Question

In reaction $N_2{O}_4\left(g\right)\to 2N{O}_2\left(g\right)$ , The observed molecular weight $80gmo{l}^{-1}$ 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 $\left[\begin{array}{cc}{N}_2{O}_4\to & 2N{O}_2\\ 1& 0\\ 1-\alpha & 2\alpha \end{array}\right]$
Mass remains same
Average molar mass $=\frac{M_{N_2{O}_4}}{\left(1+\alpha \right)}$
So, $=\frac{28+64}{\left(1+\alpha \right)}=\frac{92}{\left(1+\alpha \right)}$
$1+\alpha =\frac{92}{80}=\frac{23}{20}$
$\alpha =\frac{23}{20}-1=\frac{3}{20}$
$\%\alpha =\frac{3}{20}\times 100=15\%$