Calculate the number of aluminium ions present in0.051g of aluminium oxide
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Answered by
1
By dividing wt of Aluminum oxide with it's Molecular Mass it is 102 you will get moles that is 0.005 and multiply with ions of Al2O3 you will get 0.0020
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☺ Hello mate__ ❤
◾◾here is your answer...
1 mole of Al2O3 = 2 × 27 + 3 × 16 = 102g
i.e.,
102g of Al2O3 = 6.022 × 10^23 molecules of Al2O3
Then, 0.051 g of Al2O3 contains
=(6.022×10^23/102) ×0.051 molecules of Al2O3
= 3.011 × 10^20 molecules of Al2O3
The no. Al^3+ in one molecules of Al2O3 is 2.
Here, Al^3+ = aluminium ion .
Therefore, The number of Al^3+ present in 3.11 × 10^20 molecules (0.051g) of Al2O3
= 2 × 3.011 × 10^20
= 6.022 × 10^20
I hope, this will help you.
Thank you______❤
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