For the following equilibrium reaction,

$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$

$N{O}_2$ is $50\%$ of the total volume at a given temperature. Hence vapour density of the equilibrium mixture is:

The correct option is A $34.5$

Let's assume total moles of mixture $({NO}_2+N_2{O}_4)$ is $100$.

According to the given data,

$50\%$ by volume contains 50 moles of $N{O}_2$ in the mixture ($\because$ Volume is constant)

Similarly moles of $N_2{O}_4$ present in the mixture= 50 moles

Molar mass of ${NO}_2$ is $46gm/mol$.

Molar mass$N_2{O}_4$ is $92gm/mol$.

Total mass of the mixture $=(50\times 46)+(50\times 92)$ gm

molecular mass of mixture $=2\times V.D$ ($V.D=$ Vapour density)

$=2X$.

$\frac{\text{Total mass of the mixture}}{\text{Total moles of the mixture}}=\frac{(50\times 46)+(50\times 92)}{100}=2X$

$X=34.5$ $\Rightarrow$ Vapour density is $34.5$.

Option A is the correct answer.