`20%N_(2)O_(4)` molecules are dissociated in a sample of gas at `27^(@)C` and 760 torr. Calculate the density of the equilibrium mixture.

Answer

C

Explanation

`{:(,N_(2)O_(4)(g),hArr,2NO_(2)(g)),("Initial",1 "mole",,),("At Eq",1-0.2,,2xx0.2=0.4),(,=0.8 "mol",,):}`
Total `=0.8+0.4=1.2`
If V is the volume of the vapour per mole.
volume of vapour before dissociation =V
Hence, density `(D) prop 1/(V)`
But density before dissociation (theoretical density)
`D=("molecular weight of" N_(2)O_(4))/(2)=92/2=46`
Volume after dissociation `=1.2 V`
Density `(d) prop 1/(1.2V)`
`D/d=1/Vxx1.2 V=1.2`
`d=D/1.2=4.6/1.2=38.3`
Or use formula
`alpha=(D-d)/(d)`
`0.2=(46-d)/(d)`
`0.2d=46-d`
`0.2d+d=46`
`1.2d=46`
`d=46/1.2=38.3`