Question

Calculate the number of aluminium ions present in 0.051g of

aluminium oxide. ( At mass of Al = 27u)​

Answer 1

Answer:

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

Answer 2

Answer:

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

Step-by-step explanation:

Answer:

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

therefore, 0.051 g of Al2O3 contains 6 X 1020 Al+3 ions