In 0.051g of aluminium oxide $(A l_{2} O_{3})$ , the number of aluminium molecule present in it are:

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$6.022 \times 1 0^{23}$

$6.022 \times 1 0^{20}$

$3.011 \times 1 0^{20}$

$3.011 \times 1 0^{23}$

answer is A.

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Detailed Solution

The molecular weight of aluminium oxide $(A l_{2} O_{3})$ = (number of moles of $Al$ × molecular mass of Al) + (number of moles of O × molecular mass of O)
$\Rightarrow (2 \times 27) + (3 \times 16) g$
$\Rightarrow (54 + 48) g$
$\Rightarrow 102 g$
Now, 102g of $A l_{2} O_{3}$ contains $6.022 \times 1 0^{23}$ molecules of aluminium oxide.
Therefore, 1g of $A l_{2} O_{3}$ contains = $\frac{6.022 \times 10^{23} molecules}{102 g}$

Similarly, 0.051g of $A l_{2} O_{3}$ contains = $\frac{6.022 \times 10^{23} molecules}{102 g} \times 0.051 g$

$\Rightarrow 3.011 \times 1 0^{20}$ molecules of aluminium oxide

One molecule of aluminium oxide has 2 aluminium ions; hence the number of aluminium ions present in 0.051g of aluminium oxide is
$2 \times 3.011 \times 1 0^{20} = 6.022 \times 1 0^{20}$
Therefore, the number of molecules for aluminium is $6.022 \times 1 0^{20}$ molecules.