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Nitric Acid

For the react...

Question

For the reaction $N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)$ . If percentage of $N_{2} O_{4}$ are $25$ %, $50$ %, $75$ % and $100$ %, the the sequence of observed vapour densities will be

d_{1} > d_{2} > d_{3} > d_{4}

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d_{4} > d_{3} > d_{2} > d_{1}

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d_{1} = d_{2} = d_{3} = d_{4}

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d_{1} > d_{3} > d_{2} > d_{4}

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Solution

The correct option is A $d_{1} > d_{2} > d_{3} > d_{4}$
When dissociation increases, as well as the percentage of nitrogen dioxide, will also increase. While vapour density is decreased, it means vapour density is inversely proportional to the dissociation.
The dissociation vapour density will be high:
$d_{25} > d_{50} > d_{75} > d_{100}$
Therefore, the order will be:
$d_{1} > d_{2} > d_{3} > d_{4}$

Hence, option A is the correct answer.

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