Calculate the number of aluminium ions present
Question.
Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)

Solution:

1 mole of aluminium oxide $\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)=2 \times 27+3 \times 16=102 \mathrm{~g}$

i.e., $102 \mathrm{~g}$ of $\mathrm{Al}_{2} \mathrm{O}_{3}=6.022 \times 10^{23}$ molecules of $\mathrm{Al}_{2} \mathrm{O}_{3}$

Then, $0.051 \mathrm{~g}$ of $\mathrm{Al}_{2} \mathrm{O}_{3}$ contains

$=3.011 \times 10^{20}$ molecules of $\mathrm{Al}_{2} \mathrm{O}_{3}$

The number of aluminium ions $\left(\mathrm{Al}^{3+}\right)$ present in one molecule of aluminium oxide is 2 .

Therefore, the number of aluminium ions $\left(\mathrm{Al}^{3+}\right)$ present in $3.011 \times 10^{20}$ molecules $(0.051 \mathrm{~g})$

of aluminium oxide $\left(\mathrm{Al}, \mathrm{O}_{2}\right)=2 \times 3.011 \times 10^{20}$

$=6.022 \times 10^{20}$
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