Find the number of oxygen atom present in $0.051\text{g}$ of aluminium oxide.
(Given: atomic mass of $Al=27u$)

The correct option is B $9.033\times {10}^{20}$
Mass of one mole of $A{l}_2{O}_3=2\times 27+3\times 16=102\text{g}$

Since, $102\text{g}$ of $A{l}_2{O}_3$ $=6.022\times {10}^{23}$ molecules of $A{l}_2{O}_3$

Then, $0.051\text{g}$ 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 $A{l}_2{O}_3$

[We know, The number of atoms present in given mass = (Given mass ÷ molar mass) × Avogadro number]

$\because$ One molecule of $A{l}_2{O}_3$ contains $3$ atoms of oxygen
$\therefore 3.011\times {10}^{20}$ molecules contains $=3\times 3.011\times {10}^{20}=9.033\times {10}^{20}$ atoms of oxygen