State True or false
`F_(2)` is prepared by the electrolysis of molten `KHF_(2)` and anhydrous `HF`.

Explanation

To determine whether the statement "F₂ is prepared by the electrolysis of molten KHF₂ and anhydrous HF" is true or false, we can analyze the electrolysis process involved.

### Step-by-Step Solution:

1. **Understanding the Electrolysis Process**:
- Electrolysis involves breaking down a compound into its constituent elements by passing an electric current through it. In this case, we are considering KHF₂ (potassium hydrogen fluoride) and anhydrous HF (hydrofluoric acid).

2. **Electrolysis of KHF₂**:
- When KHF₂ is molten and subjected to electrolysis, it dissociates into potassium ions (K⁺) and fluoride ions (F⁻).
- The half-reaction at the anode involves the oxidation of fluoride ions (F⁻) to form fluorine gas (F₂):
\[ 2F^- \rightarrow F_2 + 2e^- \]
- At the cathode, potassium ions (K⁺) are reduced to form potassium metal:
\[ K^+ + e^- \rightarrow K \]

3. **Electrolysis of Anhydrous HF**:
- Anhydrous HF can also undergo electrolysis, leading to the formation of fluorine gas. The reactions are similar, where fluoride ions are oxidized at the anode to produce fluorine gas.

4. **Overall Reaction**:
- The overall reaction during the electrolysis of KHF₂ can be summarized as:
\[ KHF_2 \rightarrow K + F_2 + HF \]
- This indicates that fluorine gas (F₂) is indeed produced during the electrolysis of molten KHF₂.

5. **Conclusion**:
- Since both molten KHF₂ and anhydrous HF can lead to the production of fluorine gas (F₂) through electrolysis, the statement is **True**.

### Final Answer:
**True**