1. Identify the reaction

Sulphanilic acid undergoes diazotization with nitrous acid at $273\,\text{K}$ to form a diazonium salt, which then couples with $1$-naphthylamine in acidic medium (acetic acid) to give an azo dye.

So the product is formed by:

• diazotized sulphanilic acid
• coupling with $1$-naphthylamine

This is the well-known azo coupling product.

2. Write molecular formulas of reactants

• Sulphanilic acid: $\mathrm{H_2N{-}C_6H_4{-}SO_3H}$

Its molecular formula is: $\mathrm{C_6H_7NO_3S}$

(Check: $6C,\ 7H,\ 1N,\ 3O,\ 1S$)

• $1$-Naphthylamine: naphthalene $= \mathrm{C_{10}H_8}$, replacing one H by $\mathrm{NH_2}$ gives: $\mathrm{C_{10}H_9N}$

3. Formula of azo product

In azo coupling, the diazonium part and naphthylamine part join through $-{\rm N=N}-$, with loss of:

• one $\mathrm{NH_2}$ of sulphanilic acid becoming part of diazonium
• overall, coupling effectively removes $2$ hydrogen atoms from the two aromatic partners

Thus product formula can be obtained as: $\mathrm{C_6H_7NO_3S + C_{10}H_9N + N_2 - 2H_2O?}$

A simpler and correct way is to directly write the coupled product structure: $\mathrm{HO_3S{-}C_6H_4{-}N=N{-}C_{10}H_6{-}NH_2}$

Now count atoms:

• Carbon: $6+10=16$
• Hydrogen: $4 + 6 + 2 + 1 = 13$? Let us count carefully from groups.

From structure:

• $\mathrm{C_6H_4}$ from benzene ring
• $\mathrm{SO_3H}$ contributes $1H$
• $\mathrm{C_{10}H_6}$ from naphthalene residue
• $\mathrm{NH_2}$ contributes $2H$

Total hydrogen: $4+1+6+2=13$

Nitrogen:

• azo group: $2$
• amino group: $1$

So total formula: $\mathrm{C_{16}H_{13}N_3O_3S}$

4. Calculate molar mass

Using given atomic masses: $\mathrm{C}:12,\ H:1,\ N:14,\ O:16,\ S:32$

$M = 16(12) + 13(1) + 3(14) + 3(16) + 32$

$= 192 + 13 + 42 + 48 + 32$

$= 327\,\text{g mol}^{-1}$

5. Mass of $0.1$ mol product

$m = 0.1 \times 327 = 32.7\,\text{g}$

This is closest to: $33\,\text{g}$

6. Check options

• A: $330$ g $\times$ incorrect
• B: $33$ g $\checkmark$
• C: $343$ g $\times$
• D: $66$ g $\times$

Therefore, the correct option is B.