Question

The vapour density of a gaseous mixture containing only $Ar$ and $N_2{O}_4$ gases is $40$. When the mixture is left for some time, the vapour density is decreased and finally becomes
$\left(37.5\right)$. It happened due to the dissociation of some $N_2{O}_4$ into $N{O}_2$. $(Ar=40)$. What is the degree of dissociation of $N_2{O}_4?$

• A. $0.870$
• B. $0.780$
• C. $0.087$
• D. Data is insufficient for calculation

Answer 1

The correct option is C $0.087$

Let’s assume that the initial composition of $Ar=amole$ and of $N_2{O}_4=bmole$.
Molar mass of $Ar=40gmo{l}^{-1}$ and of $N_2{O}_4=92gmo{l}^{-1}$
Using formula:
= (Mole of component 1$\times$ molar mass) + (Mole of component 2$\times$ molar mass) = (Mole of component 1 + Mole of component 2)$\times$ molar mass of mixture
Where (molar mass of mixture = 2$\times$ vapor density)
Now, $a\times 40+b\times 92=(a+b)\times 40\times 2$
⇒ a : b = 3 : 10
$N_2{O}_4$ $\to$ $N{O}_2$
$(b–x)mole$ $2xmole$
Now, $80\times (a+b)=(a+b+x)\times 37.5\times 2$
∴ Degree of dissociation of $N_2{O}_4$ = $\frac{x}{b}=0.087$