The atomic weight of Al is `27`. When a current of `5F` is passed through a solution of `Al^(+++)` ions, the qeight of `AL` deposited is.

Options

A. 27 gm
B. 36 gm
C. 45 gm
D. 39 gm

Answer

To solve the problem of determining the weight of aluminum deposited when a current of 5 Faraday is passed through a solution of Al^(3+) ions, we can follow these steps:

### Step 1: Understand the reaction
The reduction reaction for aluminum ions can be represented as:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
This means that 3 moles of electrons (3 Faraday) are required to deposit 1 mole of aluminum.

### Step 2: Calculate the moles of aluminum deposited
We are given that a current of 5 Faraday is passed. To find out how many moles of aluminum can be deposited, we use the relationship:
\[ \text{Moles of Al deposited} = \frac{\text{Total Faraday}}{\text{Faraday required for 1 mole of Al}} \]
\[ \text{Moles of Al deposited} = \frac{5 \text{ F}}{3 \text{ F/mole}} = \frac{5}{3} \text{ moles} \]

### Step 3: Calculate the weight of aluminum deposited
We know that the atomic weight of aluminum (Al) is 27 g/mol. To find the weight of aluminum deposited, we use the formula:
\[ \text{Weight of Al} = \text{Moles of Al} \times \text{Atomic weight of Al} \]
\[ \text{Weight of Al} = \left(\frac{5}{3} \text{ moles}\right) \times 27 \text{ g/mol} \]

### Step 4: Perform the multiplication
Now, we calculate:
\[ \text{Weight of Al} = \frac{5 \times 27}{3} = \frac{135}{3} = 45 \text{ grams} \]

### Conclusion
The weight of aluminum deposited when a current of 5 Faraday is passed through the solution of Al^(3+) ions is **45 grams**.