1.

138 gm of N_2O_4(g) is placed in 8.2L container at 300 K.The equilibrium vapour density of mixture was found to be 30.67.Then (R=0.082 L atm "mol"^(-1) K^(-1))

Answer»

`ALPHA`=DEGREE of DISSOCIATION of `N_2O_4=0.25`
`K_p` of `N_2O_4 hArr 2NO_2` (g) will be 9 atm
Total pressure at equilibrium =6.75 atm
The density of equilibrium mixture will be 16.83 gm/litre.

Solution :`{:(,N_2O_4""hArr,2NO_2),(t=0,a,0),(t,a(1-alpha),2aalpha):}`
vapour density`=46/(1+alpha)=30.67`
so `1+alpha=1.5=0.5=50%`
Total pressure`=(1.5xx1.5xx0.082xx300)/8.2=6.75` atmSo, `K_p=(4alpha^2)/(1-alpha^2)P=9` atm
and for density of mixture `=138/8.2` gm/L=16.83 gm/L


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