How many isomeric forms of $C_7H_9N$ contain a benzene ring?

Match the compounds given in List-$I$ with the items given in List-$II$.

List-$I$	List-$II$
$I.$ Benzenesulphonyl Chloride	$A.$ Zwitterion
$II.$ Sulphanilic acid	$B.$ Hinsberg reagent
$III.$ Alkyl Diazonium salts	$C.$ Dyes
$IV.$ Aryl Diazonium salts	$D.$ Conversion to alcohols

What are $X$ and $Y$ respectively in the following set of reactions?

Match the following:

List-$I$ (Amine)	List-$II$ ($pK_b$ value)
$A$. $N,N$-Dimethylaniline	$I$. $9.30$
$B$. Aniline	$II$. $8.92$
$C$. $N$-Ethylethanamine	$III$. $9.38$
$D$. $N$-Methylaniline	$IV$. $3.00$

The correct answer 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 $(2)$.

Identify the incorrect statements.