The number of electrons required to deposit `1 g` atom of aluminium (At. Wt. = 27) from a solution of aluminium cholride will be (wher N is Avogadro's number)

Options

A. ` 1N`
B. ` 2N`
C. ` 3 N`
D. ` 4 N`

Explanation

To find the number of electrons required to deposit 1 gram atom of aluminum from a solution of aluminum chloride (AlCl₃), we can follow these steps:

### Step 1: Understand the Ionization of Aluminum Chloride
Aluminum chloride (AlCl₃) dissociates in solution to form aluminum ions (Al³⁺) and chloride ions (Cl⁻):
\[ \text{AlCl}_3 \rightarrow \text{Al}^{3+} + 3\text{Cl}^- \]

### Step 2: Determine the Reaction at the Cathode
At the cathode, aluminum ions (Al³⁺) gain electrons to form aluminum metal (Al):
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
This reaction shows that to deposit 1 mole of aluminum, 3 moles of electrons are required.

### Step 3: Calculate the Moles of Aluminum
Given that the atomic weight of aluminum is 27 g/mol, 1 gram atom of aluminum corresponds to 1 mole of aluminum.

### Step 4: Calculate the Total Number of Electrons Required
Since 1 mole of aluminum requires 3 moles of electrons, we can express this in terms of Avogadro's number (N):
- 1 mole of electrons = N electrons
- Therefore, 3 moles of electrons = 3N electrons

### Conclusion
The total number of electrons required to deposit 1 gram atom of aluminum from a solution of aluminum chloride is:
\[ \text{Total electrons} = 3N \]

### Final Answer
The number of electrons required to deposit 1 gram atom of aluminum from a solution of aluminum chloride is **3N**.

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