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Calculate the number of aluminium ions present in 0.051g 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 = 27u)

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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 \(=\left(6.022 \times 10^{23} / 102\right) \times 0.051\) molecules \(=3.011 \times 10^{20}\) molecules of \(\mathrm{Al}_{2} \mathrm{O}_{3}\)
The number of aluminium ions \(\left(\mathrm{Al}^{3+}\right)\) present in one molecules of aluminium oxide is 2 .
Therefore, The number of aluminium ions \(\left(\mathrm{Al}^{3+}\right)\) present in
\(3.11 \times 10^{20}\) molecules \((0.051 \mathrm{~g})\) of aluminium oxide \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)=2 \times 3.011 \times 1020=\) \(6.022 \times 10^{20}\)
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