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9 days ago

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

Chemistry

1 answer:

castortr0y [3K]9 days ago

7 0

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

Each mole of aluminum iodide yields 3 moles of iodide ions:

Thus, 0.013755 moles of aluminum iodide will provide:

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

The total 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.

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