Answer:
\[1.0\,g\] of \[O\] atoms \[=\frac{1}{16}g\] atoms of\[O=\frac{1}{16}\times 6.02\times {{10}^{23}}\]atoms\[=\frac{6.02\times {{10}^{23}}}{16}\] atoms
\[1.0\,g\] of \[{{O}_{2}}=\frac{1}{32}\]mol of\[{{O}_{2}}=\frac{1}{32}\times 6.02\times {{10}^{23}}\] molecules\[=2\times \frac{1}{32}\,\times 6.02\times {{10}^{23}}\]atoms\[=\frac{6.02\times {{10}^{23}}}{16}\] atoms
\[1.0\,g\] of \[{{O}_{3}}\]\[=\frac{1}{48}\,\] mol of \[{{O}_{3}}\]\[=\frac{1}{48}\times 6.02\times {{10}^{23}}\] molecules \[=3\times \frac{1}{48}\times 6.02\times {{10}^{23}}\]atoms\[=\frac{6.02\times {{10}^{23}}}{16}\] atoms
Thus, all of them contain equal number of atoms.
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