In each of the following pairs, indicate which has the higher concentration of \(\mathrm{Cl}^{-}\) ion: (a) \(0.10 \mathrm{MAlCl}_{3}\) solution or a $0.25 \mathrm{MLiCl}\( solution, (b) \)150 \mathrm{~mL}\(. of a \)0.05 \mathrm{MMnCl}_{3}\( solution or \)200 \mathrm{~mL}$. of \(0.10 \mathrm{M} \mathrm{KCl}\) solution, (c) a \(2.8 M H C l\) solution or a solution made by dissolving $23.5 \mathrm{~g}\( of \)\mathrm{KCl}\( in water to make \)100 \mathrm{~mL}$ of solution.

We need to find the concentration of \(\mathrm{Cl}^-\) ions in each solution. For \(0.10\,\mathrm{M}\,\mathrm{AlCl}_{3}\): A mole of \(\text{AlCl}_3\) dissociates into one \(\text{Al}^{3+}\) ion and three \(\text{Cl}^-\) ions. $$\text{AlCl}_3 \rightarrow \text{Al}^{3+} + 3\,\text{Cl}^-$$ So the concentration of \(\text{Cl}^-\) ions will be \(0.10\,\mathrm{M}\times 3=0.30\,\mathrm{M}\). For \(0.25\,\mathrm{M}\,\mathrm{LiCl}\): A mole of \(\text{LiCl}\) dissociates into one \(\text{Li}^+\) ion and one \(\text{Cl}^-\) ion. $$\text{LiCl} \rightarrow \text{Li}^+ + \text{Cl}^-$$ So the concentration of \(\text{Cl}^-\) ions will be \(0.25\,\mathrm{M}\). Comparing both concentrations, the \(0.10\,\mathrm{M}\,\mathrm{AlCl}_{3}\) solution has a higher concentration of \(\mathrm{Cl}^-\) ions.

First, find the moles of \(\mathrm{Cl}^-\) ions in each solution. For \(150\,\text{mL}\,\text{ of }\,0.05\,\text{M}\,\text{MnCl}_3\): A mole of \(\text{MnCl}_3\) dissociates into one \(\text{Mn}^{2+}\) ion and three \(\text{Cl}^-\) ions. $$\text{MnCl}_3 \rightarrow \text{Mn}^{2+} + 3\,\text{Cl}^-$$ Moles of \(\mathrm{Cl}^-\) ions = (volume in L) \(\times\) (concentration of MnCl3) \(\times\) (number of \(\mathrm{Cl}^-\) ions): $$\frac{150}{1000}\,\text{L} \times 0.05\,\text{M} \times 3 = 0.0225\,\text{ moles of }\text{Cl}^-$$ For \(200\,\text{mL}\,\text{ of }\,0.10\,\text{M}\,\text{KCl}\): A mole of \(\text{KCl}\) dissociates into one \(\text{K}^+\) ion and one \(\text{Cl}^-\) ion. $$\text{KCl} \rightarrow \text{K}^+ + \text{Cl}^-$$ Moles of \(\mathrm{Cl}^-\) ions = (volume in L) \(\times\) (concentration of KCl): $$\frac{200}{1000}\,\text{L} \times 0.10\,\text{M} = 0.020\,\text{ moles of }\text{Cl}^-$$ Comparing moles of \(\mathrm{Cl}^-\) ions, the 150 mL of the \(0.05\,\mathrm{M}\,\mathrm{MnCl}_{3}\) solution has a higher amount of \(\mathrm{Cl}^-\) ions.

First, find the concentration of \(\mathrm{Cl}^-\) ions in each solution. For \(2.8\,\mathrm{M}\,\mathrm{HCl}\): A mole of \(\mathrm{HCl}\) dissociates into one \(\mathrm{H}^+\) ion and one \(\mathrm{Cl}^-\) ion. $$\mathrm{HCl} \rightarrow \mathrm{H}^+ + \mathrm{Cl}^-$$ So the concentration of \(\mathrm{Cl}^-\) ions will be \(2.8\,\mathrm{M}\). For the solution made by dissolving \(23.5\,\mathrm{g}\) of \(\mathrm{KCl}\) in water to make \(100\,\mathrm{mL}\) of solution: Calculate the concentration of \(\mathrm{KCl}\): Molar mass of \(\mathrm{KCl} = 39.1 (\mathrm{K}) + 35.5 (\mathrm{Cl}) = 74.6\,\text{g/mol}\). Moles of \(\mathrm{KCl} = \frac{23.5\,\mathrm{g}}{74.6\,\text{g/mol}} = 0.315\,\text{moles}\). Concentration of KCl = \(\frac{0.315\,\text{moles}}{0.1\,\text{L}} = 3.15\,\mathrm{M}\). A mole of \(\mathrm{KCl}\) dissociates into one \(\text{K}^+\) ion and one \(\text{Cl}^-\) ion. $$\mathrm{KCl} \rightarrow \mathrm{K}^+ + \mathrm{Cl}^-$$ So the concentration of \(\mathrm{Cl}^-\) ions will also be \(3.15\,\text{M}\). Comparing both concentrations, the solution made by dissolving \(23.5\,\mathrm{g}\) of \(\mathrm{KCl}\) in water to make \(100\,\mathrm{mL}\) of solution has a higher concentration of \(\mathrm{Cl}^-\) ions.