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

V.D. of mixture of `NO_(2)` and `N_(2)O_(4) = 38.3`
Mol wt. of mixture of `NO_(2)` and `N_(2)O_(4) = 38.3 xx 2 = 76.6` (Mol. Wt. = `2 xx` V.D.)
Let `NO_(2)` present in 100 g of mixture = x
`N_(2)O_(4)` present in 100 g of mixture = 100 -x
Mol. wt. of `NO_(2)` = 46, Mol. wt. of `N_(2)O_(4) = 92`
Now, `x/46 + (100-x)/92 = 100/76.6`
Solving for x, we get
x=20.1 g
Moles of `NO_(2)` in the mixture `=(20.1)/46`
=0.437