Final answer:
To prepare a 350 mL of a 0.75 M solution of potassium permanganate (KMnO4), 41.46 grams of KMnO4 are required, based on its molar mass of 158.04 g/mol and using the formula for molarity.
Step-by-step explanation:
To calculate how many grams of potassium permanganate (KMnO4) are needed to make a 0.75 M solution, we need to perform the following steps:
- Determine the molar mass of KMnO4.
- Calculate the number of moles needed for the given volume of solution.
- Convert the moles of KMnO4 to grams.
Firstly, the molar mass of KMnO4 is calculated by summing the atomic masses of potassium (K), manganese (Mn), and four oxygen atoms (O), which gives us approximately 158.04 g/mol. Then, to find the number of moles needed, we use the formula:
moles KMnO4 = Molarity (M) × Volume of solution (L)
In this case:
moles KMnO4 = 0.75 M × 0.350 L = 0.2625 mol
To find out how many grams that is, we multiply the moles by the molar mass:
mass KMnO4 = 0.2625 mol × 158.04 g/mol = 41.46 g
Therefore, 41.46 grams of KMnO4 are needed to prepare a 350 mL of a 0.75 M solution.