Question

22. Calculate the number of moles (A13+) in 0.051g of aluminium (Al2O3)​

Answer 1

Answer:

⇒ n = 0.051/101.96 = 0.0005

No. of molecules = 3.022 × 10^20

Therefore, no. of Al3+ ion = 2 × 3.022 × 10^20

                                           = 6.022 × 10^20