The vapour density of a mixture containing `NO_(3)` and `N_(2)O_(4)` is 38.3 at `27^(@)C` Calcualte the moles of `NO_(2)` in 100g of the mixture.

Explanation

Mol. Mass of mixture `=2xx38.3=76.6`
No. of moles in 100g of mixture `=(100)/(76.6)`
Let a g of `NO_(2)` is present in mixture.
Moles of `NO_(2)+` moles of `N_(2)O_(4)=` Moles of mixture
`(a)/(46)+(100-a)/(92)=(100)/(76.6)` or `a=20.10g`
Moles of `NO_(2)` in mixture `=(20.10)/(46)=0.437`.