Q4.19P - Chemistry: Molecular Nature Of Matter And Change

Question

How many moles and numbers of ions of each type are present in the following aqueous solutions? 

(a) 88 mL of 1.75 M magnesium chloride 

(b) 321 mL of a solution containing 0.22 g aluminum sulfate/L 

(c) 1.65 L of a solution containing 8.83 x 10²¹ formula units of cesium nitrate per liter.

Step-by-Step Solution

Answer

ANSWER: The moles and numbers of ions of each type are present in the following aqueous solutions are:

(a) 0.154 mol Mg²⁺, 0.308 mol Cl^-

      9.27 x 10²² Mg²⁺ ions, 1.85 x 10²³ Cl^- ions

(b) 4.13 x 10^-4 mol Al³⁺, 6.19 x 10^-4 mol SO₄^2-

      2.49 x 10²⁰ Al³⁺ ions, 3.73 x 10²⁰ SO₄^2-ions

(c) 2.42 x 10^-2 mol Cs⁺, 2.42 x 10^-2 mol NO₃^-

       1.46 x 10²² Cs⁺ ions, 1.46 x 10²² NO₃^- ions

Step 1: the given values are:

(a) 88 mL of 1.75 M magnesium chloride

(b) 321 mL of a solution containing 0.22 g aluminum sulfate/L

(c) 1.65 L of a solution containing 8.83 x 10²¹ formula units of cesium nitrate per liter.

Step 2: Write down the balanced dissociation reaction and find out the moles and no. of ions as follows:

(a) Dissociation reaction of magnesium chloride is:

 $\mathit{M}\mathit{g}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\to }\mathit{M}{\mathit{g}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{C}{\mathit{l}}^{\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$

$$\begin{gathered} {\text{Moles of Mg}}^{2+}=\left(88ml\right)x\left(\frac{1L}{1000ml}\right)x\left(\frac{1.75molMgC{l}_2}{Lit}\right)\left(\frac{1molM{g}^{2+}}{1molMgC{l}_2}\right) \\ =0.154molM{g}^{2+} \\ molesofC{l}^{-}=\left(88ml\right)x\left(\frac{1L}{1000ml}\right)x\left(\frac{1.75molMgC{l}_2}{Lit}\right)\left(\frac{2molC{l}^{-}}{1molMgC{l}_2}\right) \\ =0.308molC{l}^{-} \\ M{g}^{2+}ions=\left(0.154molM{g}^{2+}\right)x\left(\frac{6.022x{10}^{23}M{g}^{2+}ion}{1molM{g}^{2+}}\right)=9.27x{10}^{22}M{g}^{2+}ions \\ C{l}^{-}ions=\left(0.308molC{l}^{-}\right)x(\frac{6.022x{10}^{23}C{l}^{-}ion}{1molC{l}^{-}} \end{gathered}$$

(b) Dissociation reaction of   $\mathit{A}{\mathit{l}}_{\mathbf{2}}\mathbf{(}{\mathbf{SO}}_{\mathbf{4}}{\mathbf{)}}_{\mathbf{3}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$is:

                $\mathit{A}{\mathit{l}}_{\mathbf{2}}\mathbf{(}\mathbf{S}{\mathbf{O}}_{\mathbf{4}}{\mathbf{)}}_{\mathbf{3}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathit{A}{\mathit{l}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{3}\mathit{S}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$           

 $$\begin{gathered} molesofA{l}^{3+}=\left(321ml\right)\left(\frac{1L}{1000ml}\right)\left(\frac{.22gmA{l}_2{\left(S{O}_4\right)}_3}{Lit}\right)\left(\frac{1molA{l}_2{\left(S{O}_4\right)}_3}{342.17gA{l}_2{\left(S{O}_4\right)}_3}\right)\left(\frac{2molA{l}^{3+}}{1molA{l}_2{\left(S{O}_4\right)}_3}\right) \\ =4.13x{10}^{-4}molA{l}^{3+} \\ molesofS{O_4}^{2-}=\left(321ml\right)\left(\frac{1L}{1000ml}\right)\left(\frac{.22gmA{l}_2{\left(S{O}_4\right)}_3}{Lit}\right)\left(\frac{1molA{l}_2{\left(S{O}_4\right)}_3}{342.17gA{l}_2{\left(S{O}_4\right)}_3}\right)\left(\frac{3molS{O_4}^{2-}}{1molA{l}_2{\left(S{O}_4\right)}_3}\right) \\ =6.19x{10}^{-4}molS{O_4}^{2-} \\ A{l}^{3+}ions=\left(4.13x{10}^{-4}molA{l}^{3+}\right)x\left(\frac{6.022x{10}^{23}A{l}^{3+}ion}{1molA{l}^{3+}}\right)=2.49x{10}^{20}A{l}^{3+}ions \\ S{O_4}^{2-}ions=\left(6.19x{10}^{-4}molS{O_4}^{2-}\right)x\left(\frac{6.022x{10}^{23}S{O_4}^{2-}ion}{1molS{O_4}^{2-}}\right)=3.73x{10}^{20}S{O_4}^{2-}ions\backslash \end{gathered}$$

 (c) Dissociation reaction of cesium nitrate is:

                                       $\mathit{C}\mathit{s}\mathit{N}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\to }\mathit{C}{\mathit{s}}^{\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathit{N}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$

 

$$\begin{gathered} C{s}^{+}ions=\left(1.65lit\right)\left(\frac{8.83x{10}^{21}F.U.CsN{O}_3}{Lit}\right)\left(\frac{1C{s}^{+}ion}{1F.U.CsN{O}_3}\right) \\ =1.46x{10}^{22}C{s}^{+}ions \\ N{O_3}^{-}ions=\left(165ml\right)\left(\frac{1L}{1000ml}\right)\left(\frac{8.83x{10}^{21}F.U.CsN{O}_3}{Lit}\right)\left(\frac{1N{O_3}^{-}ion}{1F.U.CsN{O}_3}\right)l \\ =1.46x{10}^{22}N{O_3}^{-}ions \\ \backslash \mathrm{begingathered}C{s}^{+}moles=\left(9.02x{10}^{21}C{s}^{+}ion\right)x\left(\frac{1molC{s}^{+}}{6.022x{10}^{23}C{s}^{+}ion}\right)=2.42x{10}^{-2}molC{s}^{+} \\ N{O_3}^{-}moles=\left(9.02x{10}^{21}N{O_3}^{-}ion\right)x\left(\frac{1molN{O_3}^{-}}{6.022x{10}^{23}N{O_3}^{-}ion}\right)=2.42x{10}^{-2}molN{O_3}^{-} \end{gathered}$$