Question

Calculate the number of aluminium ions present in 0.051

Answer 1

Molecular weight of Al2O3 = 102 g

102 g of Al2O3 contains 6.022X1023 molecules

0.051g of Al2O3 contain X number of molecules

X = 0.051 X 6.022X1023 /102 = 0.003 X 1020

X = 3 X 1020

1 molecule of Al2O3 contains – 2 Al+3ions

3 X 1020 molecules contains – 2 X 3 X 1020

= 6 X 1020 Al+3 ions ∴0.051 g of Al2O3 contains 6 X 1020 Al+3 ions