$HI$ was heated in a sealed tube at 4400C till the equilibrium was reached. $HI$was found to be 22% decomposed. The equilibrium constant for the dissociation of $HI$ is [$2HI\rightleftharpoons H_2+I_2$]

Consider the following reaction, $2HI\rightleftharpoons H_2+I_2$

Now, we know that Initial Concentration of [HI] = 2 moles

            After 22 % dissociation of [HI] = (2*22)/100 = 0.44 moles

Now, 0.44 moles of [HI] will form = $0.44X\frac{1}{2}$ = 0.22 moles of $H_2$ and 0.22 moles of $I_2$.

 $K_C=\frac{\left[H_2\right]\left[I_2\right]}{{\left[HI\right]}^2}$

 $K_C=\frac{(0.22/V)(0.22/V)}{{(1./56/V)}^2}$                               

$K_C=0.0199$

Hence, the correct answer is option (C) 0.0199.