How many iodide ions are present in 65.5ml of .210 m ali3 solution

are present in solution.

Explanation:

Molarity of the solution = 0.210 M

Volume of the solution = 65.5 ml = 0.0655 L

Moles of aluminum iodide= n

n = 0.013755 moles of aluminum iodide

1 mole of aluminum iodide contains 3 moles of iodide ions:

Then 0.013755 moles of aluminum iodide will contain:

of iodide ions

Number of iodide ions in 0.041265 moles:

are present in solution.