If $PCl_5$ undergoes $80\%$ dissociation at $250 \, ^\circ C$,what is its vapour density at this temperature?

Ammonium carbamate $(NH_4COONH_2)$ when heated to $200 \ ^oC$ gives a mixture of $NH_3$ and $CO_2$ with a vapour density of $26$ at equilibrium. What is the degree of dissociation of ammonium carbamate?

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:

If in the reaction $N_2O_4 \rightleftharpoons 2NO_2$,$\alpha$ is the fraction of $N_2O_4$ that dissociates,then the total number of moles at equilibrium is:

Total number of moles for the reaction $2HI \rightleftharpoons H_2 + I_2$ at equilibrium,if $\alpha$ is the degree of dissociation,is:

For the equilibrium $PCl_{5_{(g)}} \rightleftharpoons PCl_{3_{(g)}} + Cl_{2_{(g)}}$,the observed vapour density of the mixture is $80$. Given atomic masses $P = 31$ and $Cl = 35.5$,the degree of dissociation of $PCl_{5_{(g)}}$ is approximately....$\%$