Question

1 Calculate the number of Aluminium ions present in 0.005 g of Aluminium Oxide​

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:

5.904 x $10^{19}$

Explanation:

0.005 g of Aluminium Oxide contains

= (0.005 / 102) x 6.022 x $10^{23}$ molecules of Aluminium Oxide

= 2.952 x $10^{19}$

= 2 x 2.952 x $10^{19}$

= 5.904 x $10^{19}$