At 500 K and 1.00 atm, the equilibrium mixture from $PCl X_{5} (g) PCl X_{3} (g) + Cl X_{2} (g)$ has a density of 3.63 g/L. Use $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ and atomic masses P = 31, Cl = 35.5. Calculate $K_{c}$ in units of $1 0^{- 3} mol L^{- 1}$ , correct to one decimal place (give the number that multiplies $1 0^{- 3}$ ).

Numerical answer type, answer in 10^-3 mol/L, tolerance 0.1.

Answer: 4.6 10^-3 mol/L

Mean molar mass

M = ρRT / P = 3.63 \times 0.0821 \times 500/1.00 = 149.01

g/mol. Since

M = M_{0} / (1 + α)

with

M_{0} = 208.5

α = 208.5/149.01 - 1 = 0.3992

K_{p} = α^{2} P / (1 - α^{2}) = 0.1896

atm and

K_{c} = K_{p} / RT = 0.1896/41.05 = 4.619 \times 1 0^{- 3} mol L^{- 1}