Question

The mole fraction of $N_{2}O$ In a mixture of $N_{2}O$ and $NO$ is $\frac{1}{3}$ . The mole fraction of $NO$ in a mixture of $NO$ and $N{O}_2$ is $\frac{3}{4}$ . The two mixtures are mixed in $3:4$ molar ratio. the mole fraction of $N{O}_2$ in the resulting mixture is:

Answer 1

In $N_{2}O+NO$ mixture $X_{N_{2}O}=\frac{1}{3}$

$\therefore X_{N}O=1-\frac{1}{3}=\frac{2}{3}$

In $NO+N{O}_2$ mixture $X_{NO}=\frac{3}{4}$

$\therefore$ $X_{N{O}_2}=1-\frac{3}{4}=\frac{1}{4}$

The two mixtures are mixed in the ratio of $3:4$

$\therefore X_{N_{2}O}=\left(\frac{1}{3}\right)\times 3=1X_{NO}=\frac{2}{3}\times 3+\frac{3}{4}\times 4=2+3=5X_{N{O}_2}=\frac{1}{4}\times 4=1$

Mole fraction of $N{O}_2=\frac{X_{N{O}_2}}{X_{N_{2}O}+X_{NO}+X_{N{O}_2}}=\frac{1}{7}$