Calculate the number of aluminium ions $(A l^{3 +})$ present in 0.102 g of aluminium oxide $(A l_{2} O_{3}$ ).
[Molar mass of Al = 27 g/mol; Molar mass of O = 16 g/mol]

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Solution

Molar mass of $A l_{2} O_{3}$ = 102 g/mol
Moles of $A l_{2} O_{3}$ = $\frac{G i v e n m a s s (m)}{M o l a r m a s s (M)}$ = $10^{- 3}$ moles

As, 1 mole of $A l_{2} O_{3}$ contains 6.022 $\times$ $10^{23}$ molecules of $A l_{2} O_{3}$ .
Therefore, $10^{- 3}$ moles of $A l_{2} O_{3}$ contains $10^{- 3}$ $\times$ 6.022 $\times$ $10^{23}$ molecules of $A l_{2} O_{3}$ .

Also, single molecule of $A l_{2} O_{3}$ contains two $A l^{3 +}$ ions.
Therefore, $10^{- 3}$ $\times$ 6.022 $\times$ $10^{23}$ molecules of $A l_{2} O_{3}$ contains 2 $\times$ $10^{- 3}$ $\times$ 6.022 $\times$ $10^{23}$ $A l^{3 +}$ ions.

Hence, 0.102 g of aluminium oxide $(A l_{2} O_{3}$ ) contains 12.044 $\times$ $10^{20}$ number of $A l^{3 +}$ ions.

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