The vapour density of N_ 2 O_ 4 at a certain temperature is 30....

The reaction is ${N}_{2} {O}_{4} ⇌ 2 {NO}_{2}$ mol. wt. of N_{2}O_{4}_{}=14 x 2 + 16 x 4=92 $∴$ Vapour density (D) of N_{2}O_{4}=92/2=46 We know that, $\alpha=\dfrac{\text{D} - \text{d}}{\left(\text{n} - 1\right) \text{d}}=\dfrac{4 6 - 3 0}{\left(2 - 1\right)^{3} 0}=\dfrac{1 6}{3 0}$ where, D=Initial vapour density of reactant d=vapour density of equilibrium mixture $\alpha$=degree of dissociation ∴ \alpha=\text{0.533} =\text{53.3 %}