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)

1 mole of aluminium oxide = 6.022 × 10²³ molecules of aluminium oxide

1 mole of aluminium oxide Al₂O₃

= 2 × Mass of aluminium + 3 × Mass of Oxygen

= (2 × 27) + (3 × 16) = 54 + 48 = 102g

1 mole of aluminium oxide \( = 102g = 6.022 × 10^{23} \) molecules of aluminium oxide

Therefore, 0.051g of aluminium oxide has

\( = \) \( 0.051 × 6.022 × 10^{23} \over 102 \)

\( = 3.011 × 10^{20} \) molecules of aluminium oxide

One molecule of aluminium oxide has 2 aluminium ions, hence number of aluminium ions present in 0.051g of aluminium oxide

\( = 2 ×3.011 × 10^{20} \) ions of aluminium

\( = 6.022 × 10^{20} \) ions of aluminium