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09/10/2021 11:08 am
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A gaseous compound of molecular mass 82.1 dissociates on heating to 400 K as
X2Y4(g) ⇌ X2(g) + 2Y2(g)
The density of the equilibrium mixture at a pressure of 1 atm and temperature of 400K is 2.0gL-1. The percentage dissociation of the compound is
(a) 12.5%
(b) 48.5%
(c) 90.1%
(d) 25.0%
1 Answer
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09/10/2021 11:14 am
Correct answer: (a) 12.5%
Explanation:
D = \(\frac{PM}{RT}\)
= \(\frac{1 \times 82.1}{0.0821 \times 400}\) = 2.5gL-1
d = 2.0gL-1 (given)
α = \(\frac{D -d}{d(n-1)}\)
= \(\frac{2.5 - 2}{2(3-1)}\)
= 0.125 = 12.5%