- Volume \( = 4 \, L \)

Substituting these values into the equation:

\[ \text{mass (g)} = 0.05 \times 31.6 \times 4 \]

### Step 4: Calculate the mass

Now, performing the calculation:

\[ \text{mass (g)} = 0.05 \times 31.6 \times 4 = 6.32 \, g \]

### Conclusion

Thus, the mass of \( KMnO_4 \) contained in 4 liters of a 0.05 N solution is **6.32 grams**.

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