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  4. How many number of aluminium ions are present in 0.051 g of aluminium oxide?

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Q. How many number of aluminium ions are present in $0.051 \,g$ of aluminium oxide?

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Solution:

Mass of $Al_{2}O_{3} = 2\times27+3\times16 = 102$
$ 0.051 \,g$ of $Al_{2}O_{3}= \frac{0.051}{102} = 0.0005$ mol
$1$ mol of $Al_{2}O_{3}$ contains $2\times 6.023 \times 10^{23} Al^{3+}$ ions
$0.0005$ mol of $Al_{2}O_{3}$ contains $2 \times 6.023\times 6.023\times 10^{23}$
$ Al^{3+}$ ions $= 6.023\times 10^{20} Al^{3+}$ ions