(B) The dissociation of $AlCl_3$ in solution is given by: $AlCl_3 \rightarrow Al^{+3} + 3Cl^-$. From the stoichiometry of the reaction,$1 \text{ mole}

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One-third of the number of chloride ions

(B) The dissociation of $AlCl_3$ in solution is given by: $AlCl_3 \rightarrow Al^{+3} + 3Cl^-$. From the stoichiometry of the reaction,$1 \text{ mole}$ of $AlCl_3$ produces $1 \text{ mole}$ of $Al^{+3}$ ions and $3 \text{ moles}$ of $Cl^-$ ions. Therefore,the number of $Al^{+3}$ ions is $1/3$ of the number of $Cl^-$ ions.