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)

Answer 1

1 mole of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16 = 102 g

i.e., 102 g of Al₂O₃ = 6.022 × 10²³ molecules of Al₂O₃

Then, 0.051 g of Al₂O₃ contains`=(6.022xx10^(23))/102xx0.051" molecules"`

                                                 = 3.011 × 10²⁰ molecules of Al₂O₃

The number of aluminium ions (Al³⁺) present in one molecule of aluminium oxide is 2.

Therefore, the number of aluminium ions (Al³⁺) present in 3.011 × 10²⁰ molecules (0.051 g ) of aluminium oxide (Al₂O₃)

= 2 × 3.011 × 10²⁰

= 6.022 × 10²⁰