Question

$N_2{O}_4$ dissociates as $N_2{O}_{4\left(g\right)}\rightleftharpoons 2N{O}_{2\left(g\right)}$ at $273K$ and $2atm$ pressure. The equilibrium mixture has a density of $41$. What will be the degree of dissociation?

• A. $14.2\%$
• B. $16.2\%$
• C. $12.2\%$
• D. $None$

Answer 1

Answer: (C)

$12.2\%$

$\alpha =\frac{D-d}{(n-1)d}$

D= Theoretical vapour density

d=Observed vapour density

n=Number of moles obtained after dissociation after one mole

$\alpha =\frac{46-41}{(2-1)41}=0.122$

% discount= $12.2$%