Calculate the degree of dissociation $\alpha$ for $0.1 \, N \, CH_3COOH$ (dissociation constant $K_a = 1 \times 10^{-5}$).

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:

For the equilibrium reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the value of $K_p$ is $2$. Calculate the percentage dissociation of $N_2O_{4(g)}$ at a total pressure of $0.5 \ atm$. (in $\%$)

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

The degree of dissociation of $PCl_5$ $(\alpha)$ obeying the equilibrium,$PCl_5 \rightleftharpoons PCl_3 + Cl_2$,is approximately related to the pressure at equilibrium by (given $\alpha << 1$):

Total number of moles for the reaction $2HI \rightleftharpoons H_2 + I_2$ at equilibrium,if $\alpha$ is the degree of dissociation,is: