`p-Cl-C_(6)H_(4)NH_(2)` and `PhNH_(3)^(+)Cl^(-)` can be distinguished by

Options

A. `Ag(OH)`
B. `NaOH`
C. `AgNO_(3)`
D. None of these

Answer

To distinguish between para-chloroaniline (`p-Cl-C_(6)H_(4)NH_(2)`) and anilinium chloride (`PhNH_(3)^(+)Cl^(-)`), we can use a chemical test involving silver nitrate (`AgNO3`). Here’s a step-by-step solution:

### Step 1: Understand the Structures
- **Para-chloroaniline** is an aromatic amine with the formula `p-Cl-C_(6)H_(4)NH_(2)`. It has a free amino group (`-NH2`) and a chlorine atom attached to the benzene ring.
- **Anilinium chloride** is the salt formed from aniline and hydrochloric acid, represented as `PhNH_(3)^(+)Cl^(-)`. It contains a positively charged ammonium ion (`PhNH3^+`) and a chloride ion (`Cl^-`).

### Step 2: React with Silver Nitrate
- When we add silver nitrate (`AgNO3`) to **para-chloroaniline**, the chlorine atom in the para position is part of the aromatic system and is not free to react. Therefore, no reaction occurs, and no precipitate is formed.
- In contrast, when we add silver nitrate to **anilinium chloride**, the chloride ion is free and can react with silver ions (`Ag^+`) to form silver chloride (`AgCl`), which is a white precipitate.

### Step 3: Observe the Results
- **For para-chloroaniline**: No precipitate forms when treated with `AgNO3`.
- **For anilinium chloride**: A white precipitate of `AgCl` forms upon treatment with `AgNO3`.

### Step 4: Conclusion
Based on the reactions, we can distinguish between para-chloroaniline and anilinium chloride by the formation of a white precipitate of silver chloride when anilinium chloride is treated with silver nitrate, while no precipitate will form with para-chloroaniline.

### Final Answer
Para-chloroaniline and anilinium chloride can be distinguished by their reaction with silver nitrate; anilinium chloride produces a white precipitate of AgCl, while para-chloroaniline does not.