1 coulomb of electricity is passed through a solution of `AlCl_(3)`, 13.5 g of Al is deposited. The number of faradays must be

Options

A. m
B. `m xx 10^(3)`
C. `m xx 10^(-3)`
D. all wrong

Answer

To solve the problem, we need to determine the number of Faradays required to deposit 13.5 g of aluminum (Al) when 1 coulomb of electricity is passed through a solution of aluminum chloride (AlCl3).

### Step-by-Step Solution:

1. **Determine the molar mass of aluminum (Al)**:
- The atomic mass of aluminum (Al) is approximately 27 g/mol.

2. **Calculate the number of moles of aluminum deposited**:
- Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
- For 13.5 g of Al:
\[ \text{Number of moles of Al} = \frac{13.5 \text{ g}}{27 \text{ g/mol}} = 0.5 \text{ moles} \]

3. **Determine the number of electrons required to deposit aluminum**:
- Aluminum ions (Al³⁺) require 3 electrons (3 Faradays) to be reduced to aluminum metal (Al).
- Therefore, to deposit 0.5 moles of Al:
\[ \text{Total electrons required} = 0.5 \text{ moles} \times 3 \text{ electrons/mole} = 1.5 \text{ moles of electrons} \]

4. **Convert moles of electrons to Faradays**:
- Since 1 mole of electrons corresponds to 1 Faraday, 1.5 moles of electrons is equivalent to:
\[ \text{Number of Faradays} = 1.5 \text{ Faradays} \]

5. **Conclusion**:
- The number of Faradays required to deposit 13.5 g of Al is 1.5 Faradays.

### Final Answer:
The number of Faradays is **1.5 Faradays**.