One faraday of electricity will liberate one mole of metal from a molten of-

Options

A. `AuCl_3`
B. `CuSO_4`
C. `BaCl_2`
D. KCL

Answer

To solve the problem of determining which electrolyte will liberate one mole of metal when one Faraday of electricity is passed through it, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Faraday's Law**:
According to Faraday's second law of electrolysis, the amount of substance liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. One Faraday (F) of electricity corresponds to the charge of one mole of electrons (approximately 96500 coulombs).

2. **Relating Moles and Equivalent Weight**:
To liberate one mole of a metal, we need to consider the equivalent weight of the metal. The equivalent weight is defined as the molecular weight divided by the n-factor (valency factor).

3. **Determining the n-factor**:
The n-factor is the number of moles of electrons required to reduce one mole of the metal ion to its elemental form. For a metal to liberate one mole when one Faraday is passed, the n-factor must be 1. This means that the metal ion must have a valency of 1.

4. **Analyzing the Given Electrolytes**:
We need to analyze the electrolytes provided in the question:
- **AuCl3**: Gold has a valency of 3 (n-factor = 3).
- **CuSO4**: Copper has a valency of 2 (n-factor = 2).
- **BaCl2**: Barium has a valency of 2 (n-factor = 2).
- **KCl**: Potassium has a valency of 1 (n-factor = 1).

5. **Identifying the Correct Electrolyte**:
From the analysis, we can see that KCl is the only electrolyte that has an n-factor of 1. Therefore, when one Faraday of electricity is passed through KCl, it will liberate one mole of potassium metal.

### Conclusion:
The electrolyte that will liberate one mole of metal when one Faraday of electricity is passed through it is **KCl**.