The vapour density of $PCl_5$ at $25^{\circ}C$ is $100$. Calculate the degree of dissociation at this temperature.

If the density of air at $STP$ is $0.001293 \, g \, mL^{-1}$,what will be its vapor density?

$A$ $1 \ L$ vessel contains $2 \ moles$ of $PCl_5$ initially. If $K_c$ is found to be $1$,the degree of dissociation of $PCl_5$ for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ is:

The equilibrium constant at a certain temperature for the reaction $A_2 + B_2 \rightleftharpoons 2AB$ is $2$. Calculate the degree of dissociation of either $A_2$ or $B_2$.

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 temperature,$T$,a compound $AB_{2(g)}$ dissociates according to the reaction; $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$ with a degree of dissociation $x$,which is small compared with unity. The expression for $K_p$,in terms of $x$ and the total pressure,$P$ is