Question

In reaction $N_{2}O_{4}\left(g\right) \rightarrow 2NO_{2}\left(g\right)$ , The observed molecular weight $80gmol^{- 1}$ at $350$ K. The percentage 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)

Or $\begin{bmatrix} N_{2}O_{4} \rightarrow & 2NO_{2} \\ 1 & 0 \\ 1-\alpha & 2\alpha \end{bmatrix}$
Mass remains same
Average molar mass $=\frac{M_{N_{2} O_{4}}}{\left(1 + \alpha \right)}$
So, $=\frac{28 + 64}{\left(1 + \alpha \right)}=\frac{92}{\left(1 + \alpha \right)}$
$1+\alpha =\frac{92}{80}=\frac{23}{20}$
$\alpha =\frac{23}{20}-1=\frac{3}{20}$
$\%\alpha =\frac{3}{20}\times 100=15\%$