Calculate the number of aluminium ions present in 0.056 g of aluminium

1.	Calculate the number of aluminium ions present in 0.056 g of aluminium oxide with solution
Answer» Answer: Explanation: Answer: 66.2\times 10^{19}ions of Al^{3+} Explanation: Moles of Al_2O_3=\frac{\text{ GIVEN mass}}{\text{ molar mass}} Moles of Al_2O_3=\frac{0.056g}{102g/mol}=5.5\times 10^{-4}moles Al_2O_3\rightarrow 2Al^{3+}+3O^{2-} thus 1 mole of Al_2O_3 CONTAINS =2\times 6.023\times 10^{23}=12.046\times 10^{23} ions of Al^{3+} 5.5\times 10^{-4}moles of Al_2O_3 contains =\frac{12.046\times 10^{23}}{1}\times 5.5\times 10^{-4}=66.2\times 10^{19}ions of Al^{3+}

Calculate the number of aluminium ions present in 0.056 g of aluminium oxide with solution

Answer»

Answer:

Explanation:

Answer: 66.2\times 10^{19}ions of Al^{3+}

Explanation:

Moles of Al_2O_3=\frac{\text{ GIVEN mass}}{\text{ molar mass}}

Moles of Al_2O_3=\frac{0.056g}{102g/mol}=5.5\times 10^{-4}moles

Al_2O_3\rightarrow 2Al^{3+}+3O^{2-}

thus 1 mole of Al_2O_3 CONTAINS =2\times 6.023\times 10^{23}=12.046\times 10^{23} ions of Al^{3+}

5.5\times 10^{-4}moles of Al_2O_3 contains =\frac{12.046\times 10^{23}}{1}\times 5.5\times 10^{-4}=66.2\times 10^{19}ions of Al^{3+}