The vapour density of ${\mathrm{N}}_2{\mathrm{O}}_4$ at certain temperature is 30 . What is the percentage dissociation of ${\mathrm{N}}_2{\mathrm{O}}_4$ at that temperature?

[ Hint:$\alpha =(\mathrm{D}-\mathrm{d})/\mathrm{d}(\mathrm{n}-(A\left)\right]$

 The relationship between molar mass and vapour density is Molar mass $=2\times$ vapour density $=2\times 30=60$

$83.4$

 ${\mathrm{N}}_2{\mathrm{O}}_4$ &${\mathrm{NO}}_2$ 

 Initial moles  1 $\mathrm{O}$ 

 Moles at equilibrium  $1-\alpha$  $2\alpha$ 

 Mole fraction & $\left(\frac{1-\alpha }{1+\alpha }\right)$  $\left(\frac{2\alpha }{1+\alpha }\right)$

Thus$92\left(\frac{1-\alpha }{1+\alpha }\right)+46\left(\frac{2\alpha }{1+\alpha }\right)=60$

$\alpha =0.533$

The percentage dissociation of ${\mathrm{N}}_2{\mathrm{O}}_4$ is $53.3\%$.

Hence the correct answer is (A) 53.3