c
\(\mathop {\mathop {\,\,\,}\limits_{t = 0} }\limits_{t = equilibrium} \mathop {\mathop {{N_2}{O_4}(g)}\limits_1 }\limits_{1 - \alpha }  \leftrightarrow \mathop {\mathop {2N{O_2}(g)}\limits_0 }\limits_{2\alpha } \)

Where \(\alpha =\) Degree of dissociation. 

Mol. wt. of mixture

\( = \frac{{(1 - \alpha ) \times {M_{{N_2}{O_4}}} + 2\alpha  \times {M_{N{O_2}}}}}{{(1 + \alpha )}}\)

\( = \frac{{(1 - 0.2)92 + 2 \times 0.2 \times 46}}{{(1 + 0.2)}} = 76.66\)

Now, As per ideal gas equation 

\(PV = nRT\)

\(P{M_{mixture}} = dRT\)

\(\therefore d = \frac{{P{M_{mix.}}}}{{RT}} = \frac{{1 \times 76.66}}{{0.0821 \times 300}} = 3.11\,g/L\)