For the following reaction, 1 g mole of CaCO3 is enclosed in a 5 L container. CaCO3(s) ↔ CaO(s) + CO2(g). Kp = 1.16 at 1073 K, then percentage dissociation of CaCO3 is:

CaCO3(s) ↔ CaO(s) + CO2(g)
Kp = pCO2 (only gaseous molecule count)
Kp = Kc(RT)^Δn
By putting all the given values, on the above equation
1.16 = x/5 (0.0821 × 1073)^1
x = 1.16 × 5 / (0.0821 × 1073) = 0.0658
% dissociation = 0.0658 × 100 = 6.58