Question

Calculate the number of aluminium ions present in 2g of aluminium oxide(Al2O3).

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Answer 1

Answer:

Answer:

Mole of aluminium oxide (Al2O3) is

⇒ 2 x 27 + 3 x 16

Mole of aluminium oxide = 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.

Therefore, the number of aluminium ions (Al3+) present in 3.11 × 1020 molecules (0.051g) of aluminium oxide (Al2O3)

= 2 × 3.011 × 1020 = 6.022 × 1020

Answer 2

Answer:

Mole of aluminium oxide (A1203) is

⇒ 2 x 27 + 3 x 16

Mole of aluminium oxide = 102 g

i.e., 102 g of Al2O3= 6.022 x 1023 molecules of A1203

Then, 0.051 g of A1203 contains = 6.022 x 1023 / (102 x 0.051 molecules)

= 3.011 x 1020 molecules of A1203

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

Therefore, the number of aluminium ions (A13+) present in 3.11 × 1020 molecules (0.051g) of aluminium oxide (A1203)

= 2 × 3.011 × 1020 = 6.022 × 1020

Explanation:

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