$(a)\;1.09\qquad(b)\;0.956\qquad(c)\;0.437\qquad(d)\;0.0045$

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Mol.wt of mixture of $NO_2$ and $N_2O_4$=38.3$\times 2=76.6$

Let 'a' g of $NO_2$ present in 100g mixture.

$\therefore$ Mole of $NO_2$+Mole of $N_2O_4$=Mole of mixture

$\large\frac{a}{46}+\frac{100-a}{92}=\frac{100}{76.6}$

$a=20.10g$

$\therefore $ Mole of $NO_2$ in mixture =$\large\frac{20.10}{46}$

$\Rightarrow 0.437$

Hence (c) is the correct answer.

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