The vapour density of a mixture containing `NO_(2)` and `N_(2)O_(4)` is `38.3 at 27^(@)C`. Calculate the mole of `NO_(2)` in `100 g` mixture.

Explanation

Vapour density of the mixture of `NO_(2) and N_(2)O_(4)=38.3`
Molecular mass of the mixture `= 2 xx` Vapour density `= 2 xx 38.3 = 76.6 u = 76.6 g`
Mass of the mixture = 100 g
No. of moles of mixture `= (100)/(76.6)` ….(i)
Let the mass of `NO_(2)` in the mixture = x g
`:.` Mass of `N_(2)O_(4)` in the mixture `= (100 -x)g`
Molar mass of `NO_(2)=14 + 32 = 46 u = 46 g`
Molar mass of `N_(2)O_(4)=28.64 = 92 u = 92 g`
No. of moles of `N_(2)O_(4) = ((100 - x))/(92)`
Total no. of moles in the mixture `= (x)/(46)+((100-x))/(92)`....(ii)
Equating (i) and (ii), `(x)/(46)+((100-x))/(92)=(100)/(76.6)`
`92x+46 (100-x)=(100)/(76.6) xx46 xx 92 = 5524.8`
`92x-46x = 5524.8 -4600 = 924.8`
`46 x = 924.8 or x = (924.8)/(46)=20.1`
`:.` Mass of `NO_(2)=20.1` g or No. of moles of `NO_(2)=(20.1)/(46)=0.437`.