Question

The vapour density of ${N_2}{O_4}$ at a certain temperature is 30. What is percentage dissociation of ${N_2}{O_4}$ at this temperature?
(A) 46 %
(B) 53 %
(C) 75 %
(D) 92 %

Answer 1

Vapour density is given in the question so to calculate the molecular mass we have to use the formula: The molecular mass of the compound = 2 $\times$ Vapour density.

Formula used: To find degree of dissociation using formula is $x = \dfrac{{D - d}}{d}$.
D = Initial vapour density
d = Vapour density at equilibrium

Complete answer:
The reversible reaction for ${N_2O_4}$ is given,
$N_2O_4(g) \leftrightharpoons 2NO_2(g)$
Degree of dissociation for reversible reaction is determined by measuring density of reaction mixture at equilibrium. Consider the general reversible reaction
Molecular mass of ${N_2}{O_4} = \left( {28 + 64} \right) = 92$
Initial Vapour density, $D = \dfrac{{92}}{2} = 46$
Let the degree of dissociation be x.
Vapour density at equilibrium d = 30
Applying the relationship,
$x = \dfrac{{D - d}}{d} = \dfrac{{46 - 30}}{{30}} = 0.533$
Degree of dissociation$= 0.533 \times 100 = 53.3\%$

Hence correct option is (B): 53.3 %.

Note: Degree of dissociation is defined as the fraction of molecules dissociate in a given time. It is denoted by the symbol alpha. In this question degree of dissociation should be related with the density of the substance.