A mixture of `NO_(2)` and `N_(2)O_(4)` has a vapour density of 38.3 at 300 K. What is the number of moles of `NO_(2)` in 10 moles of the mixture ?

Explanation

Suppose the mixture contains x moles of `NO_(2)`.Then `N_(2)O_(4)=(10-x)` Moles.
Mole fraction of `NO_(2)=(x)/(10)`. Mole fraction of `N_(2)O_(4)=(10-x)/(10)`
Average molecular mass of the mixture`=(x)/(10)xx46+(10-x)/(10)xx92`
(Mol. Mass of `NO_(2)`=46, mol. mass of `N_(2)O_(4)=92`)
Average molecular mass from vapour density`=2xx "vapour density"=2xx38.3=76.6`
`:." " 46x+((10-x)xx92)/(10)=76.6 " or " 46x+920-92x=766`
or `" " 46x=920-766=154 " or "x=(154)/(46)=3.34` mole.