Question

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Answer 1

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Answer 2

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

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