Question

At the equilibrium of the reaction, $N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$, the observed molar mass of $N_2{O}_4$ is $77.70g$. The percentage dissociation of $N_2{O}_4$ is:

• A. $28.4$
• B. $46.7$
• C. $22.4$
• D. $18.4$

Answer 1

The correct option is D $18.4$

At the equilibrium of the reaction, $N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$, the observed molar mass of $N_2{O}_4$ is $77.70g$. The percentage dissociation of $N_2{O}_4$ is $18.4$.
The degree of dissociation of $N_2{O}_4$
$\alpha =\frac{M_0-M}{(n-1)M}$
$\alpha =\frac{92g/mol-77.70g/mol}{(2-1)\times 77.70}$
$\alpha =0.184$
The percentage dissociation of $N_2{O}_4$ is $100\times 0.184=18.4$.
Note: $n=2=$ the number of molecules obtained on dissociation of one molecule of $N_2{O}_4$.