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Types of Equillibrium

N2O3 on decom...

Question

$N_{2} O_{3}$ on decomposition gives NO and $N O_{2},$ they are found to be in equilibrium at 300 K. If the vapour density of such an equilibrium mixture is 23.75, calculate percentage by mass of $N_{2} O_{3}$ in the equilibrium mixture?

A

80%

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B

60%

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C

40%

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D

20%

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Solution

The correct option is C 40%
$N_{2} O_{3} ⇌ N O + N O_{2}$
$∵ α = \frac{D - d}{d (n - 1)}$
d=23.75
D=38
$α = \frac{38 - 23.75}{23.75 (2 - 1)} = 0.6$
Mass% of $N_{2} O_{3}$ in the equilibrium mixture $= \frac{Weight of N_{2} O_{3}}{Total weight} \times 100$
$= \frac{0.4 \times 76}{0.6 \times 30 + 0.6 \times 46 + 0.4 \times 76} \times 100$
$= 40 %$

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