At `340 K` and `1` atm pressure, `N_(2)O_(4)` is `66%` into `NO_(2)`. What volume of `10 g N_(2)O_(4)` ocuupy under these conditions?

Explanation

`{:(,N_2O_4 hArr , 2NO_2),("Mole before equilibrium" ,1, 0 ("Let" alpha " be degree of dissocation")),("Mole after equilibrium" , (1-alpha), 2alpha ) :}`
`therefore` Total moles at equilibrium =`1+alpha = 1+0.66=1.66 ( therefore alpha = 0.66)`
1 moles of `N_2O_4` is taken , mole at equilibrium = 1.66
`10/92` mole of `N_2O_4` is taken , mole at equilibrium =`(1.66xx10)/92`=0.18
`therefore PV=w/m "RT", 1 xx V = 0.18 xx 0.0821xx340`, V = 5.02 litre
Thus, volume of 10 g `N_2O_4` under above condition is 5.02 litre .