$(a)\;2.116g/l\qquad(b)\;3.116g/l\qquad(c)\;1.11g/l\qquad(d)\;0.28g/l$

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$N_2O_4\Leftrightarrow 2NO_2$

Mole present initially in $N_2O_4$ = 1

Mole present initially in $2NO_2$ = 0

Mole at equilibrium $N_2O_4$ = (1-0.2)

Mole at equilibrium $2NO_2$ = 0.4

Molecular weight of the mixture = $\large\frac{0.8\times92+0.4\times46}{1.2}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= 76.66$

Now PV =$\large\frac{w}{m}RT$

$\large\frac{w}{V} = \large\frac{Pm}{RT} $

$= \large\frac{76.66\times1}{0.082\times300}$

= 3.116 g/l

Hence answer is (b)

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