Question

Calculate the number of aluminium ions present in 0.051 g of alumininum oxide

Answer 1

molecular mass= 2*27+3*16

=102 g

102 g= 1 mole

so 0..51g

=>1/102*0.51 mole

=0.0005 mole

1 mole of ----=2 * av no.

=>0.0005mole = 2*0.0005*6.022*10 23

=6.022*10 20atoms

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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