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?

Answer

D

Explanation

`{:(,N_(2)O_(4),hArr,2NO_(2)),("moles before",1,,0),("equilibrium",,,),("moles at equilibrium",(1-alpha),,2alpha):}`
(Let `alpha` be degree of dissociation)
`:.` Total moles at equilibrium`=1+alpha( :' a=0.66)`
`=1+0.66=1.66`
`1` mol of `N_(2)O_(4)` is taken, moles at equilibrium `=1.66`
`10/92` mol of `N_(2)O_(4)` is taken, moles at equilibrium
`=(1.66xx10)/92=0.18`
`:' PV=w/m RT`
`1xxV=0.18xx0.0821xx340`
`:. V=5.04 L`