In which one of the following one faraday of electricity will liberate 1/2 gram -atom of the metal?

Options

A. `AuCl_(3)`
B. `FeCl_(3)`
C. `CuSO_(4)`
D. `NaCl`.

Explanation

To solve the question of which metal will liberate 1/2 gram-atom when one Faraday of electricity is passed, we need to analyze the relationship between the number of electrons transferred and the atomic weight of the metals involved.

### Step-by-Step Solution:

1. **Understanding Faraday's Law**:
Faraday's first law of electrolysis states that the amount of substance liberated during electrolysis is directly proportional to the quantity of electricity passed. One Faraday (F) corresponds to the charge of 1 mole of electrons (approximately 96500 coulombs).

2. **Identify the Relationship**:
For a metal with atomic weight \( A \) and valency \( n \), the weight of the metal liberated by passing one Faraday of electricity is given by:
\[ \text{Weight} = \frac{A}{n} \]
We want this weight to equal 1/2 gram-atom, which means:
\[ \frac{A}{n} = \frac{1}{2} \]

3. **Analyzing Each Metal**:
- **Gold (Au)**:
- Atomic weight = 197
- Valency = 3 (from AuCl3)
- Weight for 1 Faraday:
\[ \frac{197}{3} \approx 65.67 \text{ grams} \quad (\text{not } \frac{1}{2}) \]

- **Iron (Fe)**:
- Atomic weight = 55.8
- Valency = 3 (from FeCl3)
- Weight for 1 Faraday:
\[ \frac{55.8}{3} \approx 18.6 \text{ grams} \quad (\text{not } \frac{1}{2}) \]

- **Copper (Cu)**:
- Atomic weight = 63.5
- Valency = 2 (from CuSO4)
- Weight for 1 Faraday:
\[ \frac{63.5}{2} = 31.75 \text{ grams} \quad (\text{not } \frac{1}{2}) \]

- **Sodium (Na)**:
- Atomic weight = 23
- Valency = 1 (from NaCl)
- Weight for 1 Faraday:
\[ \frac{23}{1} = 23 \text{ grams} \quad (\text{not } \frac{1}{2}) \]

4. **Conclusion**:
None of the metals analyzed liberate 1/2 gram-atom when one Faraday of electricity is passed. However, if we are looking for a scenario where the weight liberated is half the atomic weight divided by the valency, we find that copper (Cu) comes closest, as it gives us a weight of 31.75 grams for 1 Faraday, but it does not satisfy the requirement of 1/2 gram-atom.

### Final Answer:
None of the given options liberate exactly 1/2 gram-atom of metal when one Faraday of electricity is passed.