Question

What are the respective concentrations (m) of fe3+ and i- afforded by dissolving 0.200 mol fei3 in water and diluting to 725 ml?

Answer 1

Molarity = moles / liter of solution

Given, Moles of Fe³⁺ = 0.200

Volume of solution = 725 ml = 0.725 L

Conversion factor: 1000 ml = 1L

Molarity = 0.200 / 0.725 L = 0.275 M

The dissociation of Fel₃ in water is ad follows:

Fel₃ → Fe³⁺ + 3l⁻  

1 mole of Fel₃ gives 1 mole of Fe³⁺ ions  and 3 moles of l⁻  

Since the solution is 0.275 M of Fel₃, so there are 0.275 M of Fe³⁺ ions,

and (3 x 0.275 M) = 0.825 M of l⁻ ions.

Answer 2

The concentration of $\rm Fe^3^+$ wil be 0.275 M and the concentration of $\rm I^-$ will be 0.825 M.

The concentration of $\rm FeI_3$ solution is:

Molarity = moles/liters

Molarity = 0.200 / 0.725 L

Molarity = 0.275 M

The dissolution of $\rm FeI_3$ gives:

$\rm FeI_3\;\rightarrow\;Fe^3^+\;+\;3\;I^-$

So, 1 mole of $\rm FeI_3$ gives 1 mole of $\rm Fe^3^+$ and 3 moles of $\rm I^-$.

0.275 M $\rm FeI_3$ gives:

0.275 M $\rm Fe^3^+$

0.275 $\times$ 3 M $\rm I^-$

0.825 M $\rm I^-$

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