We know,

Vapour density of single was $$ = {{\mathrm{Molar\,Mass}} \over 2}$$

and Vapour density (VD) of mixture of gas $$ = {{\mathrm{Average\,Molar\,Mass\,({M_{avg}}})} \over 2}$$

Here mixture of NO₂ and N₂O₄ gas present.

$$\therefore$$ $$\mathrm{VD} = {{{\mathrm{M_{avg}}}} \over 2}$$

Given, $$VD = 38.3$$

$$\therefore$$ $$38.3 = {{{\mathrm{M_{avg}}}} \over 2}$$

$$ \Rightarrow {\mathrm{M_{avg}}} = 76.6$$

Now let in 100 gm of mixture there are x gm of NO₂.

And molecular mass of $$N{O_2} = 14 + 32 = 46$$

And let weight of N₂O₄ in 100 gm of mixture $$ = 100 - x$$

And molecular mass of $${N_2}{O_4} = 14 \times 2 + 16 \times = 92$$

$$\therefore$$ Molos of $$N{O_2} = {x \over {46}}$$

and moles of $${N_2}{O_4} = {{100 - x} \over {92}}$$

We know,

Number of moles $$(n) = {{\mathrm{Given\,Mass\,(W)}} \over {\mathrm{Molar\,Mass\,(M)}}}$$

$$ \Rightarrow M = {W \over n}$$

Given, $$M = 76.6$$

$$W = 100$$ gm

$$n = {x \over {46}} + {{100 - x} \over {92}}$$

$$ \Rightarrow 76.6 = {{100} \over {{x \over {46}} + {{100 - x} \over {92}}}}$$

$$ \Rightarrow {x \over {46}} + {{100 - x} \over {92}} = {{100} \over {76.6}}$$

$$ \Rightarrow {{2x + 100 - x} \over {92}} = {{100} \over {76.6}}$$

$$ \Rightarrow x + 100 = {{9200} \over {76.6}}$$

$$ \Rightarrow 76.6x + 7660 = 9200$$

$$ \Rightarrow 76.6x = 1540$$

$$ \Rightarrow x = {{1540} \over {76.6}} = 20.1$$

$$\therefore$$ Weight of $$N{O_2} = 20.1$$ g

and weight of $${N_2}{O_4} = 79.9$$ g

$$\therefore$$ Moles of $$N{O_2} = {{20.1} \over {46}} = 0.437$$ moles