The $K_p$ value of ${\mathrm{N}}_2{\mathrm{O}}_4\rightleftharpoons 2{\mathrm{NO}}_2$ reaction at equilibrium is 0.6. The value of degree of dissociation of ${\mathrm{N}}_2{\mathrm{O}}_4$ at 318 K and a total pressure of 10 bar will be:

Let degree of dissociation of ${\mathrm{N}}_2{\mathrm{O}}_4=x$. Let  the vessel contains one mol of ${\mathrm{N}}_2{\mathrm{O}}_4,K_{\nu }=0.6$ 

Total pressure, P = 10 bar. Thus, for the following reaction, we have :

${\mathrm{N}}_2{\mathrm{O}}_{4 }\rightleftharpoons 2{\mathrm{NO}}_2$

Initial conc.                        1         0

 Conc. at equilibrium       1 - x       2 x

Total no. of mole at equilibrium

$\begin{array}{r}=1-x+2x=1+x\\ \therefore p_{{\mathrm{N}}_2{\mathrm{O}}_4}=\frac{10(1-x)}{1+x};p_{{\mathrm{NO}}_2}=\frac{10\times 2x}{1+x}\end{array}$

Using law of chemical equilibrium for the above reaction, we have:

$\begin{array}{l}{K}_p=\frac{{\left[{\mathrm{NO}}_2\right]}^2}{\left[{\mathrm{N}}_2{\mathrm{O}}_4\right]}=\frac{{\left(\frac{20x}{1+x}\right)}^2}{\frac{10(1-x)}{1+x}}\\ =\frac{400x^2}{(1+x)(1+x)}\times \frac{(1+x)}{10(1-x)}=\frac{40x^2}{(1+x)(1-x)}\\ 0.6=\frac{40x^2}{1-x^2};0.6-0.6x^2=40x^2\end{array}$

or $40x^2+0.6x^2=0.6$

$\begin{array}{l}\therefore 40.6x^2=0.6\text{ Or }{x}^2=\frac{0.6}{40.6}=0.0148\\ \therefore x=(0.0148{)}^{1/2}=0.121\end{array}$

So, the correct answer is, (c)