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

The degree of dissociation $(\alpha)$ of $PCl_5$ obeying the equilibrium $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is related to the pressure at equilibrium by

For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the correct relation between degree of dissociation $(\alpha)$ of $N_2O_{4(g)}$ and equilibrium constant,$K_p$ is $(P=$ total pressure of mixture $)$

At $STP$,the density of air is $0.001293 \ g \ mL^{-1}$. Its vapour density is $---$.

For the reaction $A \rightleftharpoons \frac{1}{2} B + C$,the degree of dissociation $\alpha$ in terms of vapour density $D_t$ (theoretical) and $D_o$ (observed) is given by:

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....$\%$