One mole of N2O4 (M = 92) dissociates to the extent alpha = 0.20 in N2O4 2NO2 (N = 14, O = 16). Select the correct statement(s) for the equilibrium mixture.

The average molar mass is 76.67 g mol^-1.

The average molar mass decreases as alpha increases.

The vapour density of the mixture is 38.33.

The mole fraction of NO2 is 0.40.

The average molar mass is 76.67 g mol^-1.

The average molar mass decreases as alpha increases.

The vapour density of the mixture is 38.33.

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JEE Advanced levelOne or more correct optionsDifficulty 3 of 3mc217

One mole of $N X_{2} O X_{4}$ ( $M = 92$ ) dissociates to the extent $α = 0.20$ in $N X_{2} O X_{4} 2 NO X_{2}$ (N = 14, O = 16). Select the correct statement(s) for the equilibrium mixture.

AThe mole fraction of $NO X_{2}$ is $0.40$ .
BThe average molar mass is $76.67 g mo l^{- 1}$ .
CThe average molar mass decreases as $α$ increases.
DThe vapour density of the mixture is $38.33$ .

Answer: B, C, D

Total moles = 1 + alpha = 1.20, so M(avg) = 92/1.20 = 76.67 g/mol and VD = 38.33. Moles: N2O4 = 0.80, NO2 = 0.40, so x(NO2) = 0.40/1.20 = 0.333, not 0.40. Since M(avg) = 92/(1 + alpha), it falls as alpha rises.

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One mole of NX2​OX4​ (M=92) dissociates to the extent α=0.20 in NX2​OX4​​2NOX2​ (N = 14, O = 16). Select the correct statement(s) for the equilibrium mixture.

Answer: B, C, D

More questions on Average molar mass of mixtures, vapour density

One mole of $N X_{2} O X_{4}$ ( $M = 92$ ) dissociates to the extent $α = 0.20$ in $N X_{2} O X_{4} 2 NO X_{2}$ (N = 14, O = 16). Select the correct statement(s) for the equilibrium mixture.