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

Answer 1

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.