Final answer:
There are approximately 9.50 x 10^20 sodium ions in 99.6 mg of Na2SO3, which is calculated by converting the mass of Na2SO3 to moles and then multiplying by Avogadro's number.
Step-by-step explanation:
To determine the number of sodium ions in 99.6 mg of Na₂SO₃, we need to convert the mass to moles and then use Avogadro's number to find the number of ions. First, we calculate the molar mass of Na₂SO₃ (2*22.99 + 32.06 + 3*16.00 = 126.04 g/mol). Now, we convert the mass of Na₂SO₃ to moles: 99.6 mg = 0.0996 g, so this is 0.0996 g / 126.04 g/mol = 7.90 x 10-4 moles of Na₂SO₃. Since each formula unit of Na₂SO₃ contains 2 sodium ions, the number of sodium ions is 2 * 7.90 x 10-4 moles * 6.022 x 1023 ions/mole, which equals approximately 9.50 x 1020 sodium ions.