Final answer:
To determine the amount of Cl⁻ ions in 800.0 ml of a 0.649 M solution, calculate the moles using molarity and volume, then convert the moles to grams using the molar mass. The solution contains 18.41424 grams of Cl⁻ ions.
Step-by-step explanation:
To find out how many grams of Cl⁻ ions are in an 800.0 ml aqueous solution with a molarity of 0.649 M, you need to first calculate the number of moles of Cl⁻ ions in the solution. Since molarity (M) is the number of moles of solute per liter of solution, you can calculate the moles of Cl⁻ ions as follows:
Moles of Cl⁻ = Molarity (M) × Volume (L) = 0.649 moles/L × 0.800 L = 0.5192 moles
The molar mass of Cl⁻ (chlorine ion) is approximately 35.45 g/mol. To convert moles to grams, you multiply by the molar mass:
Grams of Cl⁻ = Moles of Cl⁻ × Molar Mass = 0.5192 moles × 35.45 g/mol = 18.41424 grams of Cl⁻
Therefore, the solution contains 18.41424 grams of Cl⁻ ions.