To prepare a standard potassium permanganate solution (KMnO₄) with a concentration of 0.100 mol/dm³, we first need to determine the number of moles we can obtain from the available mass of KMnO₄, which is 3.95 grams.
Step 1: Calculate the number of moles of KMnO₄.
The molar mass of potassium permanganate (KMnO₄) can be calculated as follows:
- Potassium (K): 39.10 g/mol
- Manganese (Mn): 54.94 g/mol
- Oxygen (O): 16.00 g/mol × 4 = 64.00 g/mol
Now, adding these together gives us the molar mass of KMnO₄:
\[ \text{Molar Mass of KMnO₄} = 39.10 + 54.94 + 64.00 = 158.04 , \text{g/mol} \]
Next, we can calculate the number of moles in 3.95 grams of KMnO₄:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{3.95 , \text{g}}{158.04 , \text{g/mol}} \approx 0.0250 , \text{mol} \]
Step 2: Calculate the volume of solution needed to achieve the desired concentration.
We want to prepare a solution with a concentration of 0.100 mol/dm³. We can use the formula:
\[ \text{Concentration (C)} = \frac{\text{Number of moles (n)}}{\text{Volume (V) in dm³}} \]
Rearranging this gives:
\[ V = \frac{n}{C} = \frac{0.0250 , \text{mol}}{0.100 , \text{mol/dm³}} = 0.250 , \text{dm³} \]
Since \(1 , \text{dm³} = 1 , \text{L}\), this is equivalent to:
\[ V = 0.250 , \text{L} = 250 , \text{mL} \]
Step 3: Choose the appropriate volumetric flask.
We need to prepare 0.250 L (or 250 mL) of the solution. Among the available options, we have:
- 100 L
- 200 L
- 250 L
- 500 L
The most suitable choice is the 250 L volumetric flask, as it matches our required volume.
Summary:
To prepare a 0.100 mol/dm³ solution of KMnO₄, you should use a 250 mL volumetric flask for the 3.95 grams of solid you have, which will yield the desired concentration.