Question

The vapour density of a mixture containing $N{O}_2$ and $N_2{O}_4$ is 27.6. The mole fraction of $N{O}_2$ in the mixture will be :

• A. 0.2
• B. 0.4
• C. 0.6
• D. 0.8

Answer 1

The correct option is D 0.8

The equilibrium reaction is $N_2{O}_{4\left(g\right)}\rightleftharpoons 2N{O}_{2\left(g\right)}$.
Initially, 1 mole of $N_2{O}_4$ and 0 moles of $N{O}_2$ are present.
At equilibrium, $1-\alpha$ moles of $N_2{O}_4$ and $2\alpha$ moles of $N{O}_2$ are present. Here, $\alpha$ is the degree of dissociation of $N_2{O}_4$.
The relationship between the degree of dissociation $\alpha$, the molecular weight D of $N{O}_2$ and the vapour density d is as shown below.
$\alpha =\frac{D-d}{(n-1)d}=\frac{46-27.6}{27.6}=0.67$
The mole fraction of $N{O}_2=\frac{2\alpha }{1+\alpha }=0.80$.