Vapour density of the equilibrium mixture of ${NO}_2$ and $N_2{O}_4$ is found to be 40 for the equilibrium:

$N_2{O}_4\rightleftharpoons 2{NO}_2$.

Calculate the percentage of ${NO}_2$ in the mixture?

The correct option is B 26.08%

$N_2{O}_4\rightleftharpoons 2N{O}_2$

$t=0$ $1$

$t=t_{eq}1-\alpha 2\alpha$

Initial vapour density of N${}_2$O${}_4$ $=\frac{92}{2}=46$

Vapour density at equilibrium=40

$\frac{46}{40}=\frac{c+c\alpha }{c}$

$\frac{46}{40}=1+\alpha$

$\frac{46}{1+\alpha }=40$

$\Rightarrow 46=40+40\alpha$

$\Rightarrow \alpha -\frac{6}{40}=\frac{3}{20}$

$\therefore \%N{O}_2=\frac{2\alpha }{1+\alpha }\times 100$

$=\frac{10}{23}\times 100+\frac{6}{23}\times 100\times 26.08\%$

Hence, the correct option is $\text{C}$