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A sample of silver chromite (Ag2CrO4) has a mass of 2.8g

What is the molar mass?
a. How many formula units of silver chromate are there?
b. How many silver ions are there?
c. What is the mass of one formula unit of silver chromate?

1 Answer

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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.

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