The vapour density of a mixture containing $NO_2$ and $N_2O_4$ is $27.6$. The mole fraction of $NO_2$ in the mixture is:

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:

Phosphine decomposes at $300 \ K$ and $3 \ atm$ as
$4PH_{3(g)} \rightleftharpoons P_{4(g)} + 6H_{2(g)}$
Calculate the vapour density of phosphine if it dissociates to the extent of $30 \%$.

$A$ $1 \ L$ vessel contains $2 \ moles$ of $PCl_5$ initially. If $K_c$ is found to be $1$,the degree of dissociation of $PCl_5$ for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ is:

Gaseous $N_2O_4$ dissociates into gaseous $NO_2$ according to the reaction:
$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
At $300 \ K$ and $1 \ atm$ pressure,the degree of dissociation of $N_2O_4$ is $0.2$. If one mole of $N_2O_4$ gas is contained in a vessel,then the density of the equilibrium mixture is.......$g/L$

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: