Calculate the number of aluminium ions present in 0.0519 of aluminium

1.	Calculate the number of aluminium ions present in 0.0519 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» One mole of Aluminium oxide = 2 × 27+ 3 × 16 = 54 + 48 = 102 g. 102 g of Al₂O₃ = 6.022 × 10²³ moles (aluminium oxide) = \(\frac{6.022\times 10^{23}}{102}\times 0.051\) molecules It means Aluminium present in 0.051 gm = 3.011 × 10²⁰ Aluminium oxide molecules Number of Al ions in one mole of Al₂O₃ = 2 ∴ Number of Al Ions In 3.011 × 10²⁶ molecules 0.051 Al₂O₃ = 2 × 3.011 × Number of Al ions In 3.011 × 10²⁰ = 6.022 × 10²⁰.

Calculate the number of aluminium ions present in 0.0519 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»

One mole of Aluminium oxide

= 2 × 27+ 3 × 16

= 54 + 48

= 102 g.

102 g of Al₂O₃ = 6.022 × 10²³ moles (aluminium oxide)

= \(\frac{6.022\times 10^{23}}{102}\times 0.051\) molecules

It means Aluminium present in 0.051 gm

= 3.011 × 10²⁰ Aluminium oxide molecules

Number of Al ions in one mole of Al₂O₃ = 2

∴ Number of Al Ions In 3.011 × 10²⁶ molecules
0.051 Al₂O₃

= 2 × 3.011 × Number of Al ions In 3.011 × 10²⁰

= 6.022 × 10²⁰.