The number of aluminium ion present in $0.051 g$ of aluminium oxide is

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6.022 \times 1 0^{22}

6.022 \times 1 0^{20}

6.022 \times 1 0^{21}

6.022 \times 1 0^{23}

answer is A.

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Detailed Solution

Aluminium oxide has the formula ${Al}_{2} O_{3}$ .
$1 Mole of {Al}_{2} O_{3} = 2 \times Mass of Al + 3 \times Mass of oxygen$
$= 27 \times 2 + 3 \times 16$
$= 102 g$

Now, $1 Mole of {Al}_{2} O_{3} = 2 Mole of aluminium$
Therefore, the mass of Al in $A l_{2} O_{3} = 27 \times 2$
                                                  $= 54 g$
So, the mass of aluminium in $0.051 g of {Al}_{2} O_{3} = \frac{54}{102} \times 0.051$
                                                    $= 0.027 g$
The atomic mass of aluminium $= 27$
Number of mole of aluminium = $\frac{Given mass}{Atomic mass} = \frac{0.027}{27} 0.001 Mole$
Therefore, number of atoms of aluminium $= 6.022 \times 1 0^{23} \times 0.001$
                                                                        $= 6.022 \times 1 0^{20}$