Question

Calculate the number of Aluminium ions present in 0.051g of Aluminium oxide​

Answer 1

Mole of aluminium oxide (Al2O3) = 2 x 27 + 3 x 16 = 102 g

that is = 102 g of Al2O3= 6.022 x 1023 molecules of Al2O3

Then, 0.051 g of Al2O3 contains = 6.022 x 1023 / 102 x 0.051 molecules

= 3.011 x 1020 molecules of Al2O3

The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Therefore, the number of aluminium ions (Al3+) present in 3.11 × 1020 molecules (0.051g) of aluminium oxide (Al2O3)

= 2 × 3.011 × 1020 = 6.022 × 1020

I hope it will help you ^_^