You want to prepare a standard potassium permanganate solution of concentration 0,100mol•dm³. You have 3,95 of salt available.Determine which of these volumetric flasks you should use:100liters,200liters,250liters,500 liters.Show calculations

1 answer

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.