Question

Calculate the number of aluminium ions present in $0.051$ g of aluminium oxide. (Atomic mass of aluminium= $27u$)

Answer 1

$1$ mole of $A{l}_2{O}_3=2\times 27+3\times 16$

$=102$g

Moles of $A{l}_2{O}_3=\frac{0.051}{102}=5\times {10}^{-4}moles$

1 mole contains $6.02\times {10}^{23}molecules$

$\therefore 5\times {10}^{-4}$g of $A{l}_2{O}_3$ contains = $5\times {10}^{-4}\times 6.02\times {10}^{23}=$$3.011\times {10}^{20}$ molecule

no. of $A{l}^{3+}$ ions in $1$ molecules of $A{l}_2{O}_3$ is $2$.

$\therefore$ No. of aluminium ions will be:-

$2\times 3.011\times {10}^{20}=6.022\times {10}^{20}$.