$PCl X_{5}$ (calculated vapour density $104.25$ when undissociated) dissociates at a total equilibrium pressure of $2.0 atm$ , and the vapour density of the mixture is $62$ . $K_{p}$ for $PCl X_{5} (g) PCl X_{3} (g) + Cl X_{2} (g)$ in atm is

Answer: A

α = (104.25 - 62) /62 = 0.6815

K_{p} = α^{2} P / (1 - α^{2}) = 0.4644 \times 2.0/0.5356 = 1.734

atm. Dividing by

D

instead of

d

α

gives 0.39.