Question

How many iodide ions are present in 65.5ml of .210 m AlI3 solution

Answer 1

Answer:

$2.48\times 10^{22} ions$ 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

$Molarity=\frac{\text{Moles of compound}}{\text{Volume of the solution(L)}}$

$0.210M=\frac{n}{0.0655 L}$

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:

$3\times 0.013755 moles=0.041265 mol$ of iodide ions

Number of iodide ions in 0.041265 moles:

$0.041265 mol\times 6.022\times 10^{23}=2.48\times 10^{22} ions$

$2.48\times 10^{22} ions$ are present in solution.