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/HCl at 0°C provides P as the product. P, upon treatment with excess of HO 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.

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.

The correct answer is 9.

Let us trace the step-by-step chemical transformations starting from the initial trinitro compound to identify the structures of products P, Q, R, and S, and finally calculate the number of heteroatoms in compound R.

Step 1: Conversion of 1,3,5-tris-(4-nitrophenyl)benzene to P and Q
The starting compound is 1,3,5-tris-(4-nitrophenyl)benzene, which contains three 4-nitrophenyl groups attached to a central benzene ring at positions 1, 3, and 5. Thus, it contains three nitro (-NO₂) groups.
1. Reduction with excess Sn/HCl reduces all three nitro (-NO₂) groups to primary amino (-NH₂) groups, giving 1,3,5-tris-(4-aminophenyl)benzene.
2. Diazotization of this triamine with excess NaNO₂/HCl at 0 °C converts all three -NH₂ groups into diazonium chloride groups (-N₂⁺Cl^-). This product is compound P.
Molecular formula of P: C₂₄H₁₅N₆Cl₃.
3. Hydrolysis of compound P with excess H₂O at room temperature replaces the three diazonium groups with hydroxyl (-OH) groups, forming 1,3,5-tris-(4-hydroxyphenyl)benzene. This product is compound Q.
Molecular formula of Q: C₂₄H₁₈O₃.

Step 2: Bromination of Q to form R
Compound Q contains three phenol rings. Each -OH group is strongly activating and ortho-directing. In aqueous medium (aqueous bromination), bromination occurs at all vacant ortho-positions relative to the -OH group on each of the three outer phenyl rings.
Each of the three phenyl rings has two vacant ortho-positions (the para position is attached to the central benzene ring). Thus, a total of 6 bromine atoms (Br) are substituted onto the molecule (2 Br atoms per phenol ring).
Therefore, product R is 1,3,5-tris-(3,5-dibromo-4-hydroxyphenyl)benzene.
Molecular formula of R: C₂₄H₁₂O₃Br₆.

Step 3: Verification of Molar Mass Differences
Let us verify the molar mass difference between Q and R:

$\text{Molar mass difference}=6\times \text{Molar mass of Br}-6\times \text{Molar mass of H}$

$\text{Difference}=6\times 80-6\times 1=480-6=474\text{ g mol}{-1}^{}$

This matches the given value of 474 g mol^-1.

Let us also verify compound S formed by coupling compound P with excess phenol under basic conditions:
Coupling of 3 diazonium groups of P with 3 molecules of phenol yields an azobenzene-based dye product S (C₄₂H₃₀N₆O₃).
Molar mass of P = 24(12) + 15(1) + 6(14) + 3(35.5) = 493.5 g mol^-1.
Molar mass of S = 42(12) + 30(1) + 6(14) + 3(16) = 666 g mol^-1.
Molar mass difference = 666 - 493.5 = 172.5 g mol^-1, which matches the given data.

Step 4: Count of Heteroatoms in Compound R
Heteroatoms are defined as atoms other than carbon (C) and hydrogen (H).
The molecular formula of R is C₂₄H₁₂O₃Br₆.
The heteroatoms present in one molecule of R are:
- Oxygen (O) atoms = 3
- Bromine (Br) atoms = 6

$\text{Total number of heteroatoms}=3+6=9$

Thus, the total number of heteroatoms present in one molecule of R is 9.