20 gm of ${\mathrm{CaCO}}_3$ is allowed to dissociate in a 5.6 litres container at $819°\mathrm{C}$ if 50% ${\mathrm{CaCO}}_3$ is dissocitated at equilibrium the $\text{'}{\mathrm{K}}_{\mathrm{P}}\text{'}$value is

Amount of ${\mathrm{CaCO}}_3, \mathrm{m}=20\mathrm{g}$

Molar mass of ${\mathrm{CaCO}}_3$,MW=100g/mole

Number moles of ${\mathrm{CaCO}}_3,\mathrm{n}=\frac{\mathrm{Mass} \mathrm{of} {\mathrm{CaCO}}_3}{\mathrm{MW} \mathrm{of} {\mathrm{CaCO}}_3}$

 Then $\mathrm{n}=\frac{20\mathrm{g}}{100\mathrm{g}/\mathrm{mole}}=0.2 \mathrm{mole}$

Volume of container V =5.6L

Percent of dissociation of ${\mathrm{CaCO}}_3, \mathrm{\alpha } = 50\%$

so,$\mathrm{\alpha } = \frac{50}{100}\times 0.2=0.1 \mathrm{mole}$

The chemical equilibrium for the dissociation of ${\mathrm{CaCO}}_3 \mathrm{is}$

${\mathrm{CaCO}}_3\left(\mathrm{s}\right)\to \mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$

Initial(mole):             0 

Change(mole): +x 

Equilib(mole): x = 0.1 

From ideal gas equation PV = nRT

Then pressure of gas $\mathrm{p}=\frac{\mathrm{nRT}}{\mathrm{V}}$

so,${\mathrm{P}}_{{\mathrm{CO}}_2}=\frac{0.1\times 0.0821\times (819+273)}{5.6}=1.6 \mathrm{atm}$

Therefore, equilibrium constant of the reaction is

${\mathrm{K}}_{\mathrm{p}}={\mathrm{P}}_{{\mathrm{CO}}_2}=1.6 \mathrm{atm}$