Question

The observed molecular weight of $N_2{O}_4$ is 80 g $mo{l}^{-1}$ at 350 K. The percentage of dissociation of $N_2{O}_4\left(g\right)$ at 350 K is ?

• A. 10%
• B. 15%
• C. 20%
• D. 18%

Answer 1

Answer: (B)

15%

Given that $N_2{O}_4\rightleftharpoons 2N{O}_2$

at $t=0$ $1$ $0$

$t=t_{eq}$ $1-\alpha$ $2\alpha$

Molecular mass of $N_2{O}_4=92g/mol$

Given Molecular mass of $N_2{O}_4=80g/mol$

$i=\frac{Molecularmassof{N}_2{O}_4}{Givenmolecularmassof{N}_2{O}_4}=\frac{92}{80}=1.15$

$\Rightarrow i=\frac{1-\alpha +2\alpha }{1}=1.15$

$\Rightarrow 1+\alpha =1.15\Rightarrow \alpha =0.15$

% of dissociation of $N_2{O}_4=\alpha \times 100=0.15\times 100=15$%

Option B is correct.