Question

The vapour density of undecomposed N2O4 is 46. When heated the vapour density decreases to 24.5 due to its dissociation to NO2(g).The percentage dissociation of N2O4 is :
(1) 87.75
(2) 60
(3) 40
(4) 70

Answer 1

Is the answer first option?

Answer 2

Answer : The correct option is, (1) 87.75

Solution :

First we have to calculate the molecular mass before dissociation and after dissociation.

Formula used :

$\text{Molecular mass}=2\times \text{Vapor density}$

Molecular mass before dissociation :

$\text{Molecular mass}=2\times 46=92$

Molecular mass after dissociation :

$\text{Molecular mass}=2\times 24.5=49$

Now we have to calculate the Van't Hoff factor, $i$.

Formula used :

$i=\frac{\text{Molecular mass before dissociation}}{\text{Molecular mass after dissociation}}=\frac{92}{49}=1.8775$

Now we have to calculate the degree of dissociation.

The balanced reaction will be,

$N_2O_4\rightleftharpoons 2NO_2$

Formula used :

$\alpha=\frac{i-1}{n-1}$

where,

n = number of particles after dissociation = 2

Now put all the given values in this formula. we get the degree of dissociation.

$\alpha=\frac{1.8775-1}{2-1}=0.8775$

Now we have to calculate the percentage dissociation.

$\text{percentage dissociation}=0.8775\times 100=87.75$

Therefore, the percentage dissociation of $N_2O_4$ is, 87.75%