Vapour density of mixture of ${NO}_2$ and $N_2{O}_4$ is 34.5, then percentage abundance of ${NO}_2$ in mixture is:

The correct option is A 50%
Solution:- (A) $50\%$

Let the total moles of the mixture be $100$.

Let $x$ be the no. of moles of $N{O}_2$ present in the mixture.

Therefore,

No. of moles of $N_2{O}_4$ present in the mixture $=(100-x)$

Molecular mas of $N{O}_2=46g$

Molecular mass of $N_2{O}_4=92g$

Total mass of mixture $=\frac{x\times 46+(100-x)\times 92}{100}=\frac{9200-46x}{100}.....\left(1\right)$

Vapour density of mixture $=34.5\left(\text{Given}\right)$

$\therefore$ Molecular mass of mixture $=2\times$ Vapour density $=2\times 34.5=69g.....\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$\frac{9200-46x}{100}=69$

$\Rightarrow x=\frac{9200-6900}{46}=50$

Hence the no. of moles of $N{O}_2$ present in the mixture is $50$.

Therefore,

Percentage abundance of $N{O}_2$ in the mixture $=\frac{\text{No. of moles of}N{O}_2\text{in the mixture}}{\text{Total no. of moles in the mixture}}\times 100=\frac{50}{100}\times 100=50\%$

Hence the percentage abundance of $N{O}_2$ in the mixture is $50\%$.