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

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

Consider the reaction $X_2Y_{(g)} \rightleftharpoons X_{2(g)} + \frac{1}{2}Y_{2(g)}$. The equation representing the correct relationship between the degree of dissociation $(x)$ of $X_2Y_{(g)}$ with its equilibrium constant $Kp$ is. . . . . . . Assume $x$ to be very very small.

The vapour density of $N_2O_4$ in the equilibrium $N_2O_4 \rightleftharpoons 2NO_2$ is $40$. The degree of dissociation is: