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At $$27^\circ\text{C}$$ 4.6 grams of $$\text{N}_2\text{O}_4$$ undergoes decomposition and exerts an equilibrium pressure of 1.6 atmosphere when vapourised in a 1 lit flask. The fraction of $$\text{N}_2\text{O}_4$$ dissociated is ________. [$$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$$]
Correct Answer: 0.3
4.6 g of $$\text{N}_2\text{O}_4$$ is 0.05 mol, which alone would exert $$\frac{0.05 \times 0.0821 \times 300}{1} \approx 1.23\text{ atm}$$. If a fraction $$\alpha$$ dissociates, the total moles become $$0.05(1 + \alpha)$$, so $$1.23(1 + \alpha) = 1.6$$. This gives $$\alpha \approx 0.3$$.
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