91Ó°ÊÓ

The mass of a substance made up of an equal number of fundamental units is defined as a mole.

a)

The followingmolar mass${\mathbf{SrF}}_{\mathbf{2}}$,

$$\begin{gathered} {\mathrm{M}}_{\mathrm{Ó\neg }}\mathrm{of}\mathrm{SrF}={\mathrm{M}}_{\mathrm{Ó\neg }}\mathrm{of}\mathrm{Sr}+2×({\mathrm{M}}_{\mathrm{Ó\neg }}\mathrm{of}\mathrm{F}) \\ =87.62\frac{\mathrm{g}}{\mathrm{mol}}+2×\left(19.00\frac{\mathrm{g}}{\mathrm{mol}}\right) \\ =125.62\frac{\mathrm{g}}{\mathrm{mol}} \end{gathered}$$

The number offormula units (FU) of is calculated by multiplying the given mass of ${\mathbf{SrF}}_{\mathbf{2}}$ by thereciprocal of its molar mass and the Avogadro's number.

$$\begin{gathered} \mathbf{Number}\mathbf{}\mathbf{FU}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{SrF}}_{\mathbf{2}}\mathbf{=}\mathbf{38}\mathbf{.}\mathbf{1}\mathbf{g}\mathbf{}\mathbf{SrF}\mathbf{×}\left(\frac{\mathrm{mol}\mathrm{SrF}}{125.62g\mathrm{SrF}}\right)\mathbf{×}\left(\frac{6.022×{10}^{23}\mathrm{FU}{\mathrm{SrF}}_2}{\mathrm{mol}{\mathrm{SrF}}_2}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{83}\mathbf{×}{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathbf{FU}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{SrF}}_{\mathbf{2}} \end{gathered}$$

Because each FU of ${\mathbf{SrF}}_{\mathbf{2}}$contains three ions,the total number of ions in 38.1 gof${\mathbf{SrF}}_{\mathbf{2}}$is as follows.

$$\begin{gathered} \mathrm{Total}\mathrm{numbers}\mathrm{of}\mathrm{ions}\mathrm{of}{\mathrm{SrF}}_2=1.83×{10}^{23}\mathrm{FU}{\mathrm{SrF}}_2\left(\frac{3\mathrm{ions}}{\mathrm{FU}{\mathrm{SrF}}_2}\right) \\ =\boxed{5.49×{10}^{23}\mathrm{ions}} \end{gathered}$$

b)

The following molar mass${\mathbf{CuCl}}_{\mathbf{2}}\mathbf{}\mathbf{.}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

$$\begin{gathered} {\mathbf{M}}_{\mathbf{Ó\neg }}\mathbf{}\mathbf{of}\mathbf{}\mathbf{CuCl}\mathbf{}\mathbf{×}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{=}{\mathbf{M}}_{\mathbf{Ó\neg }}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Cu}\mathbf{}\mathbf{+}\mathbf{2}\mathbf{×}\left(M_{\mathrm{Ó\neg }}\mathrm{of}\mathrm{Cl}\right)\mathbf{+}\mathbf{4}\mathbf{×}\left(M_{\mathrm{Ó\neg }}\mathrm{of}H\right)\mathbf{+}\mathbf{2}\mathbf{×}\mathbf{(}{\mathbf{M}}_{\mathbf{Ó\neg }}\mathbf{}\mathbf{of}\mathbf{}\mathbf{O}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{63}\mathbf{.}\mathbf{55}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{2}\mathbf{×}\left(35.45\frac{g}{\mathrm{mol}}\right)\mathbf{+}\mathbf{4}\mathbf{×}\left(1.01\frac{g}{\mathrm{mol}}\right)\mathbf{+}\mathbf{2}\mathbf{×}\left(16.00\frac{g}{\mathrm{mol}}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{170}\mathbf{.}\mathbf{49}\frac{\mathbf{g}}{\mathbf{mol}} \end{gathered}$$

To calculate the mass in kilograms of 3.58 mol of ${\mathbf{CuCl}}_{\mathbf{2}}\mathbf{.}\mathbf{}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$multiply the provided number ofmoles by its molar mass and divide by 1000.

$$\begin{gathered} \mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{CuCl}}_{\mathbf{2}}\mathbf{×}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{58}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{CuCl}}_{\mathbf{2}}\mathbf{×}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{×}\left(\frac{170.49g{\mathrm{CuCl}}_2×2H_{2}O}{\mathrm{mol}{\mathrm{CuCl}}_2×2H_{2}O}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\boxed{\mathbf{0}\mathbf{.}\mathbf{610}\mathbf{}\mathbf{kg}\mathbf{}{\mathbf{CuCl}}_{\mathbf{2}}\mathbf{}\mathbf{×}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}} \end{gathered}$$

c)

The followingmolar mass $\mathbf{Bi}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}\mathbf{}\mathbf{.}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{,}$

$$\begin{gathered} {\mathbf{M}}_{\mathbf{Ó\neg }}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Bi}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}\mathbf{×}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{=}{\mathbf{M}}_{\mathbf{Ó\neg }}\mathbf{}\mathbf{of}\mathbf{}\left(M_{\mathrm{Ó\neg }}\mathrm{of}N\right)\mathbf{+}\mathbf{14}\mathbf{×}\left(M_{\mathrm{Ó\neg }}\mathrm{of}O\right)\mathbf{+}\mathbf{10}\mathbf{×}\left(M_{\mathrm{Ó\neg }}\mathrm{of}H\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{208}\mathbf{.}\mathbf{98}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{3}\mathbf{×}\left(14.01\frac{g}{\mathrm{mol}}\right)\mathbf{+}\mathbf{14}\mathbf{×}\left(16.00\frac{g}{\mathrm{mol}}\right)\mathbf{+}\mathbf{10}\mathbf{×}\left(1.01\frac{g}{\mathrm{mol}}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{485}\mathbf{.}\mathbf{11}\frac{\mathbf{g}}{\mathbf{mol}} \end{gathered}$$

The number of formula units (FU) of $\mathbf{Bi}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}\mathbf{.}\mathbf{}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$is calculated by multiplying the given mass of $\mathbf{Bi}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}\mathbf{.}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ by thereciprocal of its molar mass and the Avogadro's number.

$$\begin{gathered} \mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Bi}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}\mathbf{×}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{88}\mathbf{×}{\mathbf{10}}^{\mathbf{22}}\mathbf{FU}\mathbf{×}\left(\frac{\mathrm{mol}\mathrm{Bi}{\left({\mathrm{NO}}_3\right)}_3×5H_{2}O}{6.022×{10}^{23}\mathrm{FU}}\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{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{78}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{mol}\mathbf{}\mathbf{Bi}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}\mathbf{×}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \\ \mathrm{Molar}\mathrm{masses}\mathrm{of}\mathbf{Bi}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{,} \\ \mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Bi}\mathbf{}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{3}}\mathbf{×}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{78}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Bi}\mathbf{}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{3}}\mathbf{×}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{×}\left(\frac{485.11g\mathrm{Bi}{\left({\mathrm{NO}}_3\right)}_3×5H_{2}O}{\mathrm{mol}\mathrm{Bi}{\left({\mathrm{NO}}_3\right)}_3×5H_{2}O}\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{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\boxed{\mathbf{2}\mathbf{.}\mathbf{32}\mathbf{×}{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{mg}\mathbf{}\mathbf{Bi}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}\mathbf{×}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}} \end{gathered}$$