The vapour density of mixture consisting of `NO_2` and `N_2O_4` is 38.3 at `26.7^@C`. Calculate the number of moles of `NO_2` I `100g` of the mixture.

Explanation

Suppose `NO_(2)` present in 100 g of the mixture = x g
Then `N_(2)O_(4)` present in the mixture `=(100-x)g`
Molar mass of `NO_(2)=14+32="46 g mol"^(-1)`
`"Molar mass of "N_(2)O_(4)="92 g mol"^(-1)`
`"Molar mass of mixture "=2xxV.D.=2xx38.3="76.6 g mol"^(-1)`
Expressing all quantities in terms of moles, we should have
`(x)/(46)+(100-x)/(92)=(100)/(76.6)" or "92x+4600-46x=5524.8`
`"or "46x=924.8" or "x=20.1g`
` therefore" No. of moles of NO"_(2)" in the mixture"=(20.1)/(46)=0.437`