Question

On passing $10.0$ L of a gaseous mixture of $N{O}_2$ and $N_2$ at STP through an $NaOH$ solution, a mixture of $NaN{O}_2$ and $NaN{O}_3$ is formed. $6.32$ g of $KMn{O}_4$ is required to oxidise above $NaN{O}_2$ in $H_{2}S{O}_4$ medium. If $X$ percentage by mass of $N_2$ is present in gaseous mixture, ($N_2$ does not react with $NaOH$) then value of $100X$ is :

• A. $4282$
• B. $4000$
• C. $4500$
• D. none of these

Answer 1

The correct option is A $4282$

$2N{O}_2+2NaOH\to NaN{O}_3+NaN{O}_2+H_{2}O$

Eq of $KMn{O}_4=$ Eq of $N{O}_2^{⊝}$

$[N{O}_2^{⊝}\to N{O}_3^{⊝}(x=2\left)\right]$

Eq of $KMn{O}_4=\frac{6.32}{31.5}$ = Eq. of $N{O}_2^{⊝}$

Moles of $N{O}_2^{⊝}=\frac{6.32}{31.5}\times \frac{1}{2}=0.1$ mol

Weight of $N{O}_2^{⊝}=0.1\times 69=6.9g$

Weight of $NaN{O}_2=6.9g$

From the above equation
$0.1$ mol of $NaN{O}_2=0.2$ mol $N{O}_2$
$=0.2\times 22.4L$ at STP
$=4.48L$ $N{O}_2$

Volume of $N_2=(10-4.48)=5.52L$
$=0.246$ mol of $N_2$

Mole of $N{O}_2=0.2=0.2\times 46=9.2g$

Mole of $N_2=0.246=0.246\times 28=6.89g$

$\%$ of $N{O}_2=57.18;$ $\%$ of $N_2=42.82$

$100X=100\times 42.82=4282$