Question

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

Answer 1

1mole of aluminium oxide (AI203)
2*27+3 16
- 102 g
i.e., 102g of Al2036.022 1023
molecules of AI203
Then, 0.051 g of Al203 contains
3.011*1020 molecules of AI203
The number of aluminium ions (AI3+)
present in one molecule of aluminium
oxide is 2.
Therefore, the number of aluminium
ions (Al3+) present in 3.011* 1020
molecules (0.051g) of aluminium oxide
(AI203)2 3.011* 1020
6.022 1020

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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