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

The correct option is A 53.3%

Initially concentration of $N_2{O}_4=1$

Then at equilibrium $N_2{O}_4$ has concentration $=1-x$and $N{o}_2=2x$

Initially we have taken $N_2{o}_4=1,molarmass=92$

At equilibrium vapour density$=30$ , molecular weight$=60$

$\frac{initialconc.}{finalconc.}=\frac{initialmolarmass}{finalmolarmass}$

$\frac{1}{1-x+2x}=\frac{60}{92}$

$\frac{1}{1+x}=\frac{92}{60}$

$1+x=\frac{92}{60}$

$x=\frac{92}{60}-1=\frac{32}{60}$

$x=0.533$

% dissociation$=0.533\times 100=53.3$