To prepare `0.1M KMnO_(4)` solution in 250mL flask, the weight of `KMnO_(4)` required is:

Options

A. `4.80g`
B. `3.95g`
C. `39.5g`
D. `0.48g`

Explanation

To prepare a 0.1 M KMnO₄ solution in a 250 mL flask, we need to calculate the weight of KMnO₄ required. Here’s a step-by-step solution:

### Step 1: Understand Molarity
Molarity (M) is defined as the number of moles of solute per liter of solution. The formula for molarity is:
\[ M = \frac{n}{V} \]
where \( n \) is the number of moles of solute and \( V \) is the volume of the solution in liters.

### Step 2: Convert Volume to Liters
The volume of the solution given is 250 mL. We need to convert this to liters:
\[ V = 250 \text{ mL} = \frac{250}{1000} = 0.25 \text{ L} \]

### Step 3: Calculate Moles of KMnO₄
We know the molarity (0.1 M) and the volume (0.25 L), so we can rearrange the molarity formula to find the number of moles (\( n \)):
\[ n = M \times V = 0.1 \, \text{mol/L} \times 0.25 \, \text{L} = 0.025 \, \text{mol} \]

### Step 4: Find the Molar Mass of KMnO₄
The molar mass of KMnO₄ can be calculated as follows:
- K (Potassium) = 39.10 g/mol
- Mn (Manganese) = 54.94 g/mol
- O (Oxygen) = 16.00 g/mol × 4 = 64.00 g/mol

Adding these together:
\[ \text{Molar mass of KMnO₄} = 39.10 + 54.94 + 64.00 = 158.04 \, \text{g/mol} \]

### Step 5: Calculate the Weight of KMnO₄ Required
Now, we can calculate the weight of KMnO₄ required using the formula:
\[ \text{Weight} = n \times \text{Molar mass} \]
Substituting the values we found:
\[ \text{Weight} = 0.025 \, \text{mol} \times 158.04 \, \text{g/mol} = 3.951 \, \text{g} \]

### Step 6: Round the Answer
Rounding to two decimal places, the weight of KMnO₄ required is approximately:
\[ \text{Weight} \approx 3.95 \, \text{g} \]

### Final Answer
To prepare a 0.1 M KMnO₄ solution in a 250 mL flask, you will need **3.95 grams of KMnO₄**.

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