Calculate the number of aluminium ions present in 0.051 g ofaluminium

1.	Calculate the number of aluminium ions present in 0.051 g ofaluminium oxide.
Answer» 1 mole of aluminium oxide (Al2O3) = 227 + 3 16 = 102 g i.e., 102 g of Al2O3 = 6.022 * 1023 molecules of Al2O3 Then, 0.051 g of Al2O3 contains = = 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 thanku

Calculate the number of aluminium ions present in 0.051 g ofaluminium oxide.

Answer»

1 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 =

= 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

thanku