Question

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

Answer 1

Step 1: Calculate the formula mass of Aluminium oxide $\left({\mathrm{Al}}_2{\mathrm{O}}_3\right)$:

1. The atomic mass of aluminium $\left(\mathrm{Al}\right)$ = $27$
2. The atomic mass of oxygen $\left(\mathrm{O}\right)$= $16$
3. The formula mass of Aluminium oxide $\left({\mathrm{Al}}_2{\mathrm{O}}_3\right)$ = ( $2\times$Atomic mass of aluminium)$+$( $3\times$Atomic mass of oxygen)
4. The formula mass of Aluminium oxide = $(2\times 27)+(3\times 16)\mathrm{u}=102\mathrm{u}$

Hence, the formula mass of Aluminium oxide $\left({\mathrm{Al}}_2{\mathrm{O}}_3\right)$ = $102\mathrm{u}$

Step 2: Calculate the number of moles of aluminium oxide in $0.051\mathrm{g}$ of Aluminium oxide:

We know that,

Number of moles = (Given mass)/(Molar mass)

Number of moles = $\frac{0.051}{102}=0.0005\mathrm{mol}$

Hence, the number of moles in $0.051\mathrm{g}$ of Aluminium oxide are $0.0005\mathrm{mol}$

Step 3: Calculate the number of moles of aluminium ions in $0.0005$ moles of Aluminium oxide:

From the molecular formula $\left({\mathrm{Al}}_2{\mathrm{O}}_3\right)$,

$1\mathrm{mol}$ of ${\mathrm{Al}}_2{\mathrm{O}}_3$ = $2\mathrm{mol}$ of aluminium ions.

$0.0005\mathrm{mol}$ of ${\mathrm{Al}}_2{\mathrm{O}}_3$ = $0.0005\times 2$ = $0.001\mathrm{mol}$ of aluminium ions.

Thus, the number of moles of aluminium ions in $0.0005\mathrm{mol}$ of Aluminium oxide is $0.001\mathrm{mol}$.

Step 4: Calculate the number of aluminium ions in $0.001\mathrm{mol}$:

We know from Avogadro's number,

$1$ mole of aluminium ions = $6.022\times {10}^{23}$ aluminium ions

$0.001$ mole aluminium ions = $0.001\times 6.022\times {10}^{23}=6.022\times {10}^{20}$ aluminium ions.

Therefore, the number of aluminium ions in $0.051\mathrm{g}$ of Aluminium oxide are $6.022\times {10}^{20}$.