The vapour density of `N_(2)O_(4)` at a certain temperature is `30`. Calculate the percentage dissociation of `N_(2)O_(4)` this temperature.

Explanation

`Mol. Mass of N_(2)O_(4)=2xx14+4xx16=92`
`Vapour density=(92)/(2)=46`
Suppose `N_(2)O_(4)` dissociates as
`{:(,N_(2)O_(4),hArr,2NO_(2)),("At t=0",1,,0),("At eqm.",1-alpha,,2alpha):}`
And `(D_(0)-D_(e))/(D_(e))=(n-1)alpha`
`D_(0)=` Initial vapour density,
`D_(e)= V.D.` at equilibrium `=3`
`n=` Number of molecules formed by dissociation of one molecule of `N_(2)O_(4)=2`
and `alpha=(46-30)/(30)=(16)/(30)=0.533` or `53.3%`