Question

calculate the number of aluminium ions present in 0.051 of AL2O3​

Answer 1

Answer:

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

Answer:

Answer :

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

We know, 102 g of Aluminium Oxide (Al2O3) = 6.022 × 1023 molecules of Aluminium Oxide (Al2O3)

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)

= 3.011 × 1020 molecules of Aluminium Oxide (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