Question

calculate the number of aluminium ions present in0.051 g of aluminium oxide ​

Answer 1

Answer:

2

Step-by-step explanation:

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

= 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.