The rate constant for the reaction $2{\mathrm{N}}_2{\mathrm{O}}_5\to 4{\mathrm{NO}}_2+{\mathrm{O}}_2 \mathrm{is} 3.0\text{'} {10}^{-5}{\mathrm{S}}^{-1}.$ If the rate is $2.40\text{'} {10}^{-5}\mathrm{mol} {\mathrm{L}}^{-1}{\mathrm{S}}^{-1},$ then the concentration of ${\mathrm{N}}_2{\mathrm{O}}_3 \mathrm{is}$

For the reaction, $2{\mathrm{N}}_2{\mathrm{O}}_5\to 4{\mathrm{NO}}_2+{\mathrm{O}}_2$ 

The rate constant $\mathrm{K}=3.0\times {10}^{-5}{\mathrm{s}}^{-1}$

$\mathrm{Rate},\mathrm{r}=2.40\times {10}^{-5} \mathrm{mol} {\mathrm{L}}^{-1}.{\mathrm{S}}^{-1}$

Rate, $\mathrm{r}=\mathrm{k}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]\Rightarrow \frac{\mathrm{Rate},\mathrm{r}}{\mathrm{K}}=\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]$

$\therefore \left[{\mathrm{N}}_2{\mathrm{O}}_5\right]=\frac{2.40\times {10}^{-5} \mathrm{mol} {\mathrm{L}}^{-1}{\mathrm{S}}^{-1}}{3.0\times {10}^{-5}{\mathrm{S}}^{-1}}=0.8 \mathrm{mol}.{\mathrm{L}}^{-1}$