The density of an equilibrium mixture of $N X_{2} O X_{4}$ and $NO X_{2}$ is $2.70 g/L$ at $300 K$ and $1.00 atm$ . Take $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ and $M (N X_{2} O X_{4}) = 92$ . The degree of dissociation of $N X_{2} O X_{4}$ is about

Answer: C

M_{a vg} = ρRT / P = 2.70 \times 0.0821 \times 300/1.00 = 66.50

g/mol. With

M_{a vg} = 92/ (1 + α)

α = 92/66.50 - 1 = 0.3834

, about 0.38.