Question

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

(a) 130 mL of 0.45 M aluminum chloride

(b) 9.80 mL of a solution containing 2.59 g lithium sulfate/L

(c) 245 mL of a solution containingformula units of potassium bromide per liter.

Short answer

Answer

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

(a) 0.059 mol AI³⁺ 0.176 mol CI^-

$$\begin{gathered} \mathbf{}\mathbf{355}\mathbf{\times }{\mathbf{10}}^{22}\mathbf{}\mathit{A}{\mathit{I}}^{\mathbf{3}\mathbf{+}}\mathbf{}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{}\mathbf{,}\mathbf{1}\mathbf{.}\mathbf{06}\mathbf{\times }{\mathbf{10}}^{23}\mathbf{}\mathit{C}{\mathit{I}}^{\mathbf{-}} \\ \mathbf{}\mathbf{}{\mathbf{}}^{} \end{gathered}$$

$$\begin{gathered} \left(b\right)\mathbf{}\mathbf{4}\mathbf{.}\mathbf{62}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}\mathbf{}}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{L}{\mathit{i}}^{\mathbf{+}}\mathbf{,}\mathbf{2}\mathbf{.}\mathbf{31}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{s}{{\mathit{o}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{78}\mathbf{\times }{\mathbf{10}}^{\mathbf{20}}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{L}{\mathit{i}}^{\mathbf{+}}\mathbf{}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{}\mathbf{,}\mathbf{1}\mathbf{.}\mathbf{39}\mathbf{\times }{\mathbf{10}}^{\mathbf{20}}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{s}{{\mathbf{o}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}\mathbf{}\mathit{i}\mathit{o}\mathit{n}\mathit{s} \\ \left(c\right)\mathbf{}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}\mathbf{}}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}{\mathit{K}}^{\mathbf{+}}\mathbf{,}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}{{\mathit{B}}_{\mathbf{r}}}^{\mathbf{-}} \end{gathered}$$

Step-by-step solution

Calculate moles of ions released when  dissolve in water

The dissociation reaction of aluminium chlorideis:

$\mathit{A}\mathit{I}\mathit{C}{\mathit{I}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }\mathit{A}{\mathit{I}}^{\mathbf{3}\mathbf{+}\mathbf{}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{3}\mathit{C}{\mathit{I}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

The moles of $\mathit{A}{\mathit{I}}^{\mathbf{3}\mathbf{+}}$is calculated as:

$$\begin{gathered} \mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathit{A}{\mathit{I}}^{\mathbf{3}\mathbf{+}}\mathbf{=}\mathbf{(}\mathbf{130}\mathbf{}\mathbf{mI}\mathbf{)}\mathbf{\left(}\frac{\mathbf{1}\mathit{L}}{\mathbf{1000}\mathbf{ml}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{0}\mathbf{.}\mathbf{45}\mathbf{moI}\mathbf{}{\mathbf{AICI}}_{\mathbf{3}}}{\mathbf{Lit}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{}\mathbf{1}\mathbf{}{\mathbf{molH}}^{\mathbf{+}}}{\mathbf{1}\mathbf{mol}\mathbf{}{\mathbf{AICI}}_{\mathbf{3}}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{059}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{A}{\mathit{I}}^{\mathbf{3}\mathbf{+}} \end{gathered}$$

The moles of $\mathit{C}{\mathit{I}}^{\mathbf{-}}$is calculated as:

localid="1663932460834" $$\begin{gathered} \mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathit{A}{\mathit{I}}^{\mathbf{3}\mathbf{+}}\mathbf{=}\mathbf{(}\mathbf{130}\mathbf{}\mathbf{mI}\mathbf{)}\mathbf{\left(}\frac{\mathbf{1}\mathit{L}}{\mathbf{1000}\mathbf{ml}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{0}\mathbf{.}\mathbf{45}\mathbf{moI}\mathbf{}{\mathbf{AICI}}_{\mathbf{3}}}{\mathbf{Lit}}\mathbf{\left)}\mathbf{\right(}\frac{3molC{I}^{-}}{\mathbf{1}\mathbf{mol}\mathbf{}{\mathbf{AICI}}_{\mathbf{3}}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1769}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{C}{\mathit{I}}^{\mathbf{-}} \end{gathered}$$

Calculate number of ions of AI3+ and CI-

The number of ions of AI³⁺ is:

$\mathit{A}{\mathit{I}}^{\mathbf{3}\mathbf{+}}\mathbf{}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{059}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{A}{\mathit{I}}^{\mathbf{3}\mathbf{+}}\mathbf{)}\mathbf{(}\frac{\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathit{A}{\mathit{I}}^{\mathbf{3}\mathbf{+}\mathbf{}}\mathit{i}\mathit{o}\mathit{n}\mathbf{}\mathbf{.}}{\mathbf{1}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{A}{\mathit{I}}^{\mathbf{3}\mathbf{+}}}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{55}\mathbf{\times }{\mathbf{10}}^{\mathbf{22}}\mathbf{}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{A}{\mathit{I}}^{\mathbf{3}\mathbf{+}}\mathbf{}\mathit{i}\mathit{o}\mathit{n}\mathit{s}$

The number of ions of CI^-is:

role="math" localid="1663310371661" $$\begin{gathered} \mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathit{C}{\mathit{I}}^{\mathbf{-}}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1769}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{C}{\mathit{I}}^{\mathbf{-}}\mathbf{)}\mathbf{(}\frac{\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathit{A}{\mathit{I}}^{\mathbf{3}\mathbf{+}\mathbf{}}\mathit{i}\mathit{o}\mathit{n}\mathbf{}\mathbf{.}}{\mathbf{1}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{C}{\mathit{I}}^{\mathbf{-}}}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{06}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{C}{\mathit{I}}^{\mathbf{-}}\mathbf{}\mathit{i}\mathit{o}\mathit{n}\mathit{s} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$

Calculate moles of ions released when  dissolve in water

The dissociation reaction of lithium sulfateis:

$\mathit{L}{\mathit{i}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathit{L}{\mathit{I}}^{\mathbf{+}\mathbf{}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathit{S}{{\mathit{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

The moles of$\mathit{L}{\mathit{i}}^{\mathbf{+}}$ is calculated as:

$$\begin{gathered} \mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathit{L}{\mathit{i}}^{\mathbf{+}}\mathbf{=}\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{mI}\mathbf{)}\mathbf{\left(}\frac{\mathbf{1}\mathbf{L}}{\mathbf{1000}\mathbf{}\mathbf{ml}}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{2}\mathbf{.}\mathbf{59}\mathbf{g}\mathbf{}{\mathbf{Li}}^{\mathbf{2}}\mathbf{S}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{}}}{\mathbf{Lit}}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{1}\mathbf{}\mathbf{mole}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{Li}}^{\mathbf{+}}}{\mathbf{109}\mathbf{.}\mathbf{95}\mathbf{}\mathbf{g}\mathbf{}{\mathbf{Li}}^{\mathbf{2}}\mathbf{S}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{}}}\mathbf{\right)} \\ \mathbf{=}\mathbf{4}\mathbf{.}\mathbf{62}\mathbf{\times }{\mathbf{10}}^{-4}\mathit{L}{\mathit{i}}^{\mathbf{+}}\mathit{i}\mathit{o}\mathit{n}\mathit{s} \end{gathered}$$

The moles of$\mathit{S}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}\mathbf{}}$ is calculated as:

$$\begin{gathered} \mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{S}{{\mathit{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}\mathbf{}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\left(9.80mI\right)\left(\frac{1L}{1000ml}\right)\mathbf{\left(}\frac{\mathbf{2}\mathbf{.}\mathbf{59}\mathbf{g}\mathbf{}{\mathbf{Li}}^{\mathbf{2}}\mathbf{S}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{}}}{\mathbf{Lit}}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{1}\mathbf{}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}{\mathbf{Li}}^{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}{\mathbf{109}\mathbf{.}\mathbf{95}\mathbf{}\mathbf{g}\mathbf{}{\mathbf{Li}}^{\mathbf{2}}\mathbf{S}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{}}}\mathbf{\right)} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{31}\mathbf{\times }{\mathbf{10}}^{-4}\mathit{S}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}\mathbf{}}\mathit{i}\mathit{o}\mathit{n}\mathit{s} \end{gathered}$$

Calculate number of ions of SO42-  and  Li+

The number of Li⁺ ions of is:

$\mathit{L}{\mathit{i}}^{+}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{62}\mathbf{\times }{\mathbf{10}}^{-4}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{L}{\mathit{i}}^{\mathbf{+}}\mathbf{)}\mathbf{\left(}\frac{\mathbf{6022}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathit{C}{\mathit{I}}^{\mathbf{-}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{.}}{\mathbf{1}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{C}{\mathit{I}}^{\mathbf{-}}\mathbf{}\mathit{i}\mathit{o}\mathit{n}}\mathbf{\right)}\mathbf{2}\mathbf{.}\mathbf{78}\mathbf{\times }{\mathbf{10}}^{\mathbf{20}}\mathit{L}{\mathit{i}}^{\mathbf{+}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}$

The number of$\mathit{S}{{\mathit{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}\mathbf{}}$ ions of is:

role="math" localid="1663309282903" $$\begin{gathered} \mathit{S}{{\mathit{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}\mathbf{}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{31}\mathbf{}\mathbf{\times }{\mathbf{10}}^{-4}\mathbf{}\mathit{S}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}\mathbf{}}\mathbf{ions}\mathbf{)}\mathbf{\left(}\frac{\mathbf{6022}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathit{S}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}\mathbf{}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{.}}{\mathbf{1}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{S}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}\mathbf{}}\mathit{i}\mathit{o}\mathit{n}}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{39}\mathbf{\times }{\mathbf{10}}^{\mathbf{20}}\mathit{S}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}\mathbf{}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$

Calculate moles of ions released when  dissolve in water

The dissociation reaction of potassium bromideis:

$\mathit{K}\mathit{B}\mathit{r}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }{\mathit{K}}^{\mathbf{+}\mathbf{}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathit{B}{\mathit{r}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

The moles of K⁺is calculated as:

$$\begin{gathered} {\mathit{k}}^{\mathbf{+}}\mathbf{}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{02}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}\mathbf{}{\mathit{k}}^{\mathbf{+}}\mathbf{ions}\mathbf{)}\mathbf{\left(}\frac{1mol{\mathit{k}}^{\mathbf{+}}}{\mathbf{6022}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}{\mathit{k}}^{\mathbf{+}}\mathbf{}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{.}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{2}}\mathbf{}\mathit{m}\mathit{o}\mathit{l}{\mathit{k}}^{\mathbf{+}}\mathbf{}\mathbf{} \end{gathered}$$

The moles of Br ^-is calculated as:

role="math" localid="1663308469248" $$\begin{gathered} \mathit{B}\mathit{r}\mathbf{}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{02}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}\mathbf{}{\mathbf{Br}}^{\mathbf{-}}\mathbf{ions}\mathbf{)}\mathbf{\left(}\frac{1molb{r}^{-}}{\mathbf{6022}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathit{B}\mathit{r}\mathbf{}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{.}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{2}}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathit{B}{\mathit{r}}^{\mathbf{-}}\mathbf{}\mathbf{} \end{gathered}$$

Calculate number of ions of  and

The number of K⁺ ions of is:

$$\begin{gathered} {\mathit{K}}^{\mathbf{+}}\mathbf{}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\mathbf{(}\mathbf{245}\mathbf{}\mathbf{mI}\mathbf{)}\mathbf{\left(}\frac{\mathbf{1}\mathit{L}}{\mathbf{1000}\mathbf{ml}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{3}\mathbf{.}\mathbf{68}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{22}\mathbf{}}\mathbf{F}\mathbf{.}\mathbf{U}\mathbf{.}\mathbf{KBr}}{\mathbf{Lit}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{}\mathbf{1}\mathbf{}\mathit{B}\mathit{r}\mathbf{}\mathit{i}\mathit{o}\mathit{n}\mathit{s}}{\mathbf{1}\mathbf{F}\mathbf{.}\mathbf{U}\mathbf{.}\mathbf{KBr}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{02}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}\mathbf{}\mathbf{}{\mathit{K}}^{\mathbf{+}}\mathit{i}\mathit{o}\mathit{n}\mathit{s} \end{gathered}$$

The number of B_r^-ions of is:

$$\begin{gathered} \mathit{B}\mathit{r}\mathbf{}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\mathbf{(}\mathbf{245}\mathbf{}\mathbf{mI}\mathbf{)}\mathbf{\left(}\frac{\mathbf{1}\mathit{L}}{\mathbf{1000}\mathbf{ml}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{3}\mathbf{.}\mathbf{68}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{22}\mathbf{}}\mathbf{F}\mathbf{.}\mathbf{U}\mathbf{.}\mathbf{KBr}}{\mathbf{Lit}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{}\mathbf{1}\mathbf{}\mathit{B}\mathit{r}\mathbf{}\mathit{i}\mathit{o}\mathit{n}\mathit{s}}{\mathbf{1}\mathbf{F}\mathbf{.}\mathbf{U}\mathbf{.}\mathbf{KBr}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{02}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}\mathbf{}\mathit{B}{\mathit{r}}^{\mathbf{-}}\mathit{i}\mathit{o}\mathit{n}\mathit{s} \end{gathered}$$