How to calculate the equilibrium constant from the density of a nitrogen dioxide and dinitrogen tetroxide mixture? The density of an equilibrium mixture of $\ce{N2O4}$ and $\ce{NO2}$ at $\pu{1 atm}$ and $\pu{373.5 K}$ is $\pu{2.0 g/L}$. Calculate $K_c$ for the dissociation reaction


I did this:


Let initial amount of substance be $a$.
Initial vapor density $=(\text{mass of }\ce{N2O4})/2 =46$ 
For final vapour density I find effective molecular mass of the mixture $\ce{N2O4}$ + $\ce{NO2}$ at equilibrium using ideal gas equation which comes out to be $61.33$ and thus final vapor density is $61.33/2=30.665$.
Now supposing degree of dissociation to be $x$ , total amount of substance at equilibrium is $a\times(1+x)$.
As $\text{Vapor density} \times \text{total amount of substance} = \text{constant}$ at same temperature, $46\times a=30.665\times a\times(1+x)$
$x=0.5$ Using this in the equation 
$$K_p=\left[\frac{4x^2}{1-x^2}\right]\times P$$ for this reaction I get $K_p=\frac43\pu{ atm}$ and then $K_c=0.043$. 
The answer is 2 but how?
I confirm your answer.

The molar density of the mixture is 

$$\frac{n}{V}=\frac{p}{RT} = \pu{0.0326 mol/L}$$

So, the average molecular weight of the mixture is 

$$\frac{2}{0.0326} = \pu{61.35 g/mol}$$

If $x$ is the mole fraction of $\ce{NO2}$ and $(1-x)$ is the mole fraction of $\ce{N2O4}$, then the average molecular weight of the mixture is also

$$46x + 92 (1 - x) = 61.35$$ 

Solving for $x$ gives $x = 1/3$. So, $(1-x)=2/3$. These are also the partial pressures of $\ce{NO2}$ and $\ce{N2O4}$, respectively (in atm). This leads to the values of $K_p$ and $K_c$ that you calculated.