One mole of a mixture of `N_(2),NO_(2)` and `N_(2)O_(4)`, has a mean molar mass of `55.4`. On heating to a temperature at which `N_(2)O_(4)` may be dissociated `:N_(2)O_(4)rarr2NO_(2)`, the mean molar mass tends to the lower value of `39.6`. What is the mole ratio of `N_(2) : NO_(2) : N_(2)O_(4)` in the original mixture?

Answer

`0.5 : 0.1 : 0.4`

Explanation

Mixture `(N_(2),NO_(2),N_(2),O_(4))` has mean molar mass `=55.4`
`x" "y" "z`
Given: `{:(N_(2)O_(4)rarr2NO_(2)),(z" "2z):}`
`therefore 55.4=(28x+46(y+2z))/(x+y+z)`
`{"mean molar mass" =(wtxx"mole")/("Total mole")}`
Given : `x+y+z =1` (mole)
so `55.4 = 28x + 46 (y+2z)" "...(1)`
`therefore 39.3 = (28x + 46(y+2z))/(x+y+2z)`
`therefore 39.6(x+y+2z)=28x+46(y+2z)`
From `eq (1) & x+y+z=1`
or `1+z = (59.4)/(39.6)`
or `z=0.4`
from `eq. (1)`
`55.4 = 28 x+46(y+2z)`
`z=0.4`
` 55.4 = 28x + 46y +36.8`
`28x+46y=18.6 " "(2)`
` because x+y+z=1`
`x+y+0.4=1(beacuse z =0.4)`
`x+y=0.6 " "...(3)`
`eq (2) xx1....eq. (3) xx28`
`{:(,28x+46y=18.6),(,28x+28y=16.8),(-, " - -"),(,bar(" 18y=1.8")):}`
`y=0.1`
`because x+y+z=1 x=0.5`