How many grams of potassium permanganate

(KMnO4) should be added to a 100 mL volumetric flask to prepare a 0.02505 M solution when the flask is diluted
to the mark with water?

To determine the number of grams of potassium permanganate (KMnO4) needed to prepare a 0.02505 M solution in a 100 mL volumetric flask, we need to use the molar mass of KMnO4 and the formula for calculating molarity.

1. Find the molar mass of KMnO4:
- Potassium (K) has a molar mass of 39.10 g/mol.
- Manganese (Mn) has a molar mass of 54.94 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
- Therefore, the molar mass of KMnO4 is:
39.10 g/mol (K) + 54.94 g/mol (Mn) + 4 * 16.00 g/mol (O) = 158.03 g/mol

2. Use the formula for molarity:
Molarity (M) = moles of solute / volume of solution (in liters)

Rearrange the formula to solve for moles of solute:
Moles of solute = Molarity (M) * Volume of solution (in liters)

In this case, we know the molarity (0.02505 M) and the volume of solution (100 mL = 0.100 L).

Moles of solute = 0.02505 M * 0.100 L = 0.002505 moles of KMnO4

3. Convert moles to grams:
Multiply the number of moles by the molar mass of KMnO4.

Grams of KMnO4 = 0.002505 moles * 158.03 g/mol = 0.3957 grams of KMnO4

Therefore, approximately 0.3957 grams of potassium permanganate (KMnO4) should be added to a 100 mL volumetric flask to prepare a 0.02505 M solution when the flask is diluted to the mark with water.