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 provided and deduce the correct organic compound.

### Step-by-Step Solution:

1. **Understanding the Composition**:
- The organic compound contains about 7.8% carbon. This percentage will help us identify the possible compounds.

2. **Analyzing the Reaction with AgNO3**:
- The acidic sodium extract of the compound gives a white precipitate with AgNO3. This indicates the presence of a halide ion (Cl⁻, Br⁻, or I⁻).
- The white precipitate suggests that the halide is likely chloride (AgCl), as AgBr and AgI produce colored precipitates (yellow for AgBr and brown for AgI).

3. **Solubility in Ammonia**:
- The white precipitate formed (AgCl) is soluble in excess ammonia. This is a characteristic property of silver chloride, confirming that the halide present in the compound is indeed chloride.

4. **Calculating the Molar Mass**:
- We need to find a compound that contains chlorine and has a molar mass that corresponds to 7.8% carbon.
- For CHCl₃ (chloroform):
- Molar mass = 12 (C) + 1 (H) + 35.5 × 3 (Cl) = 119.5 g/mol
- Percentage of carbon = (12 / 119.5) × 100 = 10.08%
- For CCl₄ (carbon tetrachloride):
- Molar mass = 12 (C) + 35.5 × 4 (Cl) = 154 g/mol
- Percentage of carbon = (12 / 154) × 100 = 7.79%

5. **Conclusion**:
- Since CCl₄ gives a percentage of carbon that is approximately 7.8%, and it also reacts to form a white precipitate with AgNO3 that is soluble in ammonia, the compound must be carbon tetrachloride (CCl₄).

### Final Answer:
The compound is **CCl₄ (carbon tetrachloride)**.