Find the number of aluminium ions present in $0.051g$ of aluminium oxide$\left({Al}_2{O}_3\right)$.
(Atomic mass : $Al$ = 27 amu ; $O$ = 16 amu)

The correct option is B $6\times {10}^{20}$
Given mass of aluminium oxide = 0.051 g

molar mass of $A{l}_2{O}_3$ = 102

so, number of moles of $A{l}_2{O}_3$ = $\frac{0.051}{102}$ = 0.0005 moles

we have

$A{l}_2{O}_3\to 2A{l}^{3+}$

1 mole of $A{l}_2{O}_3$ gives 2 moles of $A{l}^{3+}$ ions,

so 0.0005 moles of $A{l}_2{O}_3$ gives $A{l}^{3+}$ ions = $0.0005\times 2=0.001moles$

1 mole of $A{l}^{3+}$ ions have number of $A{l}^{3+}$ ions = $6.023\times {10}^{23}$

so, 0.001 moles of $A{l}^{3+}$ ions contains number of $A{l}^{3+}$ ions = $6.023\times {10}^{23}\times 0.001=6\times {10}^{20}ions$