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.

Answer

B, C, D

Explanation

`{:(,N_(2)O_(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=("molecular mass of" N_(2)O_(4))/(2)=92/2=46, n=2`
So, `x=(46-38.3)/38.3=0.2`
At equilibrium, amount of `N_(2)O_(4)=1-0.2=0.8 "mol"`
and amount of `NO_(2)=2xx0.2=0.4 "mol"`
Mass of the mixture `=0.8xx92+0.4xx46`
`=73.6+18.4=92.0 g`
Since, `92 g` of the mixture contains `=0.4 "mol" NO_(2)` So
`100g` of the mixture contains `=(0.4xx100)/92`
`=0.43 "mol" NO_(2)`