Question

Calculate the number of aluminium ions present in 0.051g of aluminium oxide ​

Answer 1

Answer :

1 mole of Aluminium Oxide (Al2O3) = 2 × 27 + 3 × 16 = 102 g. ...

We know, The number of atoms present in given mass = (Given mass ÷ molar mass) × Avogadro number.

Then, 0.051 g of Aluminium Oxide (Al2O3) contains.

= (0.051 ÷ 102 ) × 6.022 × 1023 molecules of Aluminium Oxide (Al2O3)

Answer 2

Explanation:

To calculate the number of aluminium ions in 0.051g of aluminium oxide:

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

1 mole of aluminium oxide (Al2O3) = 2 x Mass of aluminium + 3 x Mass of Oxygen

= (2x 27) + (3 x16) = 54 +48 = 102g

1 mole of aluminium oxide = 102g = 6.022 x 10²³ molecules of aluminium oxide

Therefore, 0.051g of aluminium oxide has = 0.051 x 6.022 x 10²³ / 102

= 3.011 x 10²⁰ 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 x 3.011x 10²⁰molecules of aluminium oxide

= 6.022 x 10²⁰