Total charge in faraday present in 54 gm of `Al^(3+)` ion.

Explanation

To find the total charge in Faraday present in 54 grams of \( \text{Al}^{3+} \) ion, we can follow these steps:

### Step-by-step Solution:

1. **Identify the Given Data:**
- Given weight of \( \text{Al}^{3+} \) = 54 g
- Molar mass of Al = 27 g/mol
- Valency of Al = 3 (since it is \( \text{Al}^{3+} \))

2. **Calculate the Equivalent Weight of Aluminium:**
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{Valency}} = \frac{27 \, \text{g/mol}}{3} = 9 \, \text{g/equiv} \]

3. **Calculate the Number of Equivalents of Aluminium:**
\[ \text{Number of equivalents} = \frac{\text{Given weight}}{\text{Equivalent weight}} = \frac{54 \, \text{g}}{9 \, \text{g/equiv}} = 6 \, \text{equiv} \]

4. **Calculate the Total Charge in Faraday:**
- According to Faraday's law, 1 equivalent of charge corresponds to 1 Faraday (F).
- Therefore, the total charge (Q) in Faraday is equal to the number of equivalents:
\[ Q = \text{Number of equivalents} = 6 \, F \]

### Final Answer:
The total charge in Faraday present in 54 grams of \( \text{Al}^{3+} \) ion is \( 6 \, F \).

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