Question

calculate the no. of aluminium ions present in 0.056 g of aluminium oxide(AL₂O₃)

Answer 1

1mol of Al2O3 = 27*2+16*3

= 102 g

102 g of Al2O3 = 1mol of Al2O3 therefore,

0.056 g of Al2O3 = 0.056/102

= 5.49x10^-4 mol of Al2O3

1 mol of Al2O3 = 2 mol of Al3+ ions

= 2*6.023x10^23 of Al3+ ions therefore,

5.49x10^-4 mol of Al2O3 = 2*6.023x10^23*5.49x10^-4 of Al3+ ions

= 6.61x10^20 Al3+ ions