N2O4 is 25\% dissociated into NO2 at 300 K and a total pressure of 1.00 atm. Use R = 0.0821 L atm mol^-1 K^-1 and M(N2O4) = 92 g/mol. The density of the equilibrium mixture in g/L, to two decimal places, is

JEE Main levelNumerical answerDifficulty 2 of 3ce184

$N X_{2} O X_{4}$ is 25\% dissociated into $NO X_{2}$ at $300 K$ and a total pressure of $1.00 atm$ . Use $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ and $M (N X_{2} O X_{4}) = 92 g/mol$ . The density of the equilibrium mixture in $g/L$ , to two decimal places, is

Numerical answer type, answer in g/L, tolerance 0.01.

M_{a vg} = 92/1.25 = 73.6

g/mol.

ρ = P M_{a vg} / RT = (1.00 \times 73.6) / (0.0821 \times 300) = 2.9882

g/L, i.e. 2.99 g/L.