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.

Answer

0.437

Explanation

Suppose `NO_(2)` present in 100 g of the mixture = x g.
Then `N_2O_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 `=2 xx V.D. = 2 xx 38.3 = 76.6 "g mol"^(-1)`
Expressing all quantities in terms of moles, we should have-
`x/46 + (100 -x)/92 = 100/76.6`
or `46 x = 924.8`
or x = 20.1 g
`therefore` No. of moles of `NO_(2)` in the mixture.
`=20.1/46 = 0.437`