Final answer:
The molar mass of silver chromate (Ag2CrO4) is 375.87 g/mol. The sample contains 4.5 × 10^21 formula units of silver chromate. There are 9 × 10^21 silver ions in the sample.
Step-by-step explanation:
The molar mass of silver chromate (Ag2CrO4) can be calculated by adding up the atomic masses of each element in the compound. The atomic mass of silver (Ag) is 107.87 g/mol, chromium (Cr) is 52.00 g/mol, and oxygen (O) is 16.00 g/mol. So, the molar mass of Ag2CrO4 is:
(2 * 107.87) + 52.00 + (4 * 16.00) = 375.87 g/mol.
a. To find the number of formula units of silver chromate, we can use Avogadro's number (6.022 × 10^23 formula units per mole). So, the number of formula units would be:
2.8 g * (1 mol / 375.87 g) * (6.022 × 10^23 formula units / 1 mol) = 4.5 × 10^21 formula units.
b. Since there are 2 silver (Ag) ions in each formula unit of Ag2CrO4, the number of silver ions would be twice the number of formula units, which is 9 × 10^21 silver ions.
c. The mass of one formula unit of silver chromate is equal to its molar mass in grams divided by Avogadro's number:
375.87 g/mol / (6.022 × 10^23 formula units/mol) ≈ 6.24 × 10^-23 g.