An equilibrium mixture of $N X_{2} O X_{4} (g)$ and $NO X_{2} (g)$ has density $2.53 g/L$ at $350 K$ and a total pressure of $1.00 atm$ . Use $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ and $M (N X_{2} O X_{4}) = 92$ . Find the partial pressure of $NO X_{2}$ in atm, to two decimal places.

Numerical answer type, answer in atm, tolerance 0.01.

Answer: 0.42 atm

M_{a vg} = ρRT / P = 2.53 \times 0.0821 \times 350 = 72.700

g/mol.

α = 92/ M_{a vg} - 1 = 0.2655

. Mole fraction of

NO X_{2} = 2 α / (1 + α) = 0.4196

, so

p (NO X_{2}) = 0.4196

atm, i.e. 0.42.