The vapour density of undecomposed $\mathrm{N}_{2} \mathrm{O}_{4}$ is 46. When heated, vapour density decreases to $25.4$ due to its dissociation to $\mathrm{NO}_{2}$. The % dissociation of $\mathrm{N}_{2} \mathrm{O}_{4}$ at the final temperature is

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Aicharya · Accepted Answer

$\mathrm{N}_{2} \mathrm{O}_{4} ightleftharpoons 2 \mathrm{NO}_{2}$
1 $\qquad\qquad$ 0 $\qquad$ at initial
$1-\alpha \qquad 2 \alpha \qquad$ at equilibrium
$\therefore \frac{\mathrm{V} \cdot \mathrm{D}_{\text {initial }}}{\mathrm{V} \cdot \mathrm{D}_{\text {final }}}=\frac{\mathrm{n}_{\text {final }}}{\mathrm{n}_{\text {initial }}}$
$\frac{46}{25.4}=\frac{1+\alpha}{1}$
$1.8=1+\alpha \Rightarrow \alpha=0.8$ or $80 \%$

The vapour density of undecomposed $\mathrm{N}_{2} \mathrm{O}_{4}$ is 46. When heated, vapour density…

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