Question

Calculate the number of aluminium ions present in 0.051g of aluminium oxide(Al2O3).
(Atomic masses: Al= 27u;O= 16u)

Answer 1

Molecular weight of Al2O3 = 102 g

102 g of Al2O3 contains 6.022X1023 molecules

0.051g of Al2O3 contain X number of molecules

X = 0.051 X 6.022X1023 /102 = 0.003 X 1020

X = 3 X 1020

1 molecule of Al2O3 contains – 2 Al+3ions

3 X 1020 molecules contains – 2 X 3 X 1020

= 6 X 1020 Al+3 ions
∴0.051 g of Al2O3 contains 6 X 1020 Al+3 ions

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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