‘x’ lit of Nitrogen at NTP contains $3.0\times {10}^{22}$  molecules. The number of molecules in x / 2 lit of Ozone at NTP are

$\begin{array}{l}{\text{1\hspace{0.17em}\hspace{0.17em}mole\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}N}}_2=6\times {10}^{23}moleculesof{\text{\hspace{0.17em}N}}_2=22.4{\text{\hspace{0.17em}lit\hspace{0.17em}of\hspace{0.17em}N}}_2\\ 3\times {10}^{22}\text{\hspace{0.17em}molecule}\to ?{\text{\hspace{0.17em}lit\hspace{0.17em}of\hspace{0.17em}N}}_2\\ x=\frac{3\times {10}^{22}\times 22.4}{6\times {10}^{23}}\\ x=1.12{\text{\hspace{0.17em}lit\hspace{0.17em}of\hspace{0.17em}N}}_2\end{array}$

No. of molecules in  $\frac{x}{2}$ lit of Ozone at N.T.P:

$\begin{array}{l}=\frac{x}{2}{\text{lit\hspace{0.17em}of\hspace{0.17em}O}}_3\\ =\frac{1.12}{2}{\text{lit\hspace{0.17em}O}}_3\\ =0.56{\text{\hspace{0.17em}lit\hspace{0.17em}of\hspace{0.17em}O}}_3\end{array}$

$\begin{array}{l}1moleof{\text{\hspace{0.17em}\hspace{0.17em}O}}_3\to 22.4{\text{\hspace{0.17em}lit\hspace{0.17em}O}}_3\to 6\times {10}^{\text{23}}{\text{molecule of O}}_3\\ 0.56{\text{\hspace{0.17em}lit\hspace{0.17em}O}}_3\to x{\text{\hspace{0.17em}molecule\hspace{0.17em}of\hspace{0.17em}O}}_{\text{3}}\\ =\frac{0.56\times 6\times {10}^{23}}{22.4}=0.15\times {10}^{23}=1.5\times {10}^{22}\text{molecules}\end{array}$