On passing 10.0 L of a gaseous mixture of `NO_(2)` and `N_(2)` at STP, through an `NaOH` solution, a mixture of `NaNO_(2)` and `NaNO_(3)` is formed 6.32 g of `KMnO_(4)` is required to oxidise above `NaNO_(2)` in `H_(2)SO_(4)` medium. Determine the percentage by mass of gaseous mixture `(N_2` does not react with `NaOH`)

Explanation

`2NO_(2)+2NaOHtoNaNO_(3)+NaNO_(3)+NaNO_(2)+H_(2)O`
" Eq of "`KMnO_(4)=" Eq of "NO_(2)^(ɵ)`
`{NO_(2)^(ɵ)toNO_(3)^(ɵ)(x=2)]`
" Eq of "`KMnO_(4)=(6.32)/(31.5)=Eq. of NO_(2)^(ɵ)`
Moles of `NO_(2)^(ɵ)=(6.32)/(31.5)xx(1)/(2)=0.1 mol`
Weight of `NO_(2)^(ɵ)=0.1xx69=6.9g`
Weight of `NaNO_(2)=6.9g`
From the above equation
`0.1 " mol of "NaNO_(2)-=0.2 mol NO_(2)`
`-=0.2xx22.4L at STP`
`-=4.48 L NO_(2)`
Volume of `N_(2)=(10-4.48)=5.52L`
`=0.246L N_(2)`
Mole of `NO_(2)=0.2=0.2xx46=9.2g`
Mole of `N_(2)=0.246=0.246xx28=6.89g`
`% of NO_(2)=57.18,% of N_(2)=42.82g`