A1 273 K and I atm, I Lof N_(2)O_(4(g)) decomposes to NO_(2(g)) as given, N_(2)O_(4(g)) 2NO_(2(g)), At equilibrium . original volume is 25% lessthan the exisiting volume percentage decomposition of N_(2)O_(4(g)) is thus,

A1 273 K and I atm, I Lof N_(2)O_(4(g)) decomposes to NO_(2(g)) as giv

1.	A1 273 K and I atm, I Lof N_(2)O_(4(g)) decomposes to NO_(2(g)) as given, N_(2)O_(4(g)) 2NO_(2(g)), At equilibrium . original volume is 25% lessthan the exisiting volume percentage decomposition of N_(2)O_(4(g)) is thus,
Answer» 0.25 0.5 0.666 0.3333 Solution :PV=nRT, `V=(n)/(P)RT` at constant P & T , `V ALPHA n` Thus, moles can be EXPRESSED in terms of volume `underset(1-x)underset(1)(N_(2)O_(4(g))) HARR underset(2x)underset(0)(2NO_(2(g)))` Total volume at equilibrium = (1+x) Given `(75)/(100)(1+x)=1 IMPLIES x=(1)/(3)=0.333` = 33.3% dissociation