Question

In a closed container, $N{O}_2$ was taken which dimerises to give $N_2{O}_4$. After some time, mole fraction of $N_2{O}_4$ was found to be $\frac{2}{3}$ in the container, then calculate % dimerisation of $N{O}_2$. Give your answer dividing $10$.

• A. $8$
• B. $9$
• C. $10$
• D. $12$

Answer 1

The correct option is A $8$

In a closed container, $N{O}_2$ was taken which dimerises to $N_2{O}_4$. The reaction is,
$2N{O}_2\to N_2{O}_4$
$100-x$ $\frac{x}{2}$ where, x= number of moles reacted
Mole fraction of $N_2{O}_4=\frac{\frac{x}{2}}{(100-x)+\frac{x}{2}}=\frac{2}{3}$
or, $200-2x+x=\frac{3x}{2}$
or, $200=1.5x+x$
$x=\frac{200}{2.5}=80$
When divided by $10$, we get $\frac{80}{10}=8$