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JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ce208

$N X_{2} O X_{4} (g)$ ( $M = 92 g/mol$ ) is 25\% dissociated into $NO X_{2} (g)$ at constant temperature and pressure. Select the correct statements.

AThe average molar mass of the mixture is $73.6 g/mol$
BThe vapour density of the mixture (relative to $H X_{2}$ ) is $36.8$
CThe mole fraction of $NO X_{2}$ is $0.50$
DThe volume of the gas at the same $T$ and $P$ is 1.25 times that of the same amount of undissociated $N X_{2} O X_{4}$

Answer: A, B, D

M_{a vg} = 92/1.25 = 73.6

and vapour density

= 36.8

(A, B true). Per mole of

N X_{2} O X_{4}

:

N X_{2} O X_{4}

0.75,

NO X_{2}

0.50, total 1.25, so

x (NO X_{2}) = 0.40

, not 0.50 (C false). At constant

T, P

the volume is proportional to moles, so it is 1.25 times larger (D true).

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