Calculate the number of aluminium ions present in 0.051g of aluminium oxide. (Hint: The mass of an ion is the same

Monjur Alahi

Calculate the number of aluminium ions present in 0.051g 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 = 27u)


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Salim Ahmed Kawsar

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1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16 = 102g 

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

Then, 0.051 g of Al2O3 contains = (6.022×1023 /102) × 0.051 molecules 

= 3.011 × 1020 molecules of Al2O3 

The number of aluminium ions (Al3+ ) present in one molecules 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

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