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)

On passing 10.0 L of a gaseous mixture of NO_(2) and N_(2) at STP, thr

1.	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)
Answer» Solution :`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`