When $\pu{36.8 g}$ $\ce{N2O4(g)}$ is introduced into a 1.0-litre flask at $\pu{27 °C}$ the following equilibrium reaction occurs: $\ce{N2O4(g) <=> 2NO2(g)}$; $K_p = 0.1642~\mathrm{atm}$.
Find the percentage dissociation.
I tried the following:
Let $x$ be the degree of dissociation and $P$ be the initial pressure of $\ce{N2O4}$ that is $\approx 9\ \mathrm{atm}$. Then, $$\frac{9Px^2}{1-x} = 0.1642$$
Neglecting $x$ in $(1-x)$, I got $x$ as $\displaystyle \frac{0.4}{3}$ [approx]
But the answer given as $6.44~\%$.