The vapour density of a mixture consisting of `NO_(2)` and `N_(2)O_(4)` is `38.3` at `275 K`. The number of moles of `NO_(2)` in the mixture:

Explanation

`N_2O_4(g) hArr 2NO_2(g)`
At equilibrium (1-x) 2x
x(degree of dissociation) =`(D-d)/((n-1)d)`
Given , d=38.3
`D=("Mol.mass of " N_2O_4)/2=(92)/2=46,n=2`
So `x=(46-38.3)/(38.3)=0.2`
At equilibrium amount of `N_2O_4=1-0.2 =0.8` mole
and amount of `NO_2=2xx0.2=0.4` m ole
Mass of the mixture `=0.8 xx 92+0.4 xx46 =73.6 +18.4=92.0 g`
Since , 92 gram of the mixture contains `=0.4 "mole" NO_2`
So, 100 gram of the mixture contains `=(0.4xx100)/(92)=0.43 "mole" NO_2`