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System, Surroundings, Intensive, Extensive Properties

The vapour de...

Question

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 % dissociation of $N_{2} O_{4}$ is:

40

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57

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67

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87

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Solution

The correct option is D $87$
The vapour density of undissociated $N_{2} O_{4}$ and dissociated $N_{2} O_{4}$ are 46 and 24.5 respectively.

Molecular weight is twice vapour density.

For undissociated $N_{2} O_{4}$ , molecular weight $= 2 \times 46 = 92$

For dissociated $N_{2} O_{4}$ , molecular weight $= 2 \times 24.5 = 49$

$N_{2} O_{4} (g) 100 100 - x ⇌ 2 N O_{2} (g) 0 2 x$

Assume initially 100 moles of $N_{2} O_{4}$ are present and x is the percent dissociation.

x moles of $N_{2} O_{4}$ dissociate to form 2x moles of
$N O_{2}$ .
Total number of moles after dissociation $= (100 - x) + 2 x = 100 + x$
The molecular weights of pure $N_{2} O_{4}$ and pure $N O_{2}$ are 92 g/mol and 46 g/mol respectively. The molecular weight of dissociated $N_{2} O_{4}$ is 49 g/mol.
The average molecular weight after dissociation

$\Rightarrow \frac{92 (100 - x) + 46 (2 x)}{100 + x} = 49$

$\Rightarrow 92 (100 - x) + 46 (2 x) = 49 100 + x$

$\Rightarrow 9200 - 92 x + 92 x = 4900 + 49 x$

$\Rightarrow 4300 = 49 x$

$\Rightarrow x = 87$

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Similar questions

Q. The vapour density of undecomposed

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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 % dissociation of N2O4 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 % dissociation of $N_{2} O_{4}$ is: