The vapour density of undecomposed N_2O_4 is | vedclass

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(A) The relation between degree of dissociation $(\alpha)$ and vapour density is given by $\alpha = \frac{D-d}{d(n-1)}$,where $D$ is the theoretical v

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$87.75$

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(A) The relation between degree of dissociation $(\alpha)$ and vapour density is given by $\alpha = \frac{D-d}{d(n-1)}$,where $D$ is the theoretical vapour density,$d$ is the observed vapour density,and $n$ is the number of moles of products formed from one mole of reactant.For the reaction $N_2O_4(g) ightleftharpoons 2NO_2(g)$,$n = 2$.Given: $D = 46$,$d = 24.5$.Substituting the values:$\alpha = \frac{46 - 24.5}{24.5(2 - 1)}$$\alpha = \frac{21.5}{24.5} \approx 0.8775$Percentage dissociation $= \alpha 	imes 100 = 0.8775 	imes 100 = 87.75\%$.

The vapour density of undecomposed $N_2O_4$ is $46$. When heated,the vapour density decreases to $24.5$ due to its dissociation into $NO_{2(g)}$. The percentage dissociation of $N_2O_4$ is:

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