Question

calculate the total number of ions present in 0.51g of aluminium oxide.

Answer 1

Molar mass of Al2O3=2x27+3x16=54+48=102g

102g

Of Al2O3 contains 2x6.022x1023 Al3+ ions

Therefore,

0.51g of Al2O3 contains= {(2x6.022x1023)/102} x0.51

=6.022x1021 Al3+

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Answer 2

☺ 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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