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

At temperature $T$,compound $AB_{2(g)}$ dissociates as $AB_{2(g)} \rightleftharpoons AB_{(g)} + \frac{1}{2} B_{2(g)}$ having degree of dissociation $x$ (small compared to unity). The correct expression for $x$ in terms of $K_p$ and $p$ is

Vapour density of a gas is $5.6$,volume of $5.6 \ g$ of it at $STP$ is ............... $L$

The value of $K_P$ for the equilibrium reaction ${N_2}{O_4}_{(g)} \rightleftharpoons 2N{O_2}_{(g)}$ is $2$. The percentage dissociation of ${N_2}{O_4}_{(g)}$ at a pressure of $0.5 \ atm$ is

The vapour density of $PCl_5$ is $104.16$,but when heated to $230 \, ^\circ C$,its vapour density decreases to $62$. What is the degree of dissociation of $PCl_5$ at this temperature (in $\%$) ?

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: