Explanation

(d)
Number of atoms = $moles \times N_{A} \times atomicity\$
Here, $N_{A}$= Avogadro’s number
(a) Number of oxygen atoms in 1 g of O
= $\frac{1}{16} \times N_{A} \times 1$
= $\underset{16}{\overset{N_{A}}{\rightarrow}}$
(b)Number of oxygen atoms in 1 g of $O_{2}$
= $\frac{1}{32} \times N_{A} \times 2$
= $\frac{N_{A}}{16}$
(c) Number of oxygen atoms in 1 g of $O_{3}$
=$\frac{1}{48}N_{A\ } \times 3 = \frac{N_{A}}{16}$
Hence, all have the same number of oxygen atoms.