An organic compound 'A' contains nitrogen and chlorine. It dissolves readily in water to give a solution that turns litmus red. Titration of compound 'A' with standard base indicates that the molecular weight of 'A' is $$131 \pm 2$$. When a sample of 'A' is treated with aq. NaOH, a liquid separates which contains N but not Cl. Treatment of the obtained liquid with nitrous acid followed by phenol gives orange precipitate. The compound 'A' is

Explanation: The formation of an orange azo dye with nitrous acid and phenol confirms the presence of a primary aromatic amine.

Only primary aromatic amines form stable diazonium salts, which couple with phenol to give azo dyes.

Thus, the separated liquid is aniline $$C_6H_5NH_2$$.

Treatment of Compound $$A$$ with aqueous $$NaOH$$ liberates free aniline, indicating that Compound $$A$$ is the acidic salt of aniline.

Therefore, Compound $$A$$ is anilinium chloride:

$$C_6H_5NH_3^+Cl^-$$

Anilinium chloride is acidic in water because it is formed from a weak base and a strong acid, explaining the red litmus result.

Its molecular weight is:

$$6(12) + 8(1) + 14 + 35.5 = 129.5\ g/mol$$

which matches the given range $$131 \pm 2$$.

Therefore, Compound $$A$$ is anilinium chloride, and the right option is D.