When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid (273 K) and acidified with acetic acid, the mass (g) of 0.1 mole of product formed is :
(Given molar mass in g mol$$^{-1}$$ H : 1, C : 12, N : 14, O : 16, S : 32)

The reaction takes place in two steps involving diazotization followed by an azo coupling reaction.

In the first step, sulphanilic acid reacts with nitrous acid at

$$273\ \text{K}$$

to form the corresponding diazonium salt.

$$\text{Sulphanilic Acid}\xrightarrow{HNO_2,\ 273,\text{K}}\text{Benzene diazonium-4-sulfonate}$$

In the second step, the diazonium salt undergoes coupling with 1-naphthylamine in a weakly acidic medium to produce the azo dye.

$$\text{Benzene diazonium-4-sulfonate}+\text{1-Naphthylamine}\xrightarrow{H^+}\text{4-(4-Sulphophenylazo)-1-naphthylamine}$$

To determine the required mass, we first calculate the molecular formula of the product.

The sulphanilic acid fragment contributes

$$C_6H_5O_3S.$$

The azo linkage contributes

$$N_2.$$

The 1-naphthylamine fragment contributes

$$C_{10}H_8N,$$

since one hydrogen atom of the aromatic ring is replaced during coupling.

Hence, the molecular formula of the final product is

$$C_{16}H_{13}N_3O_3S.$$

Using the given atomic masses,

• Carbon:

$$16\times12=192\ \text{g mol}^{-1}$$

• Hydrogen:

$$13\times1=13\ \text{g mol}^{-1}$$

• Nitrogen:

$$3\times14=42\ \text{g mol}^{-1}$$

• Oxygen:

$$3\times16=48\ \text{g mol}^{-1}$$

• Sulfur:

$$1\times32=32\ \text{g mol}^{-1}$$

Therefore, the molar mass of the product is

$$192+13+42+48+32=327\ \text{g mol}^{-1}.$$

For

$$0.1\ \text{mol},$$

the required mass is

$$\text{Mass}=\text{Moles}\times\text{Molar Mass}$$

$$=0.1\times327$$

$$=32.7\ \text{g}.$$

Approximating to the nearest whole number,

$$32.7\ \text{g}\approx33\ \text{g}.$$

Therefore, the correct answer is

$$\boxed{\text{Option (C) }33.}$$