Chemistry - Mole Concept Question with Solution | TestHub

$16.0\mathrm{g}{\mathrm{O}}_3=\frac{16}{48}$ mole $$\begin{array}{rl}& =\frac{16}{48}\times 6.023\times 10^{23}\text{ molecules }\\ & =3\times \frac{16}{48}\times 6.023\times 10^{23}\text{ atoms }\\ & =6.023\times 10^{23}\text{ atoms }\\ & 28.0\mathrm{g}\mathrm{CO}=\frac{28}{28}\text{ mole }=1\text{ mole }\\ & =1\times 6.023\times 10^{23}\text{ molecules }\\ & =1\times 6.023\times 10^{23}\text{ atoms }\\ & =6.023\times 10^{23}\text{ atoms }\\ & 16.0\mathrm{g}{\mathrm{O}}_2=\frac{16}{16}\text{ mole }=1\mathrm{mole}\end{array}$$ $=1\times 6.023\times 10^{23}$ molecules $=1\times 6.023\times 10^{23}$ atoms $=6.023\times 10^{23}$ atoms Therefore, the ratio is $6.023\times 10^{23}:6.023\times 10^{23}:6.023\times 10^{23}$ i.e. $1:1:1$