Question

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

Answer 1

Answer:

3.011×1020 molecules of Al2O3

Explanation:

Mole of Al2O3=2×27+3×16=102g

102g of Al2O3=6.022×1023 molecules of Al2O3

Then 0.051g of Al2O3 contains 6.022×1023/102×0.051 molecules=3.011×1020 molecules of Al2O3

The no. of aluminium ions present in one molecule of Al2O3 is 2.

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Answer 2

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

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