At 350 K and 1.0 atm, N2O4 (molar mass 92) is 40% dissociated into NO2. Taking R = 0.0821 L atm K^-1 mol^-1, the density of the equilibrium mixture in g L^-1 (two decimals) is

JEE Advanced levelNumerical answerDifficulty 3 of 3ig212

At $350 K$ and $1.0 atm$ , $N X_{2} O X_{4}$ (molar mass 92) is $40%$ dissociated into $NO X_{2}$ . Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the density of the equilibrium mixture in $g L^{- 1}$ (two decimals) is

Numerical answer type, answer in g/L, tolerance 0.02.

1 mol

N X_{2} O X_{4}

gives

1 + 0.40 = 1.40 mol

of gas with total mass

92 g

\overset{ˉ}{M} = 92/1.40 = 65.71

d = \frac{P M ˉ}{RT} = \frac{1.0 \times 65.71}{0.0821 \times 350} = 2.2869 \approx 2.29 g L^{- 1}