Final answer:
To calculate the mass of potassium permanganate with 2.40 x 10^24 oxygen atoms, you divide the number of oxygen atoms by Avogadro's number to get moles of oxygen, divide that by 4 (since there are four oxygen atoms in KMnO4) to find moles of KMnO4, and then multiply by the molar mass of KMnO4 to get the mass in grams, which is 157.38 grams.
Step-by-step explanation:
To determine how many grams of potassium permanganate (KMnO4) contain 2.40 x 1024 oxygen atoms, we must first understand the molar mass of the compound and Avogadro's number, which is 6.022 x 1023 atoms/mol. Potassium permanganate is composed of one potassium (K) atom, one manganese (Mn) atom, and four oxygen (O) atoms.
The molar mass of KMnO4 is approximately 158.04 g/mol (39.10 for K, 54.94 for Mn, and 4x16.00 for O). The number of moles of oxygen atoms we have is given by the number of oxygen atoms divided by Avogadro's number, which is 2.40 x 1024 / 6.022 x 1023 mol = 3.985 moles of oxygen atoms. Since there are four oxygen atoms in each molecule of KMnO4, this corresponds to 3.985 / 4 = 0.99625 moles of KMnO4. To find the mass of KMnO4 that contains this many moles, we multiply the number of moles by the molar mass: 0.99625 moles x 158.04 g/mol = 157.38 grams of potassium permanganate.