Finding the kp with density, temperature, and pressure Consider the following reaction
$$\ce{3O2(g) &lt;=&gt; 2O3(g)}$$
At $\pu{175°C}$ and a pressure of $\pu{128Torr}$, and equilibrium mixture of $\ce{O2}$ and $\ce{O3}$ has a density of $\pu{0.168g/L}$. Calculate $K_p$ for the above reaction at $\pu{175°C}$.

I used the ideal gas law $pV=nRT,$ $K_p=K_c(RT)^n$, and $\rho=M \times c$ but I am not certain what to do after finding the total $c$ of the reaction. I was thinking of using the ICE chart to find the concentration of reactant and products which will lead to the $K_c$ and then $K_p$ value but since I don't know the initial concentration of $\ce{O2}$ I end up with an equation with two variables.
I was thinking of using the density to find the molar mass of the reaction and then somehow the  weight of $\ce{O2}$ and $\ce{O3}$ in grams but it seems like a stretch.
$$pV=nRT$$
$$T=175\ \mathrm{^\circ C}=448\ \mathrm K$$
$$p=128\ \mathrm{Torr}=0.1684\ \mathrm{atm}$$
$$0.0821=\frac nV\times\left(0.0821\ \mathrm{\frac{atm\ l}{K\ mol}}\times448\ \mathrm K\right)$$
$$0.0821=\frac nV\times36.78\ \mathrm{\frac{atm\ l}{mol}}$$
$$\frac nV=0.00457\ \mathrm{\frac{mol}l}=c$$
$$ \begin{bmatrix}
  &amp; \ce{3O2} &amp; \ce{2O3} \\
  \mathrm I &amp; y &amp; 0 \\
  \mathrm C &amp; -3x  &amp; 2x \\
  \mathrm E &amp; y-3x &amp; 2x  
 \end{bmatrix}0.00457\ \mathrm{mol/l}=(y-3x)+2x$$
You can use the formula for the ideal gas
$$\rho=\frac{p\cdot M_\mathrm{mean}}{R\cdot T}$$

As pressure, temperature and density are given, you get the mean molar mass.

$$M_\mathrm{mean}=\frac{\rho RT}{p}$$

From that, you get molar fractions.

$$x_{\ce{O2}}=\frac{ M_{\ce{O3}}-M_\mathrm{mean}}{ M_{\ce{O3}}- M_{\ce{O2}}}$$

$$x_{\ce{O3}}=1-x_{\ce{O2}}$$

From that, you get partial pressures.

$$p_{\ce{O2}}=p\cdot x_{\ce{O2}}$$

$$p_{\ce{O3}}=p-p_{\ce{O2}}$$

From that, you get $$K_\mathrm{p}=\frac{p_{\ce{O3}}^2}{p_{\ce{O2}}^3}$$