For the reaction ${\mathrm{CaCO}}_3\left(\mathrm{s}\right)\rightleftharpoons \mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right),K_n=1.16\mathrm{atm}\text{ at }1160\mathrm{K}$ if $20\mathrm{g}\text{ of }{\mathrm{CaCO}}_3\text{ in }10.0\mathrm{L}$  vessel at 1160 K, is taken the fraction of ${\mathrm{CaCO}}_3$ dissociated is 

$K_p=1.16 atm$ implies $p_{{\mathrm{CO}}_2}=1.16\mathrm{atm}$ 

The amount of ${\mathrm{CO}}_2$ is $n=\frac{pV}{RT}=\frac{\left(1.16\mathrm{atm}\right)\left(10\mathrm{L}\right)}{\left(0.082\mathrm{L}\text{ atm }{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}\right)\left(1160\mathrm{K}\right)} =\frac{1}{8.2}\mathrm{mol}$

Mass of ${\mathrm{CaCO}}_3$ dissociated, $m=nM=\left(\frac{1}{8.2}\mathrm{mol}\right)\left(100\mathrm{g}{\mathrm{mol}}^{-1}\right)=12.2\mathrm{g}$

Fraction of ${\mathrm{CaCO}}_3$ dissociated $=\frac{12.2\mathrm{g}}{20\mathrm{g}}=0.61$