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.

Options

A. `0.2`
B. `0.4`
C. `0.8`
D. `1.6`

Answer

B

Explanation

`alpha=(D-d)/(d)`
D= Vapour density before dissociation
d= vapour density after dissociation
`N_(2)O_(4) hArr 2NO_(2)`
Vapour density of `N_(2)O_(4)` before dissociation
`(D)=(14xx2+16xx4)/(2)=92/2=46`
Vapour density after dissociation `(d)=38.3`
`:. alpha=(46-38.3)/38.3=0.2`
`{:(,N_(2)O_(4),hArr,2NO_(2)),("Initial",1,,0),("At equilibrium",1-alpha,,2alpha):}`
Number of moles of `NO_(2)` at eq. `=2alpha=2xx0.2=0.4`