The vapour density of a mixture containing $NO_2$ and $N_2O_4$ is $27.6$. The mole fraction of $NO_2$ in the mixture is:

The vapour density of undecomposed $N_2O_4$ is $46$. When heated,the vapour density decreases to $24.5$ due to its dissociation into $NO_{2(g)}$. The percentage dissociation of $N_2O_4$ is:

If $D_T$ and $D_0$ are the theoretical and observed vapour densities at a definite temperature and $\alpha$ is the degree of dissociation of a substance,then $\alpha$ in terms of $D_0, D_T$ and $n$ (number of moles of product formed from $1 \, \text{mole}$ of reactant) is calculated by the formula:

At $T$ $K$,consider the following gaseous reaction,which is in equilibrium: $N_2O_5 \rightleftharpoons 2NO_2 + \frac{1}{2}O_2$. What is the fraction of $N_2O_5$ decomposed at constant volume and temperature,if the initial pressure is $300 \ mm \ Hg$ and pressure at equilibrium is $480 \ mm \ Hg$? (Assume all gases as ideal)

Degree of dissociation of an electrolyte does not depend on

For the following equilibrium $N_2O_4 \rightleftharpoons 2NO_2$ in gaseous phase,$NO_2$ is $50\%$ of the total volume when equilibrium is set up. Hence,the percentage of dissociation of $N_2O_4$ is.......$\%$