The vapour density of undecomposed $N_{2} O_{4}$ is 46. When heated, vapour density decreases to $24.5$ due to its dissociation to ${N O}_{2}$ . The percentage dissociation of $N_{2} O_{4}$ at the final temperature is:

88

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60

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40

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70

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Solution

The correct option is D $88$
The molecular weight of the substance $= 2 \times$ vapor density.

Before dissociation, $M W = 2 \times 46 = 92$

After dissociation, $M W = 2 \times 24.5 = 49$

vant half factor $(i) = \frac{MW before dissociation}{MW after dissociation} = \frac{92}{49} = 1.8775$

Also, $α = \frac{i - 1}{n - 1} = degree of dissociation which is the number of particles after dissociation.$

$N_{2} O_{4} ⟶ 2 N O_{2}; n = 2; α = \frac{1.8775 - 1}{2 - 1} = 0.8775$

$∴ % α = 87.75 % ≃ 88 %$

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The vapour density of undecomposed N2O4 is 46. When heated, vapour density decreases to 24.5 due to its dissociation to NO2. The percentage dissociation of N2O4 at the final temperature is:

The vapour density of undecomposed $N_{2} O_{4}$ is 46. When heated, vapour density decreases to $24.5$ due to its dissociation to ${N O}_{2}$ . The percentage dissociation of $N_{2} O_{4}$ at the final temperature is: