Question 11
Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
Mole of aluminium oxide (Al2O3)=2×27+3×16=102 g
i.e., 102 g of Al2O3=6.022×1023 molecules of Al2O3
Then, 0.051 g of Al2O3 contains =6.022×1023102×0.051 molecules
=3.011×1020 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 3.011×1020 molecules (0.051 g ) of aluminium oxide (Al2O3)=2×3.011×1020
=6.022×1020