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At 26.7^(@)C, the vapour density of a ga...

At `26.7^(@)C`, the vapour density of a gaseous mixture containing `NO_(2) and N_(2)O_(4)` is 38.31. calculate the number of moles of `NO_(2)` in 100g of that mixture.

Text Solution

Verified by Experts

Let the amount of `NO_(2)` in 100g of the mixture be x g
`therefore`Amount of `N_(2)O_(4)=(100-x)g`
`therefore`Number of moles of `NO_(2)` in mixture`=(x)/(46)`
and number of moles of `N_(2)O_(4)` in mixture=`((100-x))/(92)`
`[because M_(NO_(2)) =46 and M_(N_(2)O_(4))=92]`
`therefore`Total number of moles of `NO_(2)` and `N_(2)O_(4)` in the mixture
`=(x)/(46)+(100-x)/(92)=(100+x)/(92)` . . (1)
Now, molecular mass of the mixture=`(2xx38.3)=76.6`
`therefore`Total number of moles of the mixture`=(100)/(76.6)` . . . (2)
Now from (1) and (2), `(100+x)/(92)=(100)/(76.6) or, x=20.1`
So, number of moles of `NO_(2)` in mixture=`(21.1)/(46)=0.4369`.
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