Calculate the number of Aluminium ions present in 0.051 grams of Aluminium oxide

 We now that 102 g of $A{l}_2{O}_3=6.022\times {10}^{23}$  molecules of $A{l}_2{O}_3$

The number of atoms present in given mass

$=\left(Givenmass÷molarmass\right)\times N_A$

Then, 0.051 g of $A{l}_2{O}_3$  contains $=\frac{0.051}{102}\times 6.022\times {10}^{23}$  molecules of $A{l}_2{O}_3$

$=3.011\times {10}^{20}$  molecules of $A{l}_2{O}_3$

The number of $A{l}^{+3}$  ions present in one mole of $A{l}_2{O}_3$  is 2

$\therefore$  the number of $A{l}^{+3}$  ions present in $3.011\times {10}^{20}$  molecules (0.051 g) of $A{l}_2{O}_3$

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

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