Question

Calculate the number of aluminium ions present in $0.051\mathrm{g}$ of aluminium oxide?

Answer 1

Step 1: Given,

Mass of aluminium oxide, ${\mathrm{Al}}_2{\mathrm{O}}_3=0.051\mathrm{g}$

Step 2: Formula used to determine the number of particles of the substance.

$$\begin{gathered} \frac{\mathrm{m}}{\mathrm{M}}=\frac{\mathrm{N}}{{\mathrm{N}}_{\mathrm{A}}} \\ \mathrm{N}=\frac{\mathrm{m}}{\mathrm{M}}\times {\mathrm{N}}_{\mathrm{A}} \end{gathered}$$

Where,

• $\mathrm{N}$ is the number of particles of the substance
• $\mathrm{m}$ is the mass of the substance
• $\mathrm{M}$ is the molar mass of the substance
• ${\mathrm{N}}_{\mathrm{A}}$ is the Avogadro's number $=6.022\times {10}^{23}$

Step 3: To determine the molar mass of ${\mathbf{Al}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$.

$$\begin{gathered} \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{Al}}_2{\mathrm{O}}_3=(2\times \mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{Al})+(3\times \mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{O}) \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{Al}}_2{\mathrm{O}}_3=(2\times 27)+(3\times 16) \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{Al}}_2{\mathrm{O}}_3=102\mathrm{g}{\mathrm{mol}}^{-1} \end{gathered}$$

Step 4: To determine the number of particles of ${\mathbf{Al}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$.

$$\begin{gathered} \mathrm{N}=\frac{\mathrm{m}}{\mathrm{M}}\times {\mathrm{N}}_{\mathrm{A}} \\ \mathrm{N}=\frac{0.051}{102}\times 6.022\times {10}^{23} \\ \mathrm{N}=3.011\times {10}^{20} \end{gathered}$$

Step 5: To determine the number of aluminium ions.

As, 1 molecule of ${\mathrm{Al}}_2{\mathrm{O}}_3$ has 2 aluminium ions

So, $3.011\times {10}^{20}$molecule of ${\mathrm{Al}}_2{\mathrm{O}}_3$ has $2\times 3.011\times {10}^{20}=6.022\times {10}^{20}$ aluminium ions

Therefore, the number of aluminium ions present in $\mathbf{0}\mathbf{.}\mathbf{051}\mathbf{}\mathbf{g}$ of aluminium oxide are $\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{20}}$.