Question

Question 11
Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.

Answer 1

Mole of aluminium oxide $\left(A{l}_2{O}_3\right)=2\times 27+3\times 16=102g$
i.e., 102 g of $A{l}_2{O}_3=6.022\times {10}^{23}$ molecules of $A{l}_2{O}_3$
Then, 0.051 g of $A{l}_2{O}_3$ contains $=\frac{6.022\times {10}^{23}}{102\times 0.051}$ molecules
$=3.011\times {10}^{20}$ molecules of $A{l}_2{O}_3$
The number of aluminium ions $\left(A{l}^{3+}\right)$ present in one molecule of aluminium oxide is 2.
Therefore, the number of aluminium ions $\left(A{l}^{3+}\right)$ present in $3.011\times {10}^{20}$ molecules (0.051 g ) of aluminium oxide $\left(A{l}_2{O}_3\right)=2\times 3.011\times {10}^{20}$
$=6.022\times {10}^{20}$