Question

The vapour density of $N_2{O}_4$ at a certain temperature is 30. What is the percentage dissociation of $N_2{O}_4$ at this temperature?

• A. 53.3%
• B. 76.6%
• C. 26.7%
• D. None of the above

Answer 1

The correct option is A 53.3%

We know that, Molecular weight = $2\times$ Vapour density
Molecular weight of $N_2{O}_4$ is $92gmo{l}^{-1}$
if $D$ is the initial Vapour density of $N_2{O}_4$ then, $D=\frac{molecularweightof{N}_2{O}_4}{2}=\frac{92}{2}=46$
if $d$ is the Vapour density of equilibrium mixture, then $d$ = 30

we have the relation, $\alpha =\frac{(D-d)}{(n-1)d}.....\left(1\right)$
where, $\alpha$ is degree of dissociation & $n$ is number of moles of product formed.

The dissociation reaction is given by $N_2{O}_4\rightleftharpoons 2N{O}_2$
mean 2 mole of $N{O}_2$ will form from 1 mole of $N_2{O}_4$ i.e. $n=2$
putting the values in equation (1) we get,
$\alpha =\frac{(46-30)}{(2-1)30}$
$=\frac{16}{30}=0.533=53.3$
hence, the percentage dissociation of $N_2{O}_4$ is 53.3%.