A sample of $N_2{O_4}_{\left(g\right)}$, with a pressure of 1 atm is placed in a flask. When equilibrium is reached, 20% ${\mathrm{N}}_2{{\mathrm{O}}_4}_{\left(\mathrm{g}\right)}$,  has been converted to ${\mathrm{NO}}_2$,  $N_2{O_4}_{\left(g\right)}\rightleftharpoons 2N{O_2}_g$. If the original pressure is made 10% of the earlier, then percent dissociation is __________.

 It is given to find out the percent dissociation.

${\mathrm{N}}_2{{\mathrm{O}}_4}_{\left(\mathrm{g}\right)}\rightleftharpoons 2\mathrm{N}{{\mathrm{O}}_2}_{\mathrm{g}}$

 $K_c$= $\frac{{\left[N{O}_2\right]}^2}{\left[N_2{O}_4\right]}$

$K_c$= $\frac{{\left[0.2\right]}^2}{\left[1-0.2\right]}$ = 0.5

Now original pressure is made 10% of the earlier, P = 10% of 1 atm

= $\frac{10}{100}$ = 0.1

${\mathrm{N}}_2{{\mathrm{O}}_4}_{\left(\mathrm{g}\right)}\rightleftharpoons 2\mathrm{N}{{\mathrm{O}}_2}_{\mathrm{g}}$

$K_c$ = 0.5

0.5 = $\frac{(\alpha {)}^2}{0.1-\alpha }$

0.05 – $0.5\alpha$ = ${\alpha }^2$

${\alpha }^2$ + $0.5\alpha$ - 0.05 = 0

On solving we get $\alpha$ = 0.2 that is 0.2 $\times 100$= 20%

Hence, the answer is 20%.