Match List-$I$ with List-$II$.

List-$I$ (Mixture of Compounds)	List-$II$ (Reagent used to distinguish)
$A$. Diethyl amine + Ethyl amine	$I$. Bromine water
$B$. Acetaldehyde + Acetone	$II$. $CHCl_3 + KOH, \Delta$
$C$. Ethanol + Phenol	$III$. Neutral $FeCl_3$
$D$. Benzoic acid + Cinnamic acid	$IV$. Ammonical silver nitrate

The end product $C$ of the following reaction is $C_2H_5NH_2$ $\xrightarrow{HNO_2} A$ $\xrightarrow{PCl_5} B$ $\xrightarrow{NH_3 (alc.)} C$

The relative rate of reaction of the following amines with methyl iodide is:

$A$ trinitro compound,$1,3,5$-tris-($4$-nitrophenyl)benzene,on complete reaction with an excess of $Sn/HCl$ gives a major product,which on treatment with an excess of $NaNO_2/HCl$ at $0^{\circ} C$ provides $P$ as the product. $P$,upon treatment with excess of $H_2O$ at room temperature,gives the product $Q$. Bromination of $Q$ in aqueous medium furnishes the product $R$. The compound $P$ upon treatment with an excess of phenol under basic conditions gives the product $S$.
The molar mass difference between compounds $Q$ and $R$ is $474 \ g \ mol^{-1}$ and between compounds $P$ and $S$ is $172.5 \ g \ mol^{-1}$.
$(1)$ The number of heteroatoms present in one molecule of $R$ is . . . . .
[Use: Molar mass (in $g \ mol^{-1}$): $H=1, C=12, N=14, O=16, Br=80, Cl=35.5$
Atoms other than $C$ and $H$ are considered as heteroatoms]
$(2)$ The total number of carbon atoms and heteroatoms present in one molecule of $S$ . . . . . .
[Use: Molar mass in $g \ mol^{-1}$]: $H=1, C=12, N=14, O=16, Br=80, Cl=35.5$
Atoms other than $C$ and $H$ are considered as heteroatoms
Give the answer for question $(1)$ and $(2)$.

Match List-$I$ with List-$II$ and select the correct answer.

List-$I$	List-$II$
$1.$ Aniline	$a.$ Used in the preparation of azo dyes
$2.$ Nitrobenzene	$b.$ Sulpha drug
$3.$ Sulphanilamide	$c.$ Solvent in Friedel-Crafts reaction
$4.$ Trinitrotoluene	$d.$ Used as an explosive

$X \ g$ of ethanamine was subjected to reaction with $NaNO_2 / HCl$ followed by hydrolysis to liberate $N_2$ and $HCl$. The $HCl$ generated was completely neutralized by $0.2 \ mol$ of $NaOH$. $X$ is . . . .