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Gaseous $N_{2} O_{4}$ dissociates into gaseous ${NO}_{2}$ according to the reaction $N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g)$ . At 300 K and 1 atm pressure, the degree of dissociation of $N_{2} O_{4}$ is 0.2. If one mole of $N_{2} O_{4}$ gas is contained in a vessel, then the density of the equilibrium mixture is:

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3.11 g/L

6.22 g/L

1.56 g/L

4.56 g/L

answer is A.

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Detailed Solution

$\begin{array}{l} P \cdot M = d \cdot R \cdot T \\ 1 \times \frac{92}{(1 + 0.2)} \\ = d \times 0.0821 \times 300 \\ \Rightarrow d = 3.11 g / L \end{array}$

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