Question

Calculate no. Of aluminum ions in 0.056g of aluminum oxide

Answer 1

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 $Al^{3+}$

Answer 2

Given:

Given mass of aluminum oxide=0.056 g

To find:

Number of aluminum ions in 0.056 g of aluminum oxide

Solution:

Molecular mass of aluminum oxide=$2\times 27+3\times 16$

Mass of Al=27g, Mass of oxygen=16 g

Molecular mass of aluminum oxide=102 g

Number of moles of aluminum oxide=$\frac{given\;mass}{molar\;mass}$

Number of moles of aluminum oxide=$\frac{0.056}{102}=5.49\times 10^{-4} mol$

1 mole of aluminum oxide contains 2 moles of aluminum ion

$5.49\times 10^{-4} moles$ contains aluminum ion=$5.49\times 10^{-4}\times 2$ moles

1moles of aluminum ion=$6.022\times 10^{23}$ ions

$5.49\times 10^{-4}\times 2$ moles of aluminum ion=$5.49\times 10^{-4}\times 2\times 6.022\times 10^{23}$

Number of aluminum ions=$6.612\times 10^{20}$