Question

At equilibrium $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$ the observed molecular weight of $N_2{O}_4$ is 80g $mo{l}^{-1}$ at 350 K. The percentage dissociation of $N_2{O}_4$(g) at 350 K is:

• A. 10%
• B. 15%
• C. 20%
• D. 18%

Answer 1

Answer: (B)

15%

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

$t=0$, $1$ $0$

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

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

Observed molecular mass= $80g/mol$

$\therefore \text{Van't Hoff factor}i$ $=\frac{92}{80}=1.15$

$\Rightarrow i=\frac{1-\alpha +2\alpha }{1}=\frac{\text{Observed colligative property}}{\text{Calculated colligative property}}$

$\Rightarrow 1.15=\frac{1+\alpha }{1}$

$\Rightarrow \alpha =0.15$

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