Answer:
[tex]2.48\times 10^{22} ions[/tex] 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
[tex]Molarity=\frac{\text{Moles of compound}}{\text{Volume of the solution(L)}}[/tex]
[tex]0.210M=\frac{n}{0.0655 L}[/tex]
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:
[tex]3\times 0.013755 moles=0.041265 mol[/tex] of iodide ions
Number of iodide ions in 0.041265 moles:
[tex]0.041265 mol\times 6.022\times 10^{23}=2.48\times 10^{22} ions[/tex]
[tex]2.48\times 10^{22} ions[/tex] are present in solution.
