Question

calculate the no. of aluminium ions present in 0.051 g of aluminium oxide

Answer 1

How can one calculate the number of aluminum ions present in 0.051 g of aluminum oxide?

Do you mean Al2O3?

Let us compute the molar mass of the molecule first

Mass of Al = 27 AMU
Mass of O = 16 AMU
So mass of Al2O3 "molecule" = 2x27 + 3x16 = 102 AMU

So 1 mole of ALUMINIUM OXIDE will weigh 102 g
1 mole of any pure substance contains Avogrado's number of molecules, that is 6.023x10^23.
Thus 0.051g of ALUMINIUM OXIDE will contain 6.023x(10^23)x.051/102 = 6.023x(10^20)/2 "molecules "

Each molecule contains two ions of aluminium.

So total no. Of ions in the given amount of substance = 2x6.023x(10^20)/2 = 6.023x10^20

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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