Answer :

Final answer:

The number of molecules of potassium sulfate (K₂SO₄) in 99.9g is calculated by first determining the moles of K₂SO₄ present using its molar mass and then multiplying by Avogadro's number, resulting in approximately 3.45 × 10²² molecules.

Explanation:

The question asks about the number of molecules of potassium sulfate (K₂SO₄) present in 99.9g of the compound. To calculate the number of molecules, we first need to determine the number of moles present in 99.9g of K₂SO₄. This is done by using the molar mass of K₂SO₄.

The molar mass of K₂SO₄ is calculated by adding the molar masses of all the atoms in the formula: (2 × molar mass of K) + molar mass of S + (4 × molar mass of O). According to the periodic table, the respective molar masses are approximately 39.10 g/mol for K, 32.07 g/mol for S, and 16.00 g/mol for O. Therefore, the molar mass of K₂SO₄ is (2 × 39.10) + 32.07 + (4 × 16.00) = 174.27 g/mol.

To find the number of moles of K₂SO₄ in 99.9g, we divide the mass by the compound's molar mass:
99.9g ÷ 174.27g/mol ≈ 0.573 moles of K₂SO₄.

Now we need to convert moles into molecules. One mole of any substance contains Avogadro's number of particles, which is 6.022 × 10²³ molecules/mol. Therefore, to find the number of molecules in 0.573 moles of K₂SO₄, we multiply the number of moles by Avogadro's number:
0.573 moles × 6.022 × 10²³ molecules/mol ≈ 3.45 × 10²² molecules.

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