At `25^(@)C` and `1` atm, `N_(2)O_(4)` dissociates the reaction `N_(2)O_(4)(g) hArr 2NO_(2)(g)`
If it is `35%` dissociated at given condition, find the volume of above mixture will difuse if `20 mL` of pure `O_(2)` diffuse`10` minutes at same yemperature and pressure.

Explanation

For equilibrium system, `N_(2)O_(4)(g) hArr 2NO_(2)(g)`,

`rArr` Total moles at equilibrium `=a+aalpha`
`p_(N_(2)O_(4))=(a-aalpha)/(a+aalpha)P` and `p_(NO_(2))=(2aalpha)/(a+aalpha)P`
Here `P=1.0 "atm"` and `K_(p)=(p_(NO_(2))^(2))/(p_(N_(2)O_(4)))=(4alpha^(2)P)/(1-alpha^(2))[alpha=0.35]`
`rArr K_(p)=(4xx(0.35)^(2))/(1-(0.35)^(2))xx1=0.56 "atm"`
Using `(a+aalpha)/(a)=(M_(mix, i))/(M_(mix, f))=(M_(N_(2)O_(4)))/(M_(mix, f))`
`rArr 1+0.35=92/(M_(mix, f))`
`:. M_(mix, f)=(92)/(1.35)=68.15`
Let `V(mL)` volume of mixture diffused in.
From Graham's law of diffusion
`r_(O_(2))/r_(mix. f)=sqrt(M_(mix, f)/M_(O_(2)))`
`(20//10)/(V//10)=sqrt(68.15/32)rArr V=13.70 mL`