Question

calculate the number of aluminium ions which are present in 0.0051g if 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

Answer:

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) .

Solution:

Mole of aluminium oxide (Al2O3) = 2 x 27 + 3 x 16

= 102 g

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

Then, 0.051 g of Al2O3 contains = 6.022 x 1023 / 102 x 0.051 molecules

= 3.011 x 1020 molecules of Al2O3

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