= 6.022 × 1020
Explanation;
Mole of aluminium oxide (Al2O3) is
⇒ 2 x 27 + 3 x 16
Mole of aluminium oxide = 102 g
i.e., 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
hope it helps_
