Q. The vapour density of undecomposed $N_{2}O_{4}$ is $46$ . When heated, vapour density decreased to $24.5$ due to its dissociation to $NO_{2}$ . The percent dissociation of $N_{2}O_{4}$ at the final temperature is :

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A

$87$

B

$60$

C

$40$

D

$70$

Solution:

$\frac{ \left(\right. normal molar mass \left(\left.\right)_{N_{2} O_{4}}}{ \left(exp . molar mass\right)_{N_{2} O_{4}}}=i$
$i=1+\alpha $
$\frac{ normal molar mass}{ exp . molar mass}=1+\alpha $ (exp. molar mass $=2\times VD$ )
$=2\times 24.5=49$
$\frac{92}{49}=1+\alpha $
$\therefore \alpha =0.87$

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