If the density of air at $STP$ is $0.001293 \, g \, mL^{-1}$,what will be its vapor density?

The vapor density of a gas is $11.2$. At $N.T.P.$,$11.2 \ g$ of this gas occupies a volume of ............. $litres$.

For the thermal dissociation of $PCl_5$ as $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,if '$a$' moles of $PCl_5$ are taken and at equilibrium,the degree of dissociation of $PCl_5$ is $0.25$ and the total pressure is $2.0 \ atm$,then the partial pressure of $Cl_2$ at equilibrium will be: (in $atm$)

For the reaction $A_{3(g)} \rightleftharpoons 3A_{(g)}$,the initial moles of $A_3$ is $a$. If $\alpha$ is the degree of dissociation of $A_3$,then the total number of moles at equilibrium will be:

The density of air at $N.T.P.$ is $0.001293 \ g \ mL^{-1}$. Its vapor density is .....

If the degree of dissociation for the reaction,$PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is $20\%$ at $1 \ atm$ pressure,calculate $K_c$.