An organic compound contains about 7.8% carbon. Its acidic sodium extract gives a white precipitate with AgNO3. This precipitate is soluble in the excess of ammonia. The compound is:

Options

A. `CHCl_(3)`
B. `CHBr_(3)`
C. `CHl_(3)`
D. `"CCl"_(4)`

Answer

To solve the problem step by step, we will analyze the information given in the question and apply our knowledge of organic chemistry, particularly focusing on the behavior of halogenated compounds with silver nitrate.

### Step 1: Analyze the given information
The organic compound contains about 7.8% carbon. Its acidic sodium extract gives a white precipitate with AgNO3, which is soluble in excess ammonia.

### Step 2: Identify the possible compounds
The question suggests that the compound could be one of the following:
- CHCl3 (Chloroform)
- CHBr3 (Bromoform)
- CHI3 (Iodoform)
- CCl4 (Carbon Tetrachloride)

### Step 3: Understand the reaction with AgNO3
When a sodium extract containing halides (like Cl, Br, or I) reacts with AgNO3, it forms a precipitate:
- AgCl (white precipitate) for Cl
- AgBr (yellow precipitate) for Br
- AgI (yellow precipitate) for I

Since the question states that a white precipitate is formed, we can conclude that the halogen present must be chlorine (Cl).

### Step 4: Check the solubility in excess ammonia
The white precipitate (AgCl) formed is soluble in excess ammonia, confirming the presence of chlorine in the compound.

### Step 5: Calculate the percentage of carbon in the possible compounds
Now, we need to calculate the percentage of carbon in the potential candidates (CHCl3 and CCl4) to find which one has approximately 7.8% carbon.

**For CHCl3:**
- Molecular weight = C (12) + H (1) + 3 × Cl (35.5) = 12 + 1 + 106.5 = 119.5 g/mol
- Percentage of carbon = (12 / 119.5) × 100 = 10.08%

**For CCl4:**
- Molecular weight = C (12) + 4 × Cl (35.5) = 12 + 142 = 154 g/mol
- Percentage of carbon = (12 / 154) × 100 = 7.79%

### Step 6: Conclusion
Since CCl4 has a carbon percentage of approximately 7.8%, it fits the criteria given in the question. Therefore, the compound is **CCl4 (Carbon Tetrachloride)**.

### Final Answer:
The compound is **CCl4 (Carbon Tetrachloride)**.

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