The vapour density of a mixture containing `NO_(2)` and `N_(2)O_(4)` is `38.3 at 27^(@)C`. Calculate the mole of `NO_(2)` in `100 g` mixture.

The vapour density of a mixture containing `NO_(2)` and `N_(2)O_(4)` i

1.	The vapour density of a mixture containing `NO_(2)` and `N_(2)O_(4)` is `38.3 at 27^(@)C`. Calculate the mole of `NO_(2)` in `100 g` mixture.
Answer» Correct Answer - B::C::D `{:(,N_(2)O_(4)(g),hArr,2NO_(2)(g)),("At equilibrium",(1-x),,2x):}` x (degree of dissociation)=`(D-d)/((n-1)d)` Given, `d=38.3, D=("molecular mass of" N_(2)O_(4))/(2)=92/2=46, n=2` So, `x=(46-38.3)/38.3=0.2` At equilibrium, amount of `N_(2)O_(4)=1-0.2=0.8 "mol"` and amount of `NO_(2)=2xx0.2=0.4 "mol"` Mass of the mixture `=0.8xx92+0.4xx46` `=73.6+18.4=92.0 g` Since, `92 g` of the mixture contains `=0.4 "mol" NO_(2)` So `100g` of the mixture contains `=(0.4xx100)/92` `=0.43 "mol" NO_(2)`

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The vapour density of a mixture containing `NO_(2)` and `N_(2)O_(4)` is `38.3 at 27^(@)C`. Calculate the mole of `NO_(2)` in `100 g` mixture.

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