Question

calculate the number of Aluminium ions present in 0.051 gram of Aluminium oxide .

Answer 1

Molecular mass of Al2O3 = 27 x2 + 16 x3 = 102
%age composition of Al in Al2O3 =54/102 x 100 = 52.94 %
So, 0.051 gm of aluminium oxide will have 52.94 % Aluminium ion = 0.051 x 0.5294 = 0.026 gm
Hence, 27 gm of Al+3 -ion contains = 1 mole Al+3 ions = 6.022 x 1023 ions
Hence, 0.026 gm Al +3 ion will have = (6.022 x 1023 / 27 ) x 0.026 = 6.01 x 1022 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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