Question

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

• A. $53.3$%
• B. $106.6$%
• C. $26.7$%
• D. None of the above

Answer 1

Answer: (A)

$53.3$%

Vapour Density = $\frac{MolecularMass}{2}$,

If D is the initial vapour density of $N_2{O}_4$,

D = $\frac{92}{2}=46$,

If d is the vapour density at equilibrium,
$d=30$,

Also, if $\alpha$ is the degree of dissociation,
then we have the relation, $\alpha =\frac{D-d}{(n-1)d}$ = $\frac{46-30}{30}$ = $\frac{16}{30}$ = $0.53$

n=2 for $N_2{O}_4$ as 2 moles of $N{O}_2$ are formed for its 1 mole.

Substituting the values, we get $\alpha =0.533$,

Hence the % dissociation is $53.33$.