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

At temperature $T$,the dissociation equilibrium of $AB_{2(g)}$ is given as: $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$. If the degree of dissociation $x$ is very small compared to $1$,then the expression for the equilibrium constant $K_p$ in terms of $x$ and total pressure $P$ is:

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

At $200 \, ^\circ C$,the observed vapour density $(V.D.)$ of $PCl_5$ is $60$. The degree of dissociation $(\alpha)$ of $PCl_5$ is ..........$\%$ (Given: Molar mass of $PCl_5 = 208.5 \, g/mol$).

In a $100 \ L$ vessel,$3 \ moles$ of nitrogen and $3 \ moles$ of $PCl_5$ are taken and heated to $500 \ K$. The equilibrium pressure is $3.28 \ atm$. The percentage degree of dissociation of $PCl_5$ is : (Assume ideal behaviour for all gases).

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....$\%$