Question

Calculate the no. of aluminum ions in 0.051 g of $A{l}_2{O}_3$.

Answer 1

Molar mass of $A{l}_2{O}_3=2\times 27+3\times 16=54+48=102g$
$102gofA{l}_2{O}_{3}contains=2\times 6.022\times {10}^{23}A{l}^{3+}ions$
$\therefore 0.051gofA{l}_2{O}_{3}contains=\frac{2\times 6.022\times {10}^{23}}{102}\times 0.051$
$=6.022\times {10}^{20}A{l}^{3+}ions$