Question

The concentration of an aqueous solution of iron(II) chloride is 0.0550 M. What is the molarity of each ion in this solution?

Answer 1

Iron will have a +2 charge (hence the name iron(II)), and chloride ions will have a -1 charge, meaning to balance it, you need two chloride ions to each iron, making the formula FeCl2.
There is one iron ion in each molecule, meaning it is one to one with the molarity:
1 iron ion * (0.0550 moles/liter) = 0.0550 moles iron/liter = 0.0550M Fe
There are two chloride ions in each molecule, meaning it is two to one with the molarity.
2 chloride ions * (0.0550 moles/liter) = 0.110 moles chloride/liter = 0.110M Cl

Answer 2

Answer:

$[Fe^{2+} ]$=0.0550 M

$[Cl^{-} ]$=0.110 M

Explanation:

Its all due to stoichiometry: the compound is $FeCl_{2}$, thus upon ionizationm there will be two chloride anions for each iron cation:

$FeCl_{2} \rightarrow[Fe^{2+} ]+2[Cl^{-} ]$