Question

The vapour density of $N_2{O}_4$ is $46$. When heated vapour density decreases to $24.5$ due to its dissociation to $N{O}_2$ . The percentage dissociation of $N_2{O}_4$ at the final temperature is:

• A. $87$
• B. $60$
• C. $40$
• D. $70$

Answer 1

The correct option is A $87$

Vapour density of $N_2{O}_4=46$ (undecompooed)

Vapour density of $N_2{O}_4=24.5$ (after heaving $N_2{O}_4)$

Molecular weight $\left(M_w\right)=2\times$ Vapour density

Before dissociation $M_w=2\times 46=92$

After dissociation $M_w=2\times 24.5=49$

Vant Hoff faction $\left(i\right)=\frac{92}{49}=1.877=1.88$

We know that, $\alpha =\frac{i-1}{n-1}$

[Where $\alpha =$ degree of dissociation, n = no of pantloes after dissociation]

$N_2{O}_4\to 2N{O}_2$

$n=2$

$\alpha =\frac{1.88-1}{2-1}=0.88$

$\alpha \%=88\%$

So closest ans A) 87%