The vapour density of completely dissociated $NH_4Cl$ would be

Before equilibrium is set up for the chemical reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the vapour density $d$ of the gaseous mixture was measured. If $D$ is the theoretical value of vapour density,the variation of $\alpha$ with $D/d$ is given by the graph. What is the value of $D/d$ at point $A$?

At $NTP$,$5.6 \ L$ of a gas weighs $8 \ g$. The vapour density of the gas is:

For the reaction: $NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}; K_p$. The degree of dissociation $(\alpha)$ of $NH_3$ is related to total equilibrium pressure $(P^o)$ as:

At $STP$,the density of air is $0.001293 \ g \ mL^{-1}$. Its vapour density is $---$.

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: