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At $27^{0} C$ and 1 atm pressure, $N_{2} O_{4}$ is 20% dissociation into $N O_{2}$ . What is the density of equilibrium mixture of $N_{2} O_{4}$ and $N O_{2}$ at $27^{0} C$ and 1 atm?

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Detailed Solution

$α = \frac{M_{T} - M_{O}}{(n - 1) M_{O}} \Rightarrow 0.2 = \frac{92 - m_{o}}{M_{o}} \Rightarrow M_{o} = 76.66 d_{m i x t u r e} = \frac{P M_{m i x t u r e}}{R T} = \frac{1 \times 76.66}{0.821 \times 300} = 3.11 g / l i t r e$

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