Equilibrium Question 721
Question: A gaseous compound of molecular mass 82.1 dissociates on heating to 400 K as $ X_2Y_4(g)\rightarrow X_2(g)+2Y_2(g) $ The density of the equilibrium mixture at a pressure of 1 atm and temperature of 400K is $ 2.0g{L^{-1}} $ . The percentage dissociation of the compound is
Options:
A) 12.5%
B) 48.5%
C) 90.1%
D) 25.0%
Show Answer
Answer:
Correct Answer: A
Solution:
$ D=\frac{PM}{RT}=\frac{1\times 82.1}{0.0821\times 400}=2.5g{L^{-1}} $
$ d=2.0,g,L^{-1} $ (given) $ \alpha =\frac{D-d}{d(n-1)}=\frac{2.5-2}{2(3-1)}=0.125=12.5% $
BETA
