$N X_{2} O X_{4}$ ( $92 g/mol$ ) partly dissociates as $N X_{2} O X_{4} 2 NO X_{2}$ at a certain temperature, and the equilibrium mixture has vapour density $34.5$ . The degree of dissociation of $N X_{2} O X_{4}$ is

Answer: B

Initial vapour density D = 92/2 = 46. For 1 mol dissociating by

α

, moles = 1 +

α

, and mass is conserved, so d = D/(1+

α

α

= (D - d)/d = (46 - 34.5)/34.5 = 0.3333 = 33.3%.

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