The vapour density of a gas is $11.2$. The volume occupied by $11.2 \ g$ of the gas at $STP$ will be.....$L$

The dissociation equilibrium of a gas $AB_{2}$ can be represented as:
$2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$
The degree of dissociation is $x$ and is small compared to $1$. The expression relating the degree of dissociation $(x)$ with equilibrium constant $K_P$ and total pressure $P$ is:

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:

Phosphorus pentachloride dissociates as follows,in a closed reaction vessel:
$PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
If total pressure at equilibrium of the reaction mixture is $P$ and degree of dissociation of $PCl_5$ is $x,$ the partial pressure of $PCl_3$ will be:

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 $B_2$.

In a closed vessel,the dissociation of phosphorus pentachloride occurs as: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$. If the total pressure at equilibrium is $P$ and the degree of dissociation of $PCl_5$ is $x$,then the partial pressure of $PCl_3$ will be: