When electricity is passed through a solution of `AlCl_(3)` and `13.5g` of `Al` is deposited, the number of Faraday of electricity passed must be ………………….F
a.0.5
b.1.0
c.1.5
d.2.0

Options

A. `0.5`
B. `1.0`
C. `1.5`
D. `2.0`

Explanation

To solve the problem of how many Faradays of electricity are required to deposit 13.5 g of aluminum from a solution of aluminum chloride (AlCl₃), we can follow these steps:

### Step 1: Determine the molar mass of aluminum (Al).
The molar mass of aluminum (Al) is 27 g/mol.

### Step 2: Calculate the number of moles of aluminum deposited.
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
we can calculate the number of moles of aluminum deposited:
\[ \text{Number of moles of Al} = \frac{13.5 \, \text{g}}{27 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 3: Determine the number of Faradays required for the deposition of aluminum.
From the electrochemical reaction, we know that:
- 1 mole of Al requires 3 Faradays of electricity to be deposited (since Al³⁺ ions gain 3 electrons to form Al).

Thus, for 0.5 moles of Al:
\[ \text{Faradays required} = 0.5 \, \text{mol} \times 3 \, \text{Faradays/mol} = 1.5 \, \text{Faradays} \]

### Conclusion:
The number of Faradays of electricity passed must be **1.5 F**.

### Final Answer:
c. 1.5

---