$\begin{array}{1 1}(a)\;2&(b)\;.01\\(c)\;0.437&(d)\;.234\end{array}$

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

$\Rightarrow 76.6$

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

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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